Exam 2 - Sections 3.2 - 4.3
In this problem, A, B, C are subsets of a universal set U. One of the following is always true. Which one?
(A) (A ∩ B ∩ C') U (A' ∩ B' ∩ C) ⊂ A' ∩ B' ∩ C' (B) A' ∩ B' ∩ C' = {empty set} (C) A ⊂ B U A' (D) A ∩ B ⊂ B U C' (E) (A U B)' = A' U B' (F) none of the above Only D is true
A class consists of 8 sophomores and 4 juniors. Two students are selected at random to form a committee. Find the expected number of juniors on this committee.
- Make a PDF, multiply across, and add the EV (all in the figure) A: 2/3
Bob Loblaw drives past 4 stoplights on his way to work each morning. Each time he arrives at a stoplight, there is an 80% chance that the stoplight will be red. On a particular morning, what is the probability that exactly three of the four stoplights will be red?
4 stoplights/trials A: C(4, 3) (0.8)^3 (0.2)^1
A fair coin is flipped 4 times. What is the probability of getting exactly 2 heads, given that at least one head came up?
Use Bernoulli Trials - have to subtract out the one Pr that all coins are tails, because we must divide by the given: every flip contains at least one head A: 0.4 = 2/5
There are two cows and three accountants on an overloaded bus to Cleveland. Three of these five passenger are randomly selected to get off the bus, one after another and without replacement. What is the probability that the first passenger selected and the third passenger selected are both cows?
C A C = (2/5) (3/4) (1/3) = 6/60 = 1/10 A: 1/10
A multiple choice exam consists of 4 questions. Each question has 5 possible answers, exactly one of which is correct. Prstc decides to pick the answer to each question at random. What is the probability that he will get precisely two correct answers?
C(4, 2) (1/5)^2 (4/5)^2 A: 0.1536
Sue has gone to a store to purchase two dresses, a suitcase, and a diary for a trip. The store has 7 different dresses, 6 different suitcases, and 4 different diaries from which she may choose. How many different purchases of two dresses, a suitcase, and a diary are possible?
C(7, 2) x C(6, 1) x C(4, 1) A: 504 different purchases
A committee contains n members. If the total number of subcommittees with 3 member is 56, what is n?
C(n, 3) = 56 - Plugin the answer choices, and find the number that makes the statement true A: C(8, 3) = 56
The Infinite House of Pancakes prepares 100 pancakes from a batter infected with the bard virus. This virus survive cooking with a probability of .3. If the cooking of 100 pancakes are independent, what is the expected number of pancakes which will have the virus?
Cooking is independent. So, 100 cooked with a .30 Pr of being infected. 100 x 0.30 = 30 pancakes
A and F are independent subsets of U. We know that Pr(A) = 0.30 and Pr(A U F) = 0.58. What is Pr(F')?
Detailed work in Figure A: Pr(F') = 0.6
Independent Rules
Find Pr[A ∩ B] Because they are independent we know: Pr[A ∩ B] = Pr[A] x Pr[B] Pr(A U F) = Pr(A) + Pr(F) - Pr(A ∩ F) "A and B are independent" Pr[A ∩ B] = Pr[A] x Pr[B] Pr[A | B] = Pr[A] Pr[B | A] = Pr[B] Determining if Independent: Pr[O | R] = n(Odd ∩ Red) / n(Red) Pr[O | R] = 2/3 Pr[Odd] = 4/8 Those aren't the same, so no it is not independent Or, Another way to check for independence is to check to see whether Pr[ B ∩ E] = Pr[B] x Pr[E] If this statement is true, then it is independent. Pr[ B ∩ E] = 1/8 Pr[B] x Pr[E] = 1/8 x 4/8 = 1/16 Not equal, so NOT independent
Medium Challenge - Independent and Disjoint Problems
If Pr[E] = .5 and Pr[F] = .3, then what is Pr[E U F] if... What is Pr[E U F] if E and F are disjoint? .5 + .3 =0.8 What is Pr[E U F] if E and F are independent? Make a Venn Diagram - The middle has to be .5 x .3 - In the Figure Pr[E U F] = 0.65
A jury of five must decide whether Vasile will pass the Finite Math examination. In order for Vasile to pass, at least four of the jurors must be in favor. In how many different ways can vasile pass?
Order doesn't matter in the jurors saying yes. He needs AT LEAST 4 to say yes (so 5 would also work) A: 6 different ways C(5, 4) = 5 C(5, 5) = 1 5 + 1 = 6
How many 4-letter words can be formed with the letters in the word ''good''?
P(4, 4) / P(2, 2) = A: 12 words
Kartik is an excellent golfer. If this golfer's drive lands on the green, he makes a hole-in-one 15% of the time. In a golf game of 18 holes, Kartik's shots land on the green 8 times. What is the Pr that he makes at least two holes-in-one?
What is easier to find: good cases or bad cases? - bad cases - 0 success, and 1 success = C(8, 0) (.15)^0 (0.85)^8 = C(8, 1) (.15)^1 (0.85)^7 = 1 - answer of those added up will give us our answer
Flipping a coin - Bernoulli Even if you use an unfair coin, the Pr of flipping heads is the same every time you flip that coin. By convention, heads (obverse side) is always considered a success in coin flipping. So, Pr (Heads) = p, Pr(Tails) = q
What is the probability that a fair coin, flipped 6 times, comes up heads exactly 4 times? = C(n, r) p^r q^n-r (formula) = C(6, 4) p^4 q^2 = C(6, 4) (1/2)^4 (1/2)^2 = 15/64
For this problem, assume 7 males audition, one of them being Newton, 4 females audition, one of them being Barb, and 6 children audition. The casting director has 3 male roles available, 2 female roles available, and 1 child role available.
(1) How many different ways can these roles be filled from these auditioners? P(7, 3) x P(4, 2) x P(6, 1) = 15120 different ways that roles can be filled (2) Case 1: When Newton gets the role but not Barb, the number of ways = 6C2*3C2*6C1 = 270 Case 2: When Newton does not get the role but Barb does, the number of ways = 6C3*3C1*6C1 = 360 Total number of ways = 270 + 360 = 630 (3) Number of ways in which the roles can be filled = 7C3*4C2*6C1 = 1260 The required probability = 630/1260 = 0.5 (4) Number of ways in which the roles can be filled such that both Newton and Barb gets a role = 6C2*3C1*6C1 = 270 Thus, the required probability = 270/1260 = 0.2143 (5) The required probability = 0.5 + 0.2143 = 0.7143
Linguistics majors are required to take one of three foreign languages. Assume that 130 students are surveyed and every student takes at least one of the following languages. The results of the survey are as follows: 68 take Spanish. 76 take Japanese. 36 take Spanish and Japanese. 41 take Japanese and Russian. 17 take Spanish as their only foreign language. 19 take Spanish, Russian, and Japanese.
(1) How many take Russian? A: 78 (2) How many take Spanish and Japanese but not Russian? A: 17
Assume that the student has a cup with 12 writing implements: 6 pencils, 4 ball point pens, and 2 felt-tip pens.
(1) In how many ways can the student select 5 writing implements? A: 792 (2) In how many ways can the selection be made if no more than one ball point pen is selected? A: 336
A DNA sequence can be represented as a string of the letters ACTG (short for adenine, cytosine, guanine, and thymine).
(a) How many DNA sequences are exactly 33 letters long? A: 4^33 (b) Given a DNA sequence of length 33, how many one-letter mutations are possible? A: 99 (c) Given a DNA sequence of length 33, how many two-letter mutations are possible? A: 4752
A sample space S has three events, A, B, and C. The events A and C are disjoint, and the events A and B are independent. In addition, it is the case that: Pr(A) = 0.50 Pr(B) = 0.30 Pr(C) = 0.20 Pr(B ∩ C) = 0.05 Which of the following represents Pr(A' ∩ B ∩ C')?
- Draw the Venn Diagram, and use the independent and disjoint rules to solve. Venn is in the figure. A: 0.10
In a class of 40 students there are 16 freshmen, 15 sophomores, 28 Indiana residents, and 5 students who are neither freshmen nor sophomores nor Indiana residents. All freshmen in this class are Indiana residents. How many Indiana residents are neither freshmen nor sophomores?
- Make a 3 section Venn diagram. Logically no one can be a sophomore and freshman at the same time, so those values are 0. - Full Venn in the figure. A: 4 students
Imelda has 5 pairs of shoes. Four of them are worth $30 each, while the fifth is worth $2000. She selects a pair at random to walk to the midterm examination. What is the expected value of the pair selected?
- PDF show in table A: $424
Let A and B be independent events in a sample space S with Pr(A) = 0.3 and Pr(B) = 0.8. Let D be the event consisting of those outcomes that are in exactly one of the events A and B. Find Pr(D).
- This is asking in a confusing way for A and B probabilities only, without the intersection. A: 0.62
A set X with n(X ) = 50 is partitioned into three subsets X 1 , X 2 and X 3. If n(X2) = 3 n(X 1) and n(X3) = 2 n(X2), find the number of elements in subset X2 .
- This is simple a weighted probability problem in disguise. All work is in great detail shown in figure. A: X2 = 15
The Lemon brand cars have a transmission which fails with probability .6, and brakes which fail with probability .3; the two kinds of failures occur independently. What is the probability that exactly one of the failures occurs when you drive a lemon?
- draw a Venn Diagram. - The key to this problem is knowing that for independent events, Pr( A ∩ B ) = Pr(A) x Pr(B). So, 0.6 x 0.3 = 0.18, this is the center of the Venn Diagram A: 0.54 probability that exactly one of the failures occurs (simply take out the intersection for each, and then add).
A company buys 9 computers, and it plans to connect each pair of computers by a cable (one cable per pair). How many cables are needed?
- either draw it out to find the trend, or simply do a combination. A: C(9, 2) = 36
A survey of IU students reveals that 10% of the freshmen, 15% of the sophomores, 60% of the juniors, and 85% of the seniors like Finite Math. Given that 30% of all students are freshmen, 30% are sophomores, 20% are juniors, and 20% are seniors, what percentage of the student population like finite math?
- make a tree diagram with Probability on every branch. Then add up the branches that like Finite math. A: 0.365 like Finite
In an experiment, a coin is tossed until either the same side comes up twice in a row, or the coin has been tossed four times. How many outcomes are there?
- make a tree diagram, and follow the rules (in the figure) A: 8 outcomes
You and 8 others completely fill a shuttle van to the indianapolis airport. Among the others are 4 students. If you sit in the middle of the last row, so that you have a passenger on either side, what is the probability that you are sitting next to at least one student.
- make a tree diagram, with PR on branches, or use combinations. A: 11/14
A basketball team with 4 boys and 5 girls has made it to the championship game. The coach must decide which five players will start the game, and the particular order in which they are to be introduced. If the players are all equally talented, so the coach makes the choice entirely at random, determine the probability that two girls will be introduced as starters, followed by three boys.
- order matters, so we will fill in blanks using permutations. Work is in the figure. A: 2/63
Four Tennis players decide to play 'doubles' game each day (two players on each team) until all possible team combinations are used. How many days will they play?
2 teams are needed per game C(4, 2) = 6 combinations of a team with two players 6 / 2 = 3 days/games will be played to exhaust all options.
A fair six-sided die is tossed four times, each time noting which numbered face lands up. What is the probability that at least two different numbered faces land up during the four tosses?
6 x 6 x 6 x 6 = 1296 possibilities in four tosses 1 x 1 x 1 x 1 = 1 x 6 sides = 6 possibilities of all dice having the same side up for all rolls. 1296 - 6 = 1290 A: 1290/1296 = 0.99537 = 1 - 1/6^3 This is a trick question because of the answer choices. Calculate all of the answers, and select the one that matches the above answer.
Specific Examples - Only 1 way to do it A Bernoulli trial with success probability p = .2 is repeated independently 10 times. What is the probability that the second, fourth, and tenth of the tries were successes, but all the other tries were failures?
= C(n, r) p^r q^n-r (formula) = 1 (0.2)^3 (0.8)^7 There is only 1 way to have those precise tries as the only successes
Random Variable Start of Section 4.2
A random variable Y on sample space S is a real valued function of S. In math, we could write Y = f(S) e R Or, Y: S -> This means that we take an outcome in S, and we turn it into a real number. This means Y assigns one and only one real number to each outcome S. Special Case: When we have a Bernoulli process, the random variable which is the number of successes is called the binomial random variable. binomial random variable p and q stay the same and repeat over and over, if they don't you cannot use Bernoulli Y(r successes in n trials) = r
An experiment with equally likely outcomes has a sample space S and an event E, with Pr(E) = .5 and n(E) = 45. We also know that for another event F in S, Pr(F) = 0.4. Find n(F), the number of outcomes in F.
A: 36
A College Bowl team has 14 members, 8 of them surnamed Cooper and 6 of them surnamed Oglivie. Cecil is one of the Coopers, and Ben is one of the Oglivies. 5 of the team members need to be chosen (at random) to go to a prestigious event. Among the chosen 5, there are at least two Coopers and at least two Oglivies. What is the probability that Cecil Cooper is NOT chosen?
CCOOC = C(8, 3) C(6, 3) = 840 CCOOO = C(8, 2) C(6, 3) = 560 = 1400 combos with at least two Coopers and at least two Oglivies CCOOcecil = C(7, 2) C(6, 2) x 1 = 315 cecilCOOO = 1 x C(7, 1) C(6, 3) = 140 = 455 with Cecil 1400 - 455 = 945 without cecil A: 945/1400 = 27/40
Review from 4.2 #3 Rework problem 31 in section 4.2 of your text, involving the selection of two numbers. Assume that you select two 2-digit numbers at random from the set of consecutive integers from 00 through 99. The selections are made with replacement and are independent of one another. If the two numbers are the same, you make $250. If they are different, you lose $4. A random variable X is defined as your gain or loss.
How many different values are possible for the random variable X? 2 values {-$4.00, $250.00} Fill in the table below to complete the probability density function. 100 possibilities 10 instances that both numbers are the same - Probability density Function in the figure
Review from 4.2 #2 Rework problem 30 in section 4.2 of your text, involving the flipping of one of three coins. Assume that you have two fair coins and an unfair coin with Pr[H]=4/13. You randomly select one of these coins, and flip it twice. A random variable X is defined to be the number of heads you observe. How many different values are possible for the random variable X?
How many different values are possible for the random variable X? S = {0, 1, 2}, so 3 Fill in the table below to complete the probability density function. Be certain to list the values of X in ascending order. All work is shown neatly in the figure. it's important to note that there is a 2/3 chance of getting a fair coin, and 1/3 chance of a an unfair coin. This could also be solved with a tree diagram, which is in the figure as an example.
For this problem, assume there are 8 adjacent seats in a row, and 8 people the need to be seated.
How many different ways can 8 people be seated in them? A: 40320 ways
Non-negative integer solutions EQUATIONS & INEQUALITIES Template: x1 + x2 + x3 + ... + xr = n Formula: (n + r - 1)! ----------- n! (r - 1)!
How many non-negative integer solutions are there to the following equation? x1 + x2 + x3 = 7 Answer: 36 How many non-negative integer solutions are there to the following inequality? x1 + x2 + x3 < 7 Answer: 84 To get this answer, must solve for all integer r's under 7 but not negative (0 - 6).
Assume that the set S has 12 elements.
How many subsets of S have at most 2 elements? C(12, 2) = 66 C(12, 1) = 12 C(12, 0) = 1 = 79 elements
An experiment consists of rolling a fair, 4-sided die three times and noting the number of 4s that are rolled. Let the random variable F be the number of 4s that are rolled.
How many values of the random variable are possible? S = {0, 1, 2, 3} Is this Bernoulli? Yes, the Pr of rolling a 4 is the same after each roll. Use the Bernoulli formula to calculate the Pr for each value.
For this problem, assume that you are to pick 3 cars from the motor pool, which contains 9 subcompact cars, 4 compact cars, and 4 midsize cars. Detailed work in Figure
How many ways can you pick the cars so not all are the same size? A: 588 ways
Suppose a committee of three is to be selected at random from a group containing 6 women and 4 men. A random variable X is defined to be the number of women selected for the committee.
How many women could be on the committee? S = {3, 2, 1, 0} Now, lets construct the pdf of this variable. This is NOT a binomial random variable, so we need to use combinations. Overal possible combos, C(10, 3) = 120 C(6, 3) = 20 C(6, 2) + C(4, 1) = 60 C(6, 1) + C(4, 2) = 36 C(6, 0) + C(4, 3) = 4 Then, fill out the table (shown in the figure)
DETERMINING IF IT'S A BERNOULLI TRIAL
How to know if you need to use Bernoulli trials: - Can I think of the experiment as drawing balls out of a box, those balls only have two different labels on them, and every time I draw one out I put it back (the Probability of drawing balls doesn't change from one trial to the next.)
Balls are numbered 1 through 10 (including 1 and 10) in an urn. 3 are drawn simultaneously and at random. If X is 3 times the number of even balls drawn minus 2 times the number of odd balls drawn, how many distinct values for X are there? Find the probability density function of X?
If X is 3 times the number of even balls drawn minus 2 times the number of odd balls drawn, how many distinct values for X are there? 4, {EEE, EEO, EOO, OOO} Find the probability density function of X? All in the figure
QUIZ Question: Of the 200 students in a linguistics class, 80 take a course in French, 50 take a course in Swahili, and 100 take either French or Swahili. A student is chosen at random and she does not take Swahili. What is the probability that she takes French?
In the Figure A: 1/3
QUIZ Question: A fair 9-sided die is rolled twice and at least one roll shows a 6. What is Pr[sum is 10]?
In the Figure A: 2/17
QUIZ Question: A fair 8-sided die is rolled twice and the sum is 10. What is the probability that at least one roll shows a 6?
In the Figure A: 2/7
Review from 4.3 #1 Suppose that 2 fair six-sided dice are rolled and their sum is noted. What is the expected sum?
In the figure 3.5 expected sum for one die 3.5 x 2 = 7 expected sum for two dice
10 actresses, including Julia, audition for 4 different roles. If Julia is chosen, how many cast lists are possible?
JOOO OJOO OOJO OOOJ P(9, 3) x 4 = A: 2016
50% of a class are freshman, 30% sophomores, 10% juniors and 10% seniors. If 80% of freshman, 50% of sophomores, 40% of juniors and 10% of seniors are from out of state, then what percentage of students in the class are from out of state?
Make a tree, find the Pr for out of state, and then add up Pr[OS] = 60%
Spam and Lake have the mission to destroy the Star Joke. Spam's probability of success on any given try is .7, while Lake's is .8. Spam and Jake will each be able to make just one try, and they decide that Spam will be first. What is the probability that the Star Joke is in fact destroyed?
P( It is destroyed ) = P( spam's try destroyed the star jake) + P( spam is failed , now lakes turn) = 0.70 + (1-0.70)*0.80 = 0.70+0.24 = 0.94
Bayes Formula
P(Event | Information) = (P(Information | Event) * P(Event)) / P(Information) Simplest form and Slightly more complex form in the Figure
SAMPLE SPACE OF A DIE
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Suppose that a student selects all his answers to this test entirely at random among choices (A) through (E), never choosing (F). If indeed, ''none of these'' is not the correct answer to any of the questions, and the correct answers are positioned entirely at random, what is the expected score for this student? Recall that this test consists of 25 problems and each problem is worth 4 points.
Pr(right answer) = 1/5 25 questions x (1/5) = 5 5 right answers x 4 points each = A: 20 points
A rental car company has 5 Mazdas and 3 Bugattis to rent (for one day). On Monday, Matt rented a car. On Wednesday, Maxie rented a car. Let X be the number of Bugattis selected How many Bugattis do you expect M&M to get?
Probability or Expected value? - Expected Bernoulli or Combination? - Bernoulli
A rental car company has 5 Mazdas and 3 Bugattis to rent (for one day). On Monday, Matt and Maxie rented two different cars. Let X be the number of Bugattis selected How many Bugattis do you expect M&M rented?
Probability or Expected value? - Expected Bernoulli or Combination? - Combination
Probability density function (pdf)
The PDF is the function which assigns to each value of a random variable its probability. The sum of all the probabilities in the table is one, as before.
An experiment has three fair 6-sided die and two fair 10-sided die. One of the FIVE dice is selected and rolled once. The number on top is noted. What is the probability of rolling 3 or less?
The Pr of rolling a (three or less) will be found from the tree. - This problem will have the same tree as before, but the first branch probability will no longer be 1/2 (equally likely between rolling a 6 and 10 sided). Now, we have a 3/5 Pr of rolling the 6-sided, and a 2/5 for rolling a 10-sided die. The updated tree is in the figure, with the corresponding Pr.
Let [0 1/2 1/2] = P [1/2 1/4 1/4] [1/2 1/2 0] be the transition matrix for a Markov chain. For which states i and j is pij(2) largest?
The probability of a direct transition from state i to state j is denoted - P ij In the transition matrix, this is the entry in the i'th row and j'th column So, multiple P*P, and find the column and row of the largest probability. A: P 11 (2) = 0.5 i = 1 and j = 1
A restaurant offers 4 kinds of salads, 5 main courses, and 1 dessert. A meal at this restaurant consists of a salad, a main course, and possible dessert. (Dessert is optional.) How many different meals can one order at this restaurant?
The tricky part is the Desserts being optional. We must remember that a meal with a dessert, is different from the exact same meal without a dessert. So, 4 salads x 5 main courses x 1 dessert = 20 meals and, 4 salads x 5 main courses = 20 meals 20 + 20 = 40 different meals
In a baseball league of 9 teams, how many games are needed to complete the schedule if each team plays 10 games with each other?
Think about it for every team playing every other team once. You would have C(9, 2) = 36 games. Now, this is done 10 times, so 10*36 is 360 games.
Standard example of Bernoulli process - when to use the Bernoulli Process
Think of something which you could do repeatedly. It has only two outcomes, success and failure. The Pr of success is the same every time. You might bet on it
In a local restaurant, 40% of the customers were born in Indiana, 40% were born in Ohio, and 35% are IU students. In addition, 10% of the customers are IU students who were born in Indiana, and 20% are IU students who were born in Ohio. If one customer is randomly selected from the restaurant, and it is known that the customer was born in neither Indiana or Ohio, what is the probability that the customer is an IU student?
This is a trick question, because we must create a Venn and then use conditional probability. All of the work is in the figure. Pr(customer is an IU student and not from IN nor OH) = = 0.05 Pr(the customer was born in neither IN nor OH) = = 0.05 + 0.15 = 0.20 So, 0.05/0.20 A: 1/4
An unfair coin with Pr(H) = 0.90 is tossed (independently) 5 times. What is the probability that the first, third, and fifth tosses are H, and all the others are T?
Tricky Part: we are looking for the PR of a certain one trial, not all five. So, DON'T use Bernoulli. Just draw out that limb of the tree. We want HTHTH 0.9 x 0.1 x 0.9 x 0.1 x 0.9 = 0.00729 Probability
A sample space S has two independent events, A and B, with Pr(A) = 0.40, and Pr(B | A) = 0.30. Which of the following represents Pr(A' ∩ B')?
Use the independent rules to solve for the intersection of a and B, and use Pr (B | A) = Pr (B) = 0.30 A: 0.42
A sample space consists of three outcomes a, b, c. We know that Pr(a) = 1/3 and b is three times as likely as c. Find Pr(b).
We know Pr(a) takes 1/3, so that leaves 2/3 for b and c. Now, we must determine how much of that 2/3 b and c has. (solved in the figure) b = 3/4 of what is left c = 1/4 of what is left So, (3/4) x (2/3) A: b = 0.5
Four rooms are to be painted. Each room will be painted with one color of paint. There are three colors of paint which may be used. A decorating scheme is a specification of a color of paint for each room. How many different decorating schemes use at least two colors of paint?
We know, number of rooms = 4, number of colors = 3. Each room is to be painted with one colour of paint, and there are 3 options for each room. Therefore, there can be several decorating schemes, staring from painting all four rooms with the same color, to using all three colors in various possible combinations. Now, the first room can be colored in 3 different ways, as there are three different colour options. The second room can also be colored in three different ways, so can the third and the fourth. There are three possible color options for each of the four room. Therefore, total number of decorating schemes = 3*3*3*3 = 81. Now, the schemes in which less than 2 colors were used = the number of schemes in which only one color was used. One colour can be used in three possible ways. If all the four rooms are painted in one single color, it can happen in 3 possible ways as there are three types of colors. So, possible options: All 4 rooms are painted in the first color OR all 4 rooms are painted in the second color OR all 4 rooms are painted in the third colour. So, total possible outcomes where less than 2 colors are used = total possible outcomes where only one color is used = 3. So, number of decorating schemes where at least 2 colors are used = total number of schemes - number of schemes where less than 2 colours are used = 81 - 3 = 78
For this problem, assume the bag contains 13 basketballs, 5 of which have defective valves. You choose 2 out of the bag randomly.
What is the probability that at least 1 will have a defective valve? - This can be solved with a tree diagram (I prefer), or through using combinations. A: 50/78
Suppose Vanilla Ice flips a fair coin four times and counts the number of heads.
What is the sample space? {0, 1, 2, 3, 4} Is the Pr for each number the same? No, not all outcomes are equally likely How should we find the probabilities? - we could draw a tree, shown in the figure. But that would be quite long.
Katherine owns 6 different mathematics books and 6 different computer science books and wish to fill 5 positions on a shelf. If the first 2 positions are to be occupied by math books and the last 3 by computer science books, in how many ways can this be done?
Work in the Figure Positioning matters on the shelf, so we must use permutations to solve. A: 3600 ways this can be done
Cleopatra has 5 chains in a jewelry box: 4 silver chains worth $40 each, and 1 gold chain worth $2000. She selects a chain at random to wear for her midterm exam. What is the expected dollar value of the chain selected?
X = $2000 or $40 Create a PDF and multiply across and add. (pdf in the figure) A: $432
A 4-sided die is fair and the sides are numbered 1, 6, 7, 10. This die is rolled twice and the random variable is defined to be the sum. How many values of the random variable are there?
X = 2, 7, 8, 11, 12, 13, 4, 16, 17, 20 A: 10
3. A box contains 8 green and 4 blue marbles. Two marbles are selected (at once with no replacement). Find the expected number of green marbles among the selected ones.
X = {0, 1, 2} Green Marbles A: 4/3
Jack wants to buy a two-scoop sundae at Baskin-Robbins. How many choices of flavor combinations does he have if he is allowed to choose the same flavor twice? Recall that there are 31 flavors in all! ("chocolate and vanilla" is the same as "vanilla and chocolate").
You can choose C(31, 2) combinations of 2 flavors. However, since you can choose the same flavor twice you need to add another 31 possibilities. therefore, C(31, 2) + 31 = 465 + 31 A: C(31, 2) + 31, or 496 ways
Abstract examples The first outcome is A, B or C. The second outcome is X or Y. *limited given information
all work shown in figure
Mikael Gabriel, the famous Finnish rapper, does speak English, but he has never studied Finite Math. Mikael tries to take a 5-question, multiple-choice quiz (with 8 answers per question), guessing at random on every problem. What is the probability that Mikael gets only the first 2 correct?
n = 5 p = (1/8) q = (7/8) r = 2 (1/8)^2 C(3,0) (1/8)^0 (7/8)^3 = 0.0104675 A: (1/8)^2 * (7/8)^3 = 343/32768
A fair ten-sided die is rolled 9 times. What is the probability that a 2 is not rolled exactly 6 times?
n = 9 P = (9/10) q = (1/10) r = 6 C(9, 6) (9/10)^6 (1/10)^3 A: 0.044641044
How many 6-bit strings (that is, bit strings of length 6) are there which:
1. Start with sub string 101 The possible options for remaining 3 are: 2^3 ( there are two possibilities can take 3 places ) Start with sub string 101 = 8 2. Have weight 5 and start with sub string 101 If it has 5 1's then remaining three bits should be one only, the only possibility is 111 Have weight 5 and start with sub string 101 = 1 3. Either start with 101 or end with 11 (or both) Start with 101 = 8 End with 11 = 2^4 = 16 ( 2 options taking 4 places ) Both has two possibilities = 101011 and 101111 (2) Either start with 101 or end with 11 (or both) = Start with 101 + Ends with 11 - both = 8 + 16 - 2 = 22 Either start with 101 or end with 11 (or both) = 22 4. Have weight 5 and either start with 101 or end with 11 Have weight 5 and start with 101 = 1 ( 101111 ) Have weight 5 and end with 11 has 4 possibilities { 011111, 101111, 110111, 111011 } both has 1 possibility in common = 101111 Have weight 5 and either start with 101 or end with 11 = 4 + 1 - 1 = 4 Have weight 5 and either start with 101 or end with 11 = 4
5 mysteries, 3 fantasies and 4 scifi are on a shelf. Ethan takes 2 books, keeping 1 & giving 1 to Shane. How many ways?
Because Ethan is giving one of the books away, order matters in this scenario. So, P(12, 2) A: 132 ways
IHOP offers 35 different kinds of pancakes. You would like to order a stack of three pancakes. How many different choices do you have, considering that you care about the order in which the pancakes are stacked?
Because we can order the same three pancakes, and there's replacement. A: 35^3 If we couldn;t order the same pancake then the answer would be P(35, 3)
12. Consider a town in which there are two plumbers, whom we call A and B. On a certain day, four residents of the town telephone for a plumber. If each resident selects a plumber at random from the telephone directory, what is the probability that two residents will call A and two residents will call B?
Bernoulli Trials C(4, 2) (1/2)^2 (1/2)^2 A: 0.375 = 3/8
A 4-sided die is fair and the sides are numbered 1, 2, 3, 10. This die is rolled twice and the random variable is defined to be the sum. What is the expected value?
Define all of the X values, count the probabilities, make a pdf, multiply across, and add up for the EV. A: 8
A defective vending machine gives the desired product 70% of the time and gives nothing 30% of the time. An experiment tries to use the vending machine twenty-four times. What is the expected number of times that the machine will give nothing?
E(X) = np E[X] = (24) (.30) E[X] = 7.2 average (expected value) The machine is expected to give nothing 7.2 times.
Rolling a Dice - Bernoulli A fair 6-sided die is rolled 8 times. What is the probability of getting at least one 2?
First consider the fact that: Pr[at least one 2] = 1 - Pr[less than one 2, or no 2] If a fair 6-sided die is rolled 8 times, what is the Pr[no 2]? = C(8, 0) (1/6)^0 (5/6)^8 = (5/6)^8 So, the Pr[at least one 2] = 1 - (5/6)^8
Assume that 200 students are surveyed with the following results: 90 live in the west. 82 live in a large city. 92 are married. 29 live in the west in a large city. 31 are married and live in a large city. 49 are married and live in the west. 16 are married and live in a large city in the west.
How many are unmarried, do not live in a large city, and do not live in the west? A: 29 students
Assume that there are 4 types of upholstery, 6 types of wood, and 6 designs to choose from. Of the designs, 5 designs permit both a choice of wood and a choice of upholstery, but 1 of the design allows only a choice of upholstery.
How many different chairs can be made? A: 124 Different Chairs
A club is made up of 5 women and 4 men. Find the number of ways a president, vice-president, and treasurer can be selected, if these positions are to be filled by two women and one man.
MWW = 4 x 5 x 4 = 80 WMW = 80 WWM = 80 A: 240
Aniyah flips a fair coin until EITHER two Heads are flipped OR four Tails are flipped. What is the expected number of Tails flipped?
Make a Tree Probability Diagram, and define all X values. - Possible X values: 0, 1, 2, 3 A: 7/4, or 1.75 Tails
Review from 3.2 #2 Pr[A∪B] = 0.55 Pr[A] = 0.1 (1) What is Pr[B] if A and B are independent events? (2) What is Pr[B] if A and B are disjoint events?
P(A U B) = P(A) + P(B) - Pr(A ∩ B) Pr(A ∩ B) = P(A) x P(B), because they are independent So, P(A U B) = P(A) + P(B) - P(A) x P(B) 0.55 = 0.1 + P(B) - 0.1 x P(B) 0.45 = P(B) x (1 - 0.1) 0.45 = P(B) x (0.9) P(B) = 0.45/0.9 = 0.5
Using the problem above: If a Wisconsin resident owns a cheesehead hat, what is the Pr that he is a packer fan?
Pr[Fan and Hat] = 0.45 Pr[Hat] = 0.50 So, Pr[Fan | Hat] = 0.45/0.50 = 90%
Five Students A, B, C, D, E are lined up to buy tickets at the Fine Arts Theater. In how many ways can they be arranged if B and C are next to each other?
The tricky part is the order can be, B C 3 2 1 or, C B 3 2 1 Detailed work is shown in the figure A: 48
A four-digit number is formed by randomly selecting four digits, without replacement, from the set D = {1,2,3,4,5,6,7}. What is the probability that the resulting number is greater than 4200?
There are 7 digits here. Thus, there are P(7, 4) = 840 ways to choose a 4-digit number. For a number to be greater than 4200, the first digit must be 4 or more. Case 1: If the first digit is 4, the second digit must be 2 or more, there are 5 ways to do that. Then 5 ways to get the third, and 4 ways to get the fourth. There are 1*5*5*4 = 100 ways so that the first digit id 4. Case 2: If the first digit is 5 or more, there are 3 ways for the first digit, 6 ways for the second, 5 ways for the third, 4 ways for the fourth. Hence, there are 3*6*5*4 = 360. Thus, there are a total of 100+360 = 460 ways. Thus, P(greater than 4200) = 460/840 = 0.547619048 [ANSWER] = 23/42
Conditional Probability Expression Example
You are told the following facts about two subsets, R and T. - Pr[R] = 1/4 - Pr[T] = 3/5 - Pr[R U T] = 3/4 FIND Pr(R | T) 1. Make a Venn diagram to find the intersection between R and T. (Venn Diagram shown in the Figure) 2. Divide that Probability by Pr[T] - Worked out in Figure Pr(R | T) = 0.166 or 1/6 FIND Pr(T | R) = 0.10 / 0.25 Pr(T | R) = 0.4, or 2/5 Find Pr[R' | T] = 0.5 / 0.6 = .833 Find Pr[R | T'] = 0.15 / 0.40 = 0.375 or 3/8
Lucas has a helmet containing 9 slips of paper. The slips are numbered (perversely): 1, 3, 4, 7, 11, 18, 29, 47, 76. Jim takes three slips of paper from the helmet. A random variable B is defined to be three times the number of odd numbers drawn plus seven times the number of even numbers drawn.
B = 3(O) + 7(E) Create a table (shown in the figure) Once we have the table, multiply across and add down.
Review from 3.2 #5 For this problem, assume that three students are chosen from 3 freshmen, 5 sophomores, and 5 juniors. What is the probability of selecting 2 freshmen and 1 junior given that at least one freshman will be selected?
Work in the Figure A: 15/166
QUIZ Question: U partitioned into 4 Sets: A, B, C, D Pr(A, B) = 0.66 Pr(A, D) = 0.67 Pr(A, B, D) = 0.95 Which of the following is Pr(D)?
Work in the figure A: 0.29
QUIZ Question: A number between 1 and 40 is drawn at random. Given that the number was greater than 30, what is the probability that the number contained a 3?
Work in the figure A: 9/10
For this problem, assume Gwen and Harry have 6 types of cones and 9 flavors of ice cream. Detailed work in Figure
(1) In how many different ways can your order one cone and two scoops of ice cream? (As in the book, putting one flavor on top of another is different from putting them the other way around.) 9 x 9 x 6 = 486 ways (2) In how many different ways can your order one cone and two scoops of ice cream which are not the same flavor? P(9, 2) x 6 cones = 432 ways w/o same ICF
Suppose Hunter's pocket contains 6 nickels, 4 dimes, 7 quarters and 3 half dollars. Dodger, a pickpocket, reaches into the pocket and pulls out one coin. What is the expected value of the coin (in cents)?
- Make a table, calculate Pr, and then get average. (shown in the figure) - If dodger picked out three coins, instead of one, then just multiply the average for one by 3
Somewhat Specific A Bernoulli trial with failure probability q = .3 is repeated independently 10 times. what is the Probability that the first trial is a failure and there are exactly 4 successes?
= C(n, r) p^r q^n-r (formula) - first trial has to be a failure, so start with 0.3 = (0.3) C(9, 4) (.7)^4 (.3)^5 = C(9, 4) (.7)^4 (.3)^7 after the 1st trial is a success, there are 9 more trials and you need 4 more successes.
Somewhat Specific A Bernoulli trial with failure probability q = .3 is repeated independently 10 times. what is the Probability that the first trial is a success and there are exactly 4 successes?
= C(n, r) p^r q^n-r (formula) - first trial has to be a success, so start with 0.7 = (0.7) C(9, 3) (.7)^3 (.3)^6 = C(9, 3) (.7)^4 (.3)^6 after the 1st trial is a success, there are 9 more trials and you need 3 more successes.
Flipping a coin - Bernoulli What if we wanted to flip an unfair coin 6 times and get exactly 2 heads? say p = .44
= C(n, r) p^r q^n-r (formula) = C(6, 2) p^2 q^4 = C(6, 2) (.44)^2 (.56)^4 = 0.28559
Bernoulli Trials and Bernoulli Processes Start of Section 4.1
A Bernoulli trial is an experiment with two possible outcomes. - The two possible outcomes are usually called success and failure - We say that the Pr of success in a Bernoulli trial is P. - The Pr of failure in the Bernoulli trial is q. Pr(success) = p Pr(failure) = q Note: p + q = 1
Independent or Not?
A box contains three red balls (numbered 1, 2, 3), four green balls (numbered 4, 5, 6, 7), and one black ball (numbered 8). Is the event "drawing a red ball" independent of "drawing an oddball"? To check for independence we could check to see if Pr[Odd | Red] = Pr[Odd] Pr[O | R] = n(Odd ∩ Red) / n(Red) Pr[O | R] = 2/3 Pr[Odd] = 4/8 Those aren't the same, so no it is not independent Is the event "drawing a black ball" independent of "drawing an even ball"? Another way to check for independence is to check to see whether Pr[ B ∩ E] = Pr[B] x Pr[E] If this statement is true, then it is independent. Pr[ B ∩ E] = 1/8 Pr[B] x Pr[E] = 1/8 x 4/8 = 1/16 Not equal, so NOT independent
Stochastic Process - and Bernoulli process
A multi-stage experiment. First we pick A or B. Then we pick a or b. A Bernoulli process is a special stochastic process in which the same Bernoulli trial is repeated at each stage - Specifically, p = Pr(success) remains the same for all stages of the process. The Probability of r successes in n Bernoulli trials is = C(n, r) p^r q^n-r (shown in the figure) n = number of Bernoulli trials r = number of successes p = Pr(success) q = Pr(failure)
A carnival game has players randomly pick plastic fish from a tub. The tops of the fish are visible above the water, and the tops of all the fish are blue. On the bottom of the fish, though, the fish are different colors: specifically, they are pink, green, or yellow. There are 32 pink fish, each worth $0.30 There are 17 green fish, each worth $1.10 There is 1 yellow fish, worth $15.00
A player picks a fish at random. How much do you expect the fish is worth? = expected value is 0.866 - create the pdf (shown in the figure) If a player pays $1.00 to play the game, what does the player expect to win or lose on each play? = E(fish) - 1 = .866 - = -$0.134 If a player pays $1.00 to play the game, what does the carnival expect to win or lose on each play? = 1 - E(fish) = 1 - 0.866 = $0.134
QUIZ Question: You select 3 coins at random form a purse containing 6 pennies, 4 quarters, and 3 nickels, and the value of the coins is noted. What is the probability the value is greater than 25 cents?
Work in the figure A: 101/143
QUIZ Question: Universe U contains two subsets, E and F. We know Pr(E U F) = 0.8, Pr(E) = 0.7, and Pr(F) = 0.4 What is Pr(E | F)?
Work in the figure A: 3/4
The SAT used to penalize wrong answers by subtracting 1/4 a point from each incorrect answer. If each question on the SAT has 5 answer choices only one of which is correct, what is the expected number of points awarded by guessing on 50 questions?
E = (50)(1/5) = 10 correct answers, 40 incorrect answers E = (40)(1/4) = 10 points lost from right answers So, 0 average points awarded
A basketball player makes 80% of her shots inside the 3 point line and makes 30% of her shots outside the three point line. During a game, the basketball player takes 17 shots inside the 3 point line and 9 shots outside the 3 point line. What is the expected number of shots that she makes during the game?
E(inside) = (17) (.8) = 13.6 E(outside) = (9)(.30) = 2.7 13.6 + 2.7 = 16.3 is the average number of shots she makes in a game. Total points = 2(13.6) + 3(2.7) = 35.3 points on average
Recall that a 8-bit string is a bit strings of length 8, and a bit string of weight 3, say, is one with exactly three 1's.
How many 8-bit strings are there? Add up all combinations for 8. (written out in figure) A: 256 overall How many 8-bit strings have weight 0? C(8, 0) = 1 How many 8-bit strings have weight 1? C(8, 1) = 8 How many 8-bit strings have weight 4? C(8, 4) = 70 How many 8-bit strings have weight 7? C(8, 7) = 8 How many 8-bit strings have weight 8? C(8, 8) = 1 How many 8-bit strings have weight 12? C(8, 12) = 0, the weight cannot exceed more than the set
A fair 6-sided die is rolled 9 times and the resulting sequence of 9 numbers is recorded.
How many different sequences are possible? 6^9 How many different sequences consist entirely of even numbers? S = {2, 4, 6} So, 3^9 How many different sequences are possible if the first, third, and fourth numbers must be the same? - The first number is rolled, and then the third and fourth are required to be the same as the first. So two rolls are taken away from the 9. 6^7
Review from 4.2 #1 Assume that the coin is weighted so that Pr[H]= 5/9, and that the coin is flipped until a head appears, or 5 consecutive tails appear. The random variable X is defined to be the total number of flips of the coin. How many different values are possible for the random variable X? Fill in the table below to complete the probability density function. Be certain to list the values of X in ascending order.
How many different values are possible for the random variable X? 5 at most (either TTTTH, TTTTT, or less) Fill in the table below to complete the probability density function. - use a tree diagram to calculate (all in the figure)
Lew is not very decisive. He wants to move from Georgia to Canada eventually. His hockey-loving wife gives him this ultimatum: "If the Winnipeg Jets win tonight, we move immediately. If they don't win tonight but do win their next game, we move one month from today. If they lose the next two games, we move three months from today." Based on last season's results for each game, their probability of winning is .600.
How many months are expected to pass before Lew and his wife move to Canada? Create a table, and then add up the products = .72 months on average to move.
A coin is selected from one of three coins: two of the coins are fair and the other coin has two heads (p = 1). Once the coin is selected it is flipped 4 times. Let the random variable H be the number of heads that are flipped.
How many values of the random variable are possible? S = {0, 1, 2, 3, 4}, so 5 possible Is this a binomial random variable? No, because the probabilities are not equal every trial due to some coins having two heads. Pr(fair) = 2/3 Pr(2H) = 1/3 Creating the pdf, we need to calculate the Pr of each value for the fair coin and then the 2H coin. We must then multiply those probabilities by the probability of drawing a fair coin or a 2H. The addition of the two results will get the overall Pr for each value. This is basically a tree diagram, but in a table. This is all shown in the figure.
An experiment has three fair 6-sided die and two fair 10-sided die. One of the FIVE dice is selected and rolled once. The number on top is noted. What is the probability of rolling 3 or less??
If the die roll is three or less, what is the probability that a ten-sided die was picked? Pr[the roll is three or less] = 0.42 Pr[ten-sided die is picked | the roll is three or less] 0.12/0.42 = 2/7 If the die roll is more than three, what is the probability that a ten-sided die was picked? Pr[the roll is more than 3] = 0.58 Pr[ten-sided die is picked | the roll is more than 3] = = 0.28/0.58 = 48.27%
Disjoint WW Example
If the knowledge that an event A has occurred implies that a second event B cannot occur, then the events A and B are said to be A: Disjoint (b) If event A and event B are as above and event A has probability 0.5 and event B has probability 0.3, then the probability that A or B occurs is A: 0.5 + 0.3 = 0.8
Review from 3.4 #2 Rework problem 15 from section 3.4 of your text, involving the production and purchase of shoes. Assume that shoes are produced in three countries: 30 percent in Sweden, 50 percent in Finland, and 20 percent in Gabon. At the plant in each country, two types of shoes are made: racing shoes and training shoes. The production at each plant as allocated as shown in the table below. The shoes are randomly distributed to stores in the United States, and you purchase one pair from one of these stores. Sweden = Racing (0.35), Training(0.65) Finland = Racing (0.25), Training(0.75) Gabon = Racing (0.45), Training(0.55)
If you purchase a pair of training shoes selected at random, what is the probability that they were made in Sweden? A: 28.67647%
Tree Notations Applications of Conditional Probability Start of Section 3.3
In 3.3 and 3.4 tree notations will be used to show conditional probability throughout a tree diagram.
Review from 4.1 Suppose that the probability of having a boy or a girl is 50%50%. Answer the following questions: (1) What is the probability that exactly one of a family's 9 children is a boy? (2) What is the probability exactly one of your 9 children is a boy given that at least 8 of them are girls?
In Figure (1) A: 0.017578125 (2) A: 0.9
Batting Average (BA) BA = n(hits)/n(at bats) BA is the probability that a hitter will get a hit at an at-bat. A baseball team has two hitters who always bat 3rd and 4th. The 3rd batter has a BA of .400 against right-handed pitchers and .200 against left-handers. The 4th batter has a BA of .350 against left-handed pitchers and .250 against right-handers.
In a 4 game series: The 3rd batter faces righties 14 times and lefties 10 times. = 14 (.40) + 10 (.20) = 7.6 average hits The 4th place batter faces righties 10 times and lefties 8 times. = 10 (.25) + 8 (.35) = 5.3 average hits E(hits) = 12.9 on average (from both of them)
QUIZ Question: A coin purse contains 6 fair quarters and 1 double-headed quarter. A coin is taken from the purse and flipped twice. What is the probability of getting exactly one head?
In the Figure A: 3/7
QUIZ Question: 40% of customers waiting at a garage are playing games on their phones and the rest are reading. If a customer is reading, there is a 50% chance they won't respond the first time they are called. If a customer is playing games, there is a 30% chance they won't respond the first time they are called. A customer, Carlos, is called, but he does not respond the first time he is called. What is the probability Carlos was reading?
In the Figure A: 5/7
Review from 3.3 #2 Suppose that it is equally likely for a child to be a boy or a girl. Suppose a parent tells you that she or he has 6 children. What is the probability that at least one is a boy given that at least 5 of the children are girls?
P(Girl), p = 0.5 P(Boy), q = 0.5 Sample size, n = 6 Bayes' theorem: P(A | B) = P(A & B) / P(B) P(at least one of six is a boy | at least five are girls) = P(at least one boy and 5 girls) / P(at least 5 girls) = P(5 girls and 1 boy) / [P(5 girls and 1 boy) + P(6 girls)] = (C(6, 5) x 0.5(to the 5th power) x 0.5) / [(C(6, 5) x 0.5(to the 5th power) x 0.5) + 0.5 (to the 6th power)] = 0.875 or 6/7
In baseball, managers use statistics to determine many of their decisions. Two of these statistics can be used to find the expected value of a random variable. They compile batting averages of hitters facing either right-handed and left-handed pitchers. Batting Average (BA) BA = n(hits)/n(at bats) BA is the probability that a hitter will get a hit at an at-bat. Earned Run Average (ERA) ERA = 9 x (earned runs)/(inning) An ERA is the expected number of earned runs that a pitcher would allow if he pitched a complete 9-inning game.
Pitcher 1 has an ERA of 2.40 and pitches 5 innings Pitcher 2 has an ERA of 3.10 and pitches 3 innings Pitcher 3 has an ERA of 1.20 and pitches 1 inning How many earned runs is the team expected to give up? 2.40 (5/9) + 3.10 (3/9) + 1.20 (1/9) = 2.5 expected runs to be given up
A rental car company has 5 Mazdas and 3 Bugattis to rent (for one day). On Monday, Matt rented a car. On Wednesday, Maxie rented a car. Let X be the number of Bugattis selected What is the Probability that X = 1?
Probability or Expected value? - Probability Bernoulli or Combination? - Bernoulli, the cars are replaced each day, so there is the same Pr.
A rental car company has 5 Mazdas and 3 Bugattis to rent (for one day). On Monday, Matt and Maxie rented two different cars. Let X be the number of Bugattis selected What is the Probability that X = 1?
Probability or Expected value? - Probability Bernoulli or Combination? - Combination
Binomial Random Variable Example: Suppose we flipped an unfair coin (p = .40) twice. Let's find the binomial random variable for this case. Intro to the Probability density function (pdf) of the binomial random variable.
S = {0, 1, 2} Pr(r = 0) = C(2, 0) (.4)^0 (.6)^2 = .36 Pr(r = 1) = C(2, 1) (.4)^1 (.6)^1 = .48 Pr(r = 2) = C(2, 2) (.4)^2 (.6)^0 = .16 Now, based on what we've calculated, make a label to directly store the values (shown in the figure). This table is the probability density function of the binomial random variable. The probability density function means that we are looking at some table with probabilities on one side, and the values of your variable on the other side.
Review from 3.2
Problem 4 - Rework problem 5 from section 3.2 of your text, involving sets A and B. Suppose for this problem that Pr[A|B] = 1/2, Pr[A] = 9/20, and Pr[B′] = 45. (1) What is Pr[B|A]? Pr(A|B) = Pr(A ∩ B) / Pr(B) = 0.5 So, Pr(A ∩ B) = 0.5 Pr(B) --- this is the interesction Pr(B | A) = 2/20 / 9/20 = 2/9 (2) What is Pr[B | A′]? 2/20 / 11/20 = 2/11
Since Gretchen is a lifeguard, she knows she has to be prepared. On hand, she has 10 bottles of sunscreen and 3 pairs of sunglasses. One day, in a rush to go to the beach, Gretchen takes five of these items at random. A random variable Y is defined to be the number of bottles of sunscreen she takes. Find the pdf of Y.
S = {5S & 0G, 4S & 1G, 3S & 2G, 2S & 3G} Y = {5, 4, 3, 2} Can we use Bernoulli? No, there is no first choice, second choice, etc. She is picking at random all at once, so we use combinations. Total = C(13, 5) = 1287 All combination calculations are shown in the figure, within the PDF.
Mimi rolls a fair 4-sided die twice The random variable S is defined to be the sum of the two rolls. Find the probability density function of S.
S = {8, 7, 6, 5, 4, 3, 2} Is S a binomial random variable? No, we don't have successes and failures. Draw a table, and fill it out based on the die chart (all shown in the figure). - there are 16 outcomes overall The table is the PDF.
Stochastic Processes
Some experiments are best thought of as occurring in sequence - We represent the main experiment as a sequence of smaller (sub) experiments in stages. - Finding the probabilities for each of the stages, we are able to find the probabilities for the larger experiment. (multiplying along the branches of the tree) - These probabilities are found on a tree diagram.
Independent Example
Suppose A and B are independent events, with Pr[A] = 0.2 and Pr[B] = 0.7 Find Pr[A ∩ B] Because they are independent we know: Pr[A ∩ B] = Pr[A] x Pr[B] 0.2 x 0.7 = 0.14 Find Pr[A' ∩ B] = Pr[A'] x Pr[B] = 0.8 x 0.7 = 0.56 Find Pr[A' | B] = Pr[A'] = 0.8 Suppose A and B are disjoint events, with Pr[A] = 0.2 and Pr[B] = 0.7 Find Pr[A' | B] = Pr[A' ∩ B] / Pr[B] = 0.7 / 0.7 = 1
Expected value of Random Variables (mean) Start of 4.3
Suppose a random variable X takes the values x1, x2, ...xk, with probabilities associated to these values p1, p2, ...pk (listed in our PDF) When we have a binomial random variable, E(X) = np Then, the expected value of X is defined to be: E(X) = x1p1 + x2p2 + ... xkpk Example pdf shown in the figure
Conditional Probability and Independence Start of 3.2
The conditional probability of A given B is the probability that A happens if it is known that B happens. - Notation: Pr(A|B) - Read As: "probability of A given B" Let A and B be two nonempty events. The conditional probability of A given B can be found using the formula that follows (also in the figure): - Pr(A|B) = Pr(A ∩ B) / Pr(B)
Independent Events If A & B are independent, Pr[A ∩ B] = Pr[A] x Pr[B]
The outcome of one event does not affect the outcome of the second event - this does not mean that they are disjoint if the Pr[E | F] = Pr[E] and Pr [F | E] = Pr[F] then this means that E has no effect on F, and E has no effect on F. Suppose E and F are independent events (and non-empty). we know through independence that: Pr[E | F] = Pr[E] conditional probability gives us: Pr[E | F] = Pr[E ∩ F] / Pr[F] Putting those together: Pr[E] = Pr[E ∩ F] / Pr[F] Which can also be written as: Pr[E] x Pr[F] = [E ∩ F] Independent vs. Disjoint "A and B are independent" Pr[A ∩ B] = Pr[A] x Pr[B] Pr[A | B] = Pr[A] Pr[B | A] = Pr[B] "A and B are Disjoint" PrA ∩ B] = 0 Pr[A | B] = 0 Pr[ B | A] = 0
Expected Value Continued
This is the same way you might find the average (or mean). As a result, E(X) is also called the mean of X.
75% of the residents of Wisconsin cheer for the Green Bay Packers. Among the packer fans, 60% own a cheesehead hat. Among those who do not cheer for the packers, 20% own a cheesehead hat.
What is the Pr that a randomly selected Wisconsin resident owns a cheesehead hat? Setup the tree. - The first choice on the tree needs to be Fan/not Fan because we know what percentage are fans - The second branch will break into, hat or not hat Take the calculated Pr and add up the two Pr of having a hat Pr[Hat] = 0.50 Tree Diagram in the figure
Review from 3.4 #1 Problem 16: Assume that there are 6 cars in the parking lot: 3 Chevrolets, 2 Fords, and 1 Toyota. Two cars leave at random, one after the other. Draw a tree diagram and use it to answer the following questions.
Tree in Figure (1) What is the probability that the Toyota leaves first? 1/6 (2) What is the probability that the two cars which leave are Chevrolets? 6/30 (3) What is the probability that the Toyota is the second car to leave? 5/30 (4) What is the probability that the Toyota is the second car to leave given that one of the two cars is the Toyota? 5/10 (5) What is the probability that a Ford is the last to leave given that the Toyota leaves first? 2/5
Christina rolls a fair 6-sided die four times. The random variable T is defined to be the number of twos and threes rolled. Find the probability density function of T.
What are the possible values of T? {0, 1, 2, 3, 4} Is T a binomial random variable? Yes So, we can use Bernoulli Trials Make a PDF of T, shown in the figure.
An urn contains balls numbered 1 through 11. A player pays $4 to pick 3 balls from the urn. The player receives money for each ball he picks: each odd ball is worth $0.80 and each even ball is worth $2.00. The random variable Z is defined to be the player's net gain on each play.
What are the possible values of Z? {-1.60, -0.40, 0.80, 2.00} Is this Bernoulli? No, we aren't replacing the balls. So, we must use combinations. Total = C(11, 3) = 165 For Z = 0.80, we have 1 odd and 2 even balls chosen. Pr(z = 0.80) = [C(6, 1)*C(5, 2)]/C(11, 3) = 60/165 Repeat this for everything else in the sample space, and then create the PDF (shown in figure)
David has a bucket and an urn. In the urn there are two red eyeballs and two blue eyeballs. In the bucket there are three blue eyeballs and one red eyeball. The second experiment consists of drawing two eyeballs. Draw an eyeball from the bucket. Put the eyeball in the urn. Take an eyeball from the urn. Note the color of the eyeball you draw.
What is the Pr of drawing a blue eyeball? 2/20 x 9/20 = Pr[blue eyeball] = 11/20 If the eyeball from the urn is blue, what is the Pr that the eyeball from the bucket was red? Pr[the eyeball from bucket is red and the urn eyeball is blue] = 2/20 Pr[eyeball from urn is blue] = 11/20 So, 2/11
30% earned an A or B on the first exam. If a student earned an A or B on the first exam. the probability of getting an A or B on the midterm exam was 80%. If a student did not earn an A or B on the first exam, the Pr of an A or B on the midterm was 20%.
What is the Pr that a randomly selected student has an A or B recorded for his midterm exam grade? - make a tree diagram (shown in the figure), add up for what they're asking Pr[A or B on Midterm] = 38% If Kaylie Q has an A or B recorded for the midterm exam, what is the Pr that she earned an A or B on the 1st exam? Pr[Ab on MT] = 0.38 Pr[A or B on exam 1 | Pr of A or B on the Midterm] = 0.24/0.38 = 12/19 = 63.15%
Connie makes rice every day. Unfortunately, she is an indifferent cook, and only makes it correctly 1/3 of the time.
What is the Pr that she makes rice correctly exactly 3 times in a typical week? = C(n, r) p^r q^n-r (formula) = C(7, 3) p^3 q^4 = C(7, 3) (1/3)^3 (2/3)^4 = 560/2187
A test has 14 multiple choice questions (with 4 choices for each question) and 6 true or false questions. Every question has exactly one correct answer. Scott answers each question at random.
What is the expected number of correctly answered questions? - this is two binomials E(multiple choice) = (14) (1/4) = 3.5 E(TF) = (6)(1/2) =3 So, E(correct) = 6.5 questions on average If each question is worth 5 points, what is Scott's expected score? = 5 x 6.5 = 32.5 average score
First Stochastic Process Problem A single die is rolled, and you note that the result is a 3 or less. An experiment has a fair 6-sided die and a fair 10-sided die. One of the two dice is selected and rolled once. The number on the top is noted.
What is the probability of rolling a 3 or less? - think of this as a tree (shown in the figure) - Pr for 6 sided die = 3/6 = 1/2 (added on second stage branches) - Pr for a 10 sided die = 3/10 (added on second stage branches) - the probability of selecting each of the dice is 1/2 each. This is added to the first branches of the tree because it is the first step in the experiment. Now, use the tree and multiply out each branch to determine its Pr. All branches should add up to 1, or 100%
Marcy chooses a die from a box containing 6 orange dice and 3 blue dice. If the die is orange, she rolls it twice, recording the sum of the two rolls. If the die is blue, she rolls it once, recording the result.
What is the probability that the result is a 5? - make a tree diagram, and then add what they are asking for. = 7/54 If the result is a 5, what is the Pr Marcy picked a blue die? Pr[the result is a 5] = 7/54 Pr[Marcy picks a blue die | the result is a 5] = = 1/18 / 7/54 = 3/7 = 42.85% What is the Pr that the result is an 8? - draw a new tree diagram. We can't get 8 from just one dice so that Pr is 0. Pr[a result of 8] = 10/108 = 9.2% If the result is an 8, what is the probability Marcy picked a blue die? - only one blue die, that can get to as high as 6. So, the Pr[Picking a blue die | the result is 8] = 0
Grounded Example: Jocelyn often gets into trouble for staying out late. Indeed, she stays out late 4/7 of the week. If she stays out late, she gets grounded 49% of the time. If she does not stay out late, she gets grounded 14% of the time.
What percent of the time is Jocelyn grounded? work in the figure = 34% she is grounded If Jocelyn is grounded, what is the Pr it was punishment for staying out late? 196/700 / 238/700 = 196/238 = 82.35%
A pregnancy test claims to be 99% accurate. This means: if a woman is pregnant, then the test will be positive 99% of the time. if a woman is not pregnant, then the test will be negative 99% of the time. For a clinical trial, a large group of women take a pregnancy test. It is known that 10% of the women are pregnant.
What percentage of women get positive results from the test? Using Bayes formula = (.99)(.10) + (.01)(.90) = .099 + .009 = .108 If a woman in the clinical trial gets a positive result, what is the Pr that she is, in fact, pregnant? - solved in the figure, showing Bayes formula
David has a bucket and an urn. In the urn there are two red eyeballs and two blue eyeballs. In the bucket there are three blue eyeballs and one red eyeball. The first experiment consists of drawing a card from a standard deck of 52 cards. If the card is a heart, take an eyeball from the bucket. If the card is any other suit, take an eyeball from the urn. Note the color of the eyeball. Start of 3.4
What's the Pr of drawing a red eyeball? 0.25 x 0.25 = 1/16 0.75 x 0.5 = 6/16 So, 7/16 If the eyeball is red, what is the Pr it came from the earn? Pr[red and from earn] = 6/16 Pr[red] = 7/16 So, 6/7
Review from 3.2 #6 Suppose for this problem that Pr[E] = 5/8 and Pr[F]=7/12. Just as in the book, Pr[(E ∪ F)′] = 0 (1) What is Pr[E | F]? (2) What is Pr[F | E]?
Work in Figure 1) A: 5/14 2) A: 1/3
Review from 4.1 Rework problem 11 from section 4.1 of your text involving the flipping of a weighted coin. Assume that the coin is weighted so that a tail is 3 times as likely as a head. The coin is flipped 8 times. What is the probability that both heads and tails occur?
Work in Figure A: 0.8998718262
Review from 3.3 #1 Assume that Pr[E] = 35 Pr[F | E] = 11/18 and Pr[E′ ∩ F′] = 4/15 Find, Pr[F] = Find, Pr[E ∩ F] =
Work in Figure Pr[F] = 0.5 Pr[E ∩ F] = 0.36666666
Review from 4.3 #2 Rework problem 15 from section 4.3 of your text, involving the expected number of heads when flipping one of two coins. One coin is fair and one is weighted so that Pr[H]=1/4. You randomly select one of the two coins, and flip it 5 times, noting the result of each flip. What is the expected number of heads?
Work in the Figure A: 1.875
QUIZ Question: A fair coin is tossed 5 times. What is the probability that both heads and tails occur at least once?
Work in the figure A: 30/32 = 15/16
Review from 3.2 #3 (1 point) Rework problem 21 from section 3.2 of your text, involving the motor pool. Assume for this problem that the company has 6 Jaguars and 5 Yugos, and two cars are selected randomly and given to sales representatives. What is the probability of both cars being Jaguars, given that both are of the same make?
Work is in the picture A: 3/5 probability that both cars are jags and given to the two sales representatives
Review from 3.2 #4 Box Contains: 7 Red 5 Blue 5 White Two are chosen at random
Work shown in the figure A: 5/33
Conditional Probability word problem example
You are told the following facts about Kenosha County. 73% of the residents are adults (18+) 51% of the residents are female 38% of the residents are adults and female If a Kenosha County resident is female, what is the probability she is an adult? Pr[A | F] = 0.38 / 0.51 = 0.745 or 74.5% If a Kenosha County resident is an adult, what is the probability they are female? Pr[F | A] = 0.38 / 0.73 = 0.521 or 52.1%