Finite Math Lectures 12-19
a number of restaurants in NYC offer a 3 course prix fixe lunch for 29 If a restaurant has 3 appetizers, 4 entrees and 1 dessert for the prix-fixe menu, in how many different ways can a 3-course lunch be chosen? What if a restaurant has 3 appetizers, 4 entrees and 2 desserts? How do you solve?
(a) (3)(4) = 12 (b) (3)(4)(2) = 24
A number is drawn at random from the integers 1,2,3,...,20. Let A be the event that a number selected at random is even and let B be the event that a number selected at random is greater than 12. (a) Find the sample space S of this experiment. (b) Find P(A), the probability of the event A that the number drawn is even. (c) Find P(B), the probability of the event B that the number drawn is greater than 12. How to solve?
(a) The sample space, usually denoted by S, is the set of all possible outcomes of an experiment. Here, the possible outcomes are 1, 2, 3, 4,. . . , 19, 20. Therefore, the sample space is S = {1,2,3,4,...,19,20}. (b) An event is a subset of the sample space S. We say an event A occurs if a trial yields an outcome in A. In this particular problem, a number is drawn at random, so we can assume all outcomes are equally likely. Since there are 10 even numbers in the sample space S, P(A) = P(even) = n(even)/n(S) = 10/20 = 1/2 (c) Similarly, since {13, 14, . . . , 20} is the complete list of numbers in S that are greater than 12, P(B) = P(greater than 12) = n(greater than 12)/n(S) = 8/20= 2/5
A baseball team consists of 15 players. (a) If the team manager wants to make a batting order for tonight's game, how many different possibilities are there? (b) What if the team manager wants to simply choose 9 players who will play in tonight's game? How to solve?
(a) There are 15 ways to choose the first batter, 14 ways to choose the second batter, 13 ways to choose the third batter, and so on. Hence, the number of possible batting orders is P (15, 9) = (15)(14)(13)(12)(11)(10)(9)(8)(7). (b) The order doesn't matter, so the argument given in Lecture Question 13.1 implies that the answer to (a) would overcount the correct answer by a factor of 9!. Hence, the correct answer is P(15,9)/9!, which is usually expressed as C(15,9) or (15 on top of 9) see 13.2 if unclear.
If events A and B are disjoint
A and B cannot happen simultaneously (A and B have no outcomes in common), then P (A u B) = P (A) + P (B).
About the term "with" or "without" replacement
After we draw a ball from the bag, we can put the ball aside or we can put it back into the bag. If we put the ball back in the bag, it may be selected again in the next draw, in which case, we say that two balls are drawn with replacement. On the other hand, if we put the selected ball aside, it cannot be selected in the next draw, in which case, we say that two balls are drawn without replacement.
A number is drawn at random from the integers 1, 2, 3, . . . , 20. Let A be the event that a number selected at random is even and let B be the event that a number selected at random is greater than 12. Find P(A n B), P(A u B) and P(B') How to Solve?
Be sure to draw a Venn diagram to see which region each event represents. In this specific example, P(A n B)= P(even and > 12)= P({14,16,18,20})= 4/20 =1/5. P(A u B)=P(even or > 12)= P({2,4,6,8,10,12,13,14,15,16,17,18,19,20)=14/20= 7/10. P(B')= P(not > 12) = P(less than or equal 12) = P({1,2,3,4,...,11,12}) = 12/20 = 3/5.
Consider an election with two candidates, Candidate A and Candidate B. Every voter is invited to participate in an exit poll, where they are asked whom they voted for; some accept and some refuse. For a randomly selected voter, let A be the event that he or she voted for A, and W be the event that he or she is willing to participate in the exit poll. Suppose that P (W | A) = 0.7 but P (W | A0) = 0.3. In the exit poll, 60% of the respondents say they voted for A (assume that they are all honest), suggesting a comfortable victory for A. Find P(A), the true proportion of people who voted for A. How to Solve?
By Bayes' rule together with LOTP P(A|W) = P(W |A)P(A)/P(W)= P(W |A)P(A)/P(W |A)P(A) + P(W |A')P(A') Letting p = P (A), we obtain 0.6= 0.7p/0.7p + 0.3(1-p) = 7p/4p+3 Solve for P p= 9/23= 0.391 Hence, A actually received fewer than half of the votes. This example shows that you should NOTbelieve results of exit polls.
Suppose A and B are two events such that P (A) = 0.7, P (B) = 0.4, and P (A u B)=0.9. Find P(A|B) and P(B|A').
By inclusion-exclusion theorem P (A u B) = P (A) + P (B)-P (A n B), 0.9=0.7+0.4-P(A n B) =) P(A n B)=0.2.
How many different outcomes are there in rolling a six-sided die 3 times? r times? How do you solve?
Each roll results in one of the six outcomes, so if the die is rolled 3 times, there are (6)(6)(6) = 216 outcomes. Similarly, if rolled r times, there are (6)(6) · · · (6) = 6 to the r power outcomes.
Note 15.2
How about P(A' u B')? It is always a good practice to draw a Venn diagram. In this case, drawing a Venn diagram (or using one of De Morgan's laws) allows us to observe P(A' u B')= P(A n B')= 1 P(A n B)=1-0.09=0.91.
A professor gives a multiple-choice quiz with eight questions. Each question has four possible answers. A student wildly guesses the answers. What is the probability that he passes the test if at least four correct answers are needed to pass? How to Solve
I think its easier to understand in our notes but here is the example from the lecture he posted online Let X be the number of correct answers. Then X has a binomial distribution with parameters n = 8 and p = 1/4, so P(X greater than or equal to 4)= P(X =4)+P(X =5)= C(8,4)p^4q^4 +C(8,5)p^5q^3 = 70(1/4)^4 times (3/4)^4 + 56(1/4)^5 times (3/4)^3 = 0.1096
Note from 17.1
If A and B are independent, then so areA' and B, AandB', and A' and B'. Independence (P(A n B)=P(A)P(B)) and disjointness (A and B do not overlap) are completely different concepts.
General Formula from 14.1 about equally likely events. Helps with 14.1(b) and (c)
If all outcomes in a (finite) sample space S have the same probability of occurring (that is, the outcomes are equally likely), then the probability that an event A occurs is given by P(A) = n(A)/n(S)= # of elements in A/# of elements in S
Note 15.1
In this example, P(A)+P(B)-P(A n B) gives the same probability as P(AuB). Moreover, the answer to (c) can also be obtained as P(B')=1-P(B)=1-8/20=3/5.
A laboratory blood test is 98% e↵ective in detecting a certain disease when it is, in fact, present. However, the test also yields a "false positive" result for 5% of the healthy persons tested. (That is, if a healthy person is tested, then, with probability 0.05, the test result will imply he or she has the disease.) If 3% of the population actually has the disease, what is the probability that a person has the disease given that the test result is positive? How to Solve?
In this problem, we need to find the conditional probability P (D | +), where D = disease and "+" = positive. By Bayes' rule together with LOTP, P(D|+) = P(+|D)P(D)/P(+)= P(+|D)P(D)/ P(+|D)P(D) + P(+|D')P(D')= (0.98)(0.03)/(0.98)(0.03)+ (0.05)(0.97)= 0.377 Hence, the prior probability 0.03 of having the disease has been updated to the posterior probability 0.377 based on the evidence that the test result was positive. It is a little surprising that even if the test result is positive, there is still a big chance that the person may not have the disease.
Note from 16.4
LOTP also allows us to calculate the probability P(Cn) that the nth card selected is a club,which turns out to be P(Cn)=1/4 for any n. This suggests that the order of picking lottery tickets does not affect the chance of your winning (or losing).
A survey of randomly selected 10000 college faculty members shows that 50% of the faculty own a car, 25% own a bicycle, and 9% own both. Find the probability that a faculty member chosen at random owns a car or a bicycle (or both). How to Solve?
Let A be the event that a faculty member owns a car and B be the event that a faculty member owns a bicycle. We have P (A) = 0.50, P (B) = 0.25, P (A n B) = 0.09. By the inclusion-exclusion theorem, P (A u B) = P (A) + P (B)-P (A n B) = 0.50 + 0.25- 0.09 = 0.66.
A machine needs two components to be operating. The components will fail during the course of a month with probabilities 0.7 and 0.2, respectively. Assuming that the failures are independent, what is the chance that the machine will still be working at the end of the month? How to Solve?
Let A be the event that the first component will fail during the month and let B be the event that the second will fail during the month. We want to find P(A' n B'), but by independence (together with NOTE in Question 17.1), this equals P(A' n B') = P(A')P(B') = (1-0.7)(1-0.2) = 0.24.
Cards are selected one by one from a standard deck of 52. Find the probability that the second card selected is a club. How to Solve?
Let C1 and C2 be the event that the first card selected is a club and the event that the second card selected is a club, respectively. Then by LOTP with the partition {C1, C1'}, P(C2) = P(C1)P (C2 | C1 ) + P (C1' )P (C2 | C1' ) = 13/52 · 12/52 + 39/52 · 13/51 = 1/4 .
Suppose that a bag contains 8 red balls and 4 white balls. We draw 2 balls from the bag without replacement. If we assume that at each draw each ball in the bag is equally likely to be chosen, what is the probability that both balls drawn are red? How to Solve?
Let R1 and R2 be the event that the first ball selected is red and the event that the second ball selected is red, respectively. As in Question 15.3, we can calculate the probability as P(R1 n R2)= C(8,2)/ C(12,2) =14/33 On the other hand, it is possible to use the multiplication rule, which follows from the definition of conditional probability: For any events A and B (with P (A) > 0), the equality P (A n B) = P (A)P (B | A) holds P(R1 n R2)=P(R1)P(R2|R1)= 8/12 · 7/11 =14/33
A spam filter is designed by looking at commonly occurring phrases in spam. Suppose that 80% of email is spam. In 10% of the spam emails, the phrase "free money" is used, whereas this phrase is only used in 1% of non-spam emails. A new email has just arrived, which does mention "free money." What is the probability that it is spam? How to Solve?
Let S be the event that a randomly selected email is spam and F be the event that a randomly se- lected email contains the phrase "free money." We need to find the conditional probability P (S | F ). By Bayes' rule together with LOTP, P(S|F)= P(F|S)P(S)/P(F) =P(F|S)P(S)/P(F|S)P(S)+ P(F|S')P(S')= (0.1)(0.8)/(0.1)(0.8)+ (0.01)(0.2)= 0.976. Hence, the prior probability 0.8 of spam has been updated to the posterior probability 0.976 based on the evidence that the email contains "free money."
A single fair die is rolled seven times. Find the probability that a 3 turns up exactly twice How to Solve?
Let X be the number of times a 3 turns up. Then X has a binomial distribution with parameters n = 7 and p = 1/6, so P(X = 2) = C(7,2)p^2q^5 = 21(1/6)^2 times (5/6)^5 = 0.2344.
The multiplication rule yields the law of total probability (LOTP)
Let {A,B} be a partition of the sample space S (i.e. A n B = and A u B = S). Then for any event C, P (C) = P (C n A) + P (C n B) = P(C | A)P(A) + P (C | B)P(B).
A basketball player makes 80% of her free throws. She takes 3 free throws in a game. Let X be the count of her successful free throws. Assume that the shots are independent of each other. Find the probability P (X = k) for all possible values of k. How to Solve?
Look at 18.1 too complicated
Two doctors A and B each performed 100 surgeries in the past few years. Their overall surgery success rates are 80% and 83%, respectively. Based on this information, is it safe to say that Dr. B is a better doctor?
Look at 19.3 too complicated
An instant lottery game gives you probability 0.02 of winning on any one play. Plays are independent of each other. If you play 3 times, find the probability that you win on at least one play. How to Solve?
NOTE: The pairwise independence (the first condition) does not imply mutual independence. In this specific problem, P(at least one W) = 1-P(all L), but by independence, this equals 1-P(L on first)P(L on second)P(L on third) = 1-(1-0.02)^3=0.0588
From a group of 6 women and 7 men, how many different committees consisting of 2 women and 3 men can be formed? How to solve?
Note that the order doesn't matter in this question. The number of different ways to choose 2 women from the 6 women is (6 on top of 2) = P(6,2)/2! = 15 The number of di↵erent ways to choose 3 men from the 7 men is (7 on top of 2) = P(7,3)/3! = 35. Hence, by the multiplication rule (Lecture Question 12.1), (15)(35) = 525 different committees can be formed.
A natural generalization of the concept of independence to more than two events is the following: Events A, B and C are said to be (mutually) independent if both of the following are satisfied:
P(A n B) = P(A)P(B), P(B n C) = P(B)P(C), P(C n A) = P(C)P(A) P(A n B n C)=P(A)P(B)P(C)
inclusion-exclusion theorem
P(A u B)=P(A)+P(B)-P(A n B)
complement theorem
P(A')=1-P(A), or equivalently, P(A)= 1-P(A')
The conditional probability of an event A given an event B is defined by
P(A|B)=P(A n B)/P(B), providedP(B)>0.
Three people from a group of seven are chosen for a photograph. How many distinct groups are possible (if we do not distinguish different orders)? How to solve?
Since we do not distinguish different orders, the following 6 orders for three people a, b and c are considered to represent 1 possible group for a photograph: abc, acb, bac, bca, cab, cba Therefore, the calculation (7)(6)(5) = 210 in Question 12.3 would overcount the correct answer by a factor of 6. Hence, the number of possible distinct groups is 210/6= 35
A box has 10 good and 3 defective light bulbs; 2 are selected at random. What is the probability that at least 1 is good? How to Solve?
Solution 1: By the complement theorem P(A) = 1-P(A'), P (at least 1 is good) = 1-P (both are defective). Let S be the set of all outcomes (all different ways to choose 2 from the 13 bulbs), and let A be the event that 2 defective bulbs are chosen. Then P(at least 1 is good)=1-P(A)=1 n(A)/n(S)=1-[C(3,2)/C(13,2)] =1-(1/26) =25/26. Solution 2: The above probability can also be calculated in the following way: P (at least 1 is good) = P ("both are good" OR "1 is good and 1 is defective") = P (both are good) + P (1 is good and 1 is defective) = C(10, 2)/C(13, 2)+ [C(10, 1)C(3, 1)]/ C(13, 2)= 25/26
How many different 6-letter words can be formed using the letters PEPPER? How to solve? Solution 1
Solution 1: If we could distinguish the 6 letters, the answer would be 6!. However, three P's cannot be distinguished, and two E's cannot be distinguished, so we need to adjust for overcounting by dividing by 3!*2!. Therefore, the correct answer is 6!/3!*2!=60
Three people from a group of seven line up for a photograph. How many distinct line-ups are possible? How to solve
Suppose there are three seats. • There are 7 ways to choose a person who sits in the left seat. • There are 6 ways to choose a person who sits in the middle seat. • There are 5 ways to choose a person who sits in the right seat there are (7)(6)(5) = 210 distinct line-ups. The above answer can be expressed as P (7, 3) = (7)(6)(5).
There are 9 football teams. How many football games are played during a season if the teams all play each other exactly once? How to solve?
The answer coincides with the number of different ways to choose 2 teams out of 9 teams, so C(9,2) = P(9,2)/2! = 36.
13.2 general equation for combination
The combination of size k from a set of size n represents the number of different ways to choose k objects from n distinct objects (but without arranging them into an order). The combination is given by the binomial coeffcient C(n,k) = (n on top of k) = P(n,k)/k! (read as "n choose k").
The definition of conditional probability together with the multiplication rule yields Bayes' rule:
The equality P (A | B) = P (A)P(B|A)/P(B) holds provided P(A) > 0 and P (B) > 0.
Note about 13.1 and over counting
The factor 6 was obtained above by listing all different letter arrangements of a, b and c. It is also possible to obtain it simply as 3! = (3)(2)(1) = 6. Why did we need to divide the answer to Question 12.3 by 3! to get the correct answer? Mathematically speaking, the process of seating three people as in Question 12.4 can be divided into the two activities: (i) selecting a group of three people and (ii) arrange those three into an order. The multiplication rule then yields [answer to 12.3] = [answer to 13.1]times [# of distinct seating arrangements for three]. This gives us the correct answer as [answer to 13.1] = [answer to 12.3] divided by [# of distinct seating arrangements for three] = P (7, 3)/3!
Note 15.4
The first value of k for which the probability of at least one match exceeds 0.5 is k = 23, so in this classroom, there is a better than 50% chance that two or more of us have the same birthday. By the time we reach k = 57, the probability of at least one match exceeds 99%.
know this: permutation
The number of different ways to select k objects from a set with n distinct objects without replacement (i.e. choosing a certain object precludes it from being chosen again) and arrange them into an order is given by the so-called permutation P(n,k)=nPk =n(n 1)(n 2)···(n k+1).
An auto dealer sold 120 hybrid cars. His records show that during the first year • 70 cars required minor repairs, • 28 cars required major repairs, a • the other cars required no repairs. Based on this information, estimate the probability that a hybrid car a woman purchased from the dealer will require no repairs during the first year. How to solve?
The relative frequency of cars with no repairs is # of occurrences of cars with no repairs divided by total # of cars sold = 120-(70 + 28)/120=0.1833, which is considered to be an estimate for the exact probability (which no one knows)
Note from 14.2
The same argument shows that P (exactly two T's) is also 3/8. If H appears more frequently than T (in which case the coin is unfair), then we expect P (exactly two H's) > P (exactly two T's)
8 people will be seated in a row at random. Find the probability that three people a, b and c sit next to each other. How to solve?
The sample space S consists of all different line-ups of 8 people, so n(S) = 8!. Let A be the event that a, b and c sit next to each other. To find n(A): • Group a, b and c together and regard them as 1 person. • Then there are "6 people" (with a, b and c regarded as 1 person) who will be seated in a row, for which there are 6! different ways. • With the "6 people" seated, a, b and c are automatically next to each other. • But the 3 people can be seated in 3! ways. Therefore, by the multiplication rule: n(A) = (6!)(3!). Consequently, P(A)= n(A)/n(S) = (6!)(3!)/8!=3/28
A committee of 5 is to be selected from a group of 7 men and 6 women. If the selection is made randomly, what is the probability that the committee consists of 3 men and 2 women? How to Solve?
The sample space S consists of all different ways of forming a committee of 5 out of 13 people, so n(S) = C(13,5) = P(13,5)/5! = (13)(12)(11)(10)(9)/(5)(4)(3)(2)(1)= 1287. If A denotes the event that a committee consisting of 3 men and 2 women is formed, then by Question 13.4, n(A) = 525. Therefore, P(A)=n(A)/N(S)= 525/1287= 0.4079.
A pair of 4-sided fair dice, one blue and one white with faces 1, 2, 3 and 4, is rolled. Find the probability that the two numbers that turn up add to 4. How to Solve?
The sample space consists of pairs (i,j) of two numbers i and j, where i = 1,2,3,4 and j = 1,2,3,4: S = {(1,1), (1,2), (1,3), (1,4) (2,1), (2,2), (2,3), (2,4) (3,1), (3,2), (3,3), (3,4) (4,1), (4,2), (4,3), (4,4)}. Since the dice are fair, all the outcomes are equally likely. Only the three outcomes underlined above yield two numbers whose sum is 4, so P (two numbers adding to 4) = n(two numbers adding to 4)/n(S) = 3/16
Find the probability of exactly two heads appearing in a sequence of three tosses of a fair coin. How to Solve?
The sample space contains the following 8 outcomes: S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. These outcomes are equally likely since the coin is assumed to be fair. Only the three underlined outcomes contain exactly two H's, so P (exactly two H's) = n(exactly two H's)/n(S) = 3/8
Assume that you need to choose a 7-character password for a certain website, where each character must be a lowercase letter, an uppercase letter or a number. (a) How many possibilities are there if you are allowed to use each character only once? (b) How many possibilities are there if you are allowed to use each character more than once but are required to have at least one uppercase letter in your password? How to solve?
There are 62 (= 26 + 26 + 10) letters available, so P (62, 7) = (62)(61)(60)(59)(58)(57)(56) = 2,478,652,606,080= 2.478*10 to the 12th power. There are 62 (= 26 + 26 + 10) letters available and repetitions are allowed, so if there are no restrictions, there are 627 possible passwords. However, we have to exclude those consisting only of numbers and lowercase letters. There are 367 such passwords. Therefore, if we are required to have at least one uppercase letter in our password, the number of possible passwords is 62^7-36^7 = 3,521,614,606,208-78,364,164,096 which equals 3.443*10^12.
In 1998, Sally Clark was tried for murder after two of her sons died shortly after birth. During the trial, an expert witness for the prosecution testified that the probability of a newborn dying of sudden infant death syndrome (SIDS) was 1/8500, so the probability of two deaths due to SIDS in one family was (1/8500)2, or about one in 73 million. Therefore, he continued, the probability of Clark's innocence was one in 73 million. Is there any problem in what the prosecutor said? How to Solve?
There are at least two major problems. • The expert assumed that deaths due to SIDS were independent. This is very unlikely to be a correct assumption. • The expert confused P (innocence | two deaths) with P (two deaths | innocence). He should have discussed the former conditional probability, but he instead looked at the latter.
Four people line up for a photograph. How many distinct line-ups of all four are possible? how to solve?
This can be identified with the number of distinct arrangements of four different objects, which is 4! = P (4, 4) = (4)(3)(2)(1) = 24.
There are k people in a room. Assume each person's birthday is equally likely to be any of the 365 days of the year (we exclude February 29), and that the k people's birthdays are independent (we assume there are no twins in the room). What is the probability that two or more people in the group have the same birthday?
This is called the birthday problem. The sample space S consists of all birthday combinations of k people. For each person, there are 365 possible birthdays, so n(S) = 365^k. Consider the event that there is no match between the k birthdays. There are 365 possible birthdays for the 1st person, 364 possible birthdays for the 2nd person, 363 possible birthdays for the 3rd person, and so on. Therefore, n(no match) = P (365, k) By assumption, all outcomes in the sample space are equally likely. By the complement theorem, P (at least one match) = 1-P (no match) = 1-n(no match)/ n(S)= 1-P (365, k)/365^k .
How many ways can we award a 1st, 2nd and 3rd place prize (Gold, Silver and Bronze) among eight contestants?
We are choosing 3 people out of 8 and arrange them into an order, so P (8, 3) = (8)(7)(6) = 336.
How many different 6-letter words can be formed using the letters PEPPER? How to solve? Solution 2
We need to decide where to put the three P's from the 6 available positions, then where to put the two E's from the remaining 3 positions. (Then the R is placed in the remaining 1 position.)Therefore (6 on top of three)(3 on top of 2) = 60. Fun Fact: if the letters were all different it would just be 6!7
Give a definition of independence of two events A and B.
We say two events A and B are independent if learning that B occurred gives us no information that would change the probability of A occurring (and vice versa), i.e. P(A|B) = P(A) (and P (B | A) = P (B)). This leads to the following general definition: Events A and B are said to be independent if P(A n B) = P(A)P(B).
union P(A u B)
is the event consisting of those outcomes that are in A or B (or both).
intersection (A n B)
is the event consisting of those outcomes that are in both A and B.
multiplication rule
if two activities 1 and 2 can be performed in m and n different ways respectively then the compound activity "1 followed by 2" can be performed in m*n different ways
complement A'
is the event consisting of those outcomes in S that are not in A.
If k = n, then the notation n! (read as "n factorial") is used:
n! = P (n, n) = n(n-1)(n-2) · · · (3)(2)(1).
know this: stuff about replacement
the number of different ways to select k objects from a set with n distinct objects with replacement (i.e. choosing a certain object does not preclude it from being chosen again) and arrange them into an order is given by n to the k power.
note from 13.6
• The relative frequency is sometimes referred to as the empirical probability as opposed to the theoretical probability. Empirical probabilities are calculated based on real data and used to estimate theoretical probabilities (which are unknown in many real-life applications). • In general, as more and more trials of an experiment are repeated, the relative frequency of an outcome approaches the theoretical probability of the outcome. This concept is called the law of large numbers. • Probabilities can take values between 0 and 1.