Chapter 5- Discrete Probability

Consider an experiment in which a fair coin is flipped 3 times. Define the random variable A to be the number of flips that come up heads. Which choice corresponds to the distribution over A? (3,1/8), (2,3/8), (1, 3/8), (0,1/8) (3,1/4), (2,1/4), (1, 1/4), (0,1/4) (3,1/8), (2,3/8), (1, 3/8)

(3,1/8), (2,3/8), (1, 3/8), (0,1/8) A has a value of 3 for only one outcome (HHH), so p(A = 3) = 1/8. A has a value of 2 for three outcomes (HHT, HTH, THH), so p(A = 2) = 3/8. A has a value of 1 for three outcomes (HTT, THT, TTH), so p(A = 1) = 3/8. A has a value of 0 for only one outcome (TTT), so p(A = 0) = 1/8.

What is the number of outcomes from dealing a 5-card hand? What about if four of the cards were jacks and one was any random card?

(52 / 5) 48

Consider the sample space of all outcomes of a roll of a blue and a red die. Define M to be the value obtained by subtracting the number on the red die from the number on the blue die. What is the value of M(1, 5)? 4 -4 6 Consider the sample space of all outcomes of a roll of a blue and a red die. Define M to be the value obtained by subtracting the number on the red die from the number on the blue die. What is the range of M? {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} {-5, -4, -3, -2, -1, 0} Consider an experiment in which a fair coin is tossed 3 times. Define the random variable A to be the number of flips that come up heads. What is the value of A(HTH)? 1 2 3 Consider an experiment in which a fair coin is tossed 3 times. Define the random variable A to be the number of flips that come up heads. What is the range of A? {0, 1, 2, 3} {1, 2, 3} {0, 1, 2}

-4 The number obtained by subtracting the value on the red die (5) from the value on the blue die (1) is -4. {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} The value of M(1, 6) is -5. The value of M(6, 1) is 5. Every other integer from -5 to 5 is in the range of M. 2 The number of heads in the sequence HTH is 2. Therefore A(HTH) = 2. {0, 1, 2, 3} There are 3 flips of the coin. The number of heads can range from 0 through 3.

If a red and a blue die are thrown, then what is the probability that the numbers on the two dice are the same? 5 / 36 1 / 2 1 / 6 What is the probability that a random 5-card hand has all four aces? 48 / (52 / 5) 4⋅48 / (52 / 5) 1 / (48 / 5)

1 / 6 The event that the two dice have the same number is {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}. The number of outcomes in the event is 6. The size of the sample space is 36. Therefore the probability is 6/36 = 1/6. 58 / (52 / 5) The size of the sample space is (52 / 5). The number of 5-card hands that contain all 4 aces is 48. Four of the cards in the hand must be the only 4 aces in the deck, so there are 52 - 4 = 48 possibilities for the fifth card. Therefore the number of outcomes in the event is 48.

A fair coin is flipped 3 times. Define the following events: F: the first flip comes up heads. E: the number of heads is even. (Zero is an even number.) W: at least two of the three flips come up heads. What is p(F)? What is p(E)? What is the probability that the first flip comes up heads and there is an even number of heads? (What is p(F ∩ E)?) Are E and F independent? What is the probability that at least two flips come up heads? (What is p(W)?) What is the probability that the first flip is heads and there are at least two flips that are heads? Are F and W independent?

1/2 F = {HHH, HHT, HTH, HTT}. Therefore p(F) = |F|/8 = 1/2. 1/2 E = {HHT, HTH, THH, TTT}. Therefore p(E) = |E|/8 = 1/2. 1/4 F ∩ E = {HHT, HTH}. p(F ∩ E) = 2/8 = 1/4. yes Yes, p(F ∩ E) = 1/4 = (1/2)·(1/2) = p(F)·p(E). 1/2 W = {HHH, HHT, HTH, THH}. p(W) = |W|/8 = 1/2. 3/8 F ∩ W = {HHH, HHT, HTH}. p(F ∩ W) = |F ∩ W|/8 = 3/8. No No, p(F ∩ W) = 3/8 ≠ (1/2)·(1/2) = p(F)·p(W).

The following questions pertain to an experiment in which a blue and red die are tossed. Express probabilities as a simplified fraction What is the probability that the sum is at least 10 given that the two die have the same number? What is the probability that the sum is less than 10 given that the two die have the same number? What is the probability that the sum is at least 10 given that the blue die comes up 2? What is the probability that the sum is less than 10 given that the blue die comes up 2?

1/3 There are two outcomes in which the sum is at least 10 and the two dice have the same number: (5, 5) and (6, 6). There are 6 outcomes in which the dice have the same number. Therefore, the probability is 2/6 = 1/3. 2/3 Event T: The sum of the two dice is at least 10 Event |T: The sum of the two dice is less than 10 Event N: The two dice have the same number The previous question showed that p(T|N) = 1/3. Since p(T|N) + p(T|N) = 1, then p(T|N) = 1 - 1/3 = 2/3. 0 There are no outcomes in which the blue die comes up 2 and the sum is at least 10. Therefore, the probability is 0. 1 Event T: The sum of the two dice is at least 10 Event |T: The sum of the two dice is less than 10 Event W: The blue die comes up 2 The previous question showed that p(T|W) = 0. Since p(T|W) + p(T|W) = 1, then p(T|W) = 1 - 0 = 1.

An experiment consists of three consecutive flips of a coin. The outcome is the sequence of outcomes of the three flips. There are eight possible outcomes in the sample space which are listed below. "H" stands for heads and "T" stands for tails: HHH HHT HTH HTT THH THT TTH TTT If the coin is a fair coin, then all eight outcomes are equally likely. The event that all three flips come out the same is E = {HHH, TTT} What is the probability of this event?

1/4

Give your answer as an integer or a fraction, simplified to its lowest terms, (e.g., 3/4 instead of 9/12). In an experiment consisting of three flips of a fair coin, what is the probability that the first two flips are both heads? In an experiment consisting of three flips of a fair coin, what is the probability that the first two flips are the same? In an experiment consisting of a roll of a red and blue die, what is the probability that the red die is one more than the blue die?

1/4 The event is the set E = {HHH, HHT}. |E|/|S| = 2/8 = 1/4. 1/2 The event is the set E = {HHH, HHT, TTH, TTT}. |E|/|S| = 4/8 = 1/2. 5/36 The event is the set E = {(1,2), (2,3), (3,4), (4,5), (5,6)}. |E|/|S| = 5/36.

Enter your answer as a simplified fraction Consider the sample space of all outcomes of a roll of a blue and a red die. Define M to be the value obtained by subtracting the number on the red die from the number on the blue die. What is the probability that M = 2?Remember that in the ordered pair, the value of the blue die is first and the value of the red die is second. Consider an experiment in which a fair coin is tossed 3 times. Define the random variable A to be the number of flips that come up heads. What is the probability that A = 1?

1/9 The event that M = 2 is {(6, 4), (5, 3), (4, 2), (3,1 )}. Therefore the probability that M = 2 is 4/36 = 1/9. 3/8 The event that A = 1 is {HTT, THT, TTH}. The size of the sample space is 8, so p(A = 1) = 3/8.

Answer the questions below using the picture on the other side of the card What is the size of the sample space? What is the probability that the two balls chosen are both red? What is the probability that one ball is red and one ball is blue? What is the probability that both chosen balls are blue? What is the expected number of red balls chosen?

10 The outcomes are all possible ways of selecting 2 balls from the five which is(5 / 2) = (5⋅4) / (2⋅1) = 10. 1/10 There is only one way to pick two red balls since there are only two red balls in the box: they both have to be chosen. Therefore, the probability is 1/10. 3/5 There are 2 ways to select a red ball and 3 ways to select a blue ball. Therefore there are 2 ⋅ 3 = 6 ways to select one red and one blue ball. Therefore the probability is 6/10. 3/10 There are(3 / 2) = (3⋅2⋅1) / (2⋅1) = 3ways to select 2 of the 3 blue balls. Therefore, the probability is 3/10. 4/5 E[R] = 2⋅p(R=2) + 1 ⋅ p(R=1) + 0 ⋅ p(R=0) =2(1/10) + 1(6/10) =8/10 =4/5

Express answer as an integer or a fraction in lowest terms (e.g., 3/4, not 9/12) If a fair coin is flipped three times, what is the probability that there is at least one heads? If a fair coin is flipped ten times, what is the probability that there is at least one heads?

7/8 The event that there are no heads is E = {TTT}. The probability that there is at least one heads is1 - p(E) = 1 - (1/8) = 7/8. 1023/1024 or 1 - 1/1024 The event that there are no heads is E = {TTTTTTTTTT}, so P(E) = 1/1024. The probability that there is at least one heads is 1 - p(E) = 1 - (1/1024) = 1023/1024.

Express the expectation as a simplified fraction A fair coin is flipped. Define a random variable X that is 1 if the outcome of the flip is heads and 3 if the outcome of the flip is tails. X(H) = 1, X(T) = 3. What is E[X]? A fair die is tossed. Define random variable Y to be 1 if the outcome is a six and 0 otherwise. What is E[Y]? A single share of a stock is purchased for $200. With probability 0.3, the company will meet the deadline for their new product and the value of the stock will go to $300 in the next week. With probability 0.7, the company will miss the deadline and the value of the stock will go down to $100. What is the expected value of the stock next week?

2 The sample space S = {H,T}. p(H) = p(T) = 1/2. E[X] = X(H)p(H) + X(T)p(T) = 1 · (1/2) + 3 · (1/2) = 2. 1/6 The sample space is {1, 2, 3, 4, 5, 6}. Each outcome has probability 1/6. Y(6) = 1 and Y = 0 for all the other outcomes. E[Y] = 1 · (1/6) + 0 · (5/6) = 1/6. 160 Expected value is 300(0.3) + 100(0.7) = 90 + 70 = 160.

Two dice are rolled. Enter the size of the set that corresponds to the event that the sum of the dice is 5 Enter the size of the sample space when two 6-sided dice are rolled A fair coin is flipped three times. Enter the probability that all three flips are the same

4 36 1/4

The following questions pertain to the experiment in which a fair coin is tossed 3 times. Let W be the event that there are at least two heads. What is |W|? Let F be the event that the first flip comes up heads. What is |W ∩ F|? What is p(F|W)?

4 W = {HHH, HHT, HTH, THH}, so |W| = 4. 3 W ∩ F = {HHH, HHT, HTH}. Therefore, |W ∩ F| = 3. 3/4 p(F|W) = |F ∩ W|/|W| = 3/4.

Consider the example with the fair and loaded die. The gambler selects one of the two die at random and rolls the die twice. The outcomes of the two rolls are independent. The first outcome is not a 6 and the second outcome is a 6. Define Y to be the event that two rolls of the die result in the first roll not coming up 6 and the second roll coming up 6. Express probabilities as a simplified fraction What is p(Y|F)? What is p(Y|F)? Apply the results from the previous two questions to Bayes' Theorem to determine p(F|Y).

5/36 If the die is fair, the probability of not getting a six is 5/6. The probability of getting a 6 is 1/6. Since the two rolls are independent, the two probabilities can be multiplied to get 5/36. 10/49 If the gambler picks the loaded die, the probability of not getting six is 5/7. The probability of getting a 6 is 2/7. Since the two rolls are independent, the two probabilities can be multiplied to get 10/49. 49/121 The numerator is (5/36)(1/2). The denominator is (5/36)(1/2) + (10/49)(1/2).

A loaded die is twice as likely to come up 6 than any of the other five possibilities. The probability distribution over the outcomes of a single roll of the die is defined by: P(1) = P(2) = P(3) = P(4) = P(5) = 1 / 7 P(6) = 2 / 7 The loaded die is rolled twice. The outcomes of the two rolls are independent, meaning that any event that only depends on the outcome of the first roll is independent of any event that only depends on the outcome of the second roll. Express probabilities as a simplified fraction What is the probability that the first roll does not come up 6? What is the probability that the first roll does not come up 6 and the second roll does come up 6? What is the probability that exactly one of the two rolls comes up 6?

5/7 p(1) + p(2) + p(3) + p(4) + p(5) = 5/7 10/49 The probability that the first roll does not come up 6 is 5/7. The probability that the second roll does come up 6 is 2/7. The two events (that the first is not 6 and that the second is 6) are independent. Therefore the probability that the first roll does not come up 6 and the second roll does come up 6 is (5/7)(2/7) = 10/49. 20/49 Either the first roll is 6 and the second is not 6, or the first roll is not 6 and the second is 6.The two possibilities are mutually exclusive and each has a probability of 10/49. Therefore the probability that exactly one roll comes up 6 is 10/49 + 10/49 = 20/49.

A blue die and a red die are thrown. B is the event that the blue comes up an even number. E is the event that both dice come up even. Enter the sizes of the sets |E ∩ B| and |B| |E ∩ B| = ? |B| = ? A blue die and a red die are thrown. B is the event that the blue comes up with a 5. E is the event that the sum of the dice is 6 Enter the value for the probability of event E given event B. p(E|B) =

9 18 1/6

What is the distribution of a random variable?

A list of possible values of a variable together with how often each value occurs. Since a random variable has some value for every outcome in the sample space, the sum of the values p(X = r), over all r ∈ X(S), must equal 1.

What is a random variable?

A variable whose value depends on the outcome of a random event

With experiments and outcomes, what is a sample space and event?

An experiment is a procedure that results in one out of a number of possible outcomes. The set of all possible outcomes is called the sample space of the experiment. A subset of the sample space is called an event.

What is Bayes' Theorem? What is the formula?

Bayes' Theorem is the cornerstone of many algorithms in machine learning in which the goal is to determine the likelihood of some event based on data obtained from observations.

We calculated earlier that when the blue and red die are thrown, the event E that the two dice are the same has probability 1/6. Now suppose we consider the probability of E conditioned on the event F that the blue die comes up 5. What is p(E|F)? The event E ∩ F is the event that the first die comes up 5 and the two dice are the same. There is only one outcome then in E ∩ F: (5, 5)

Conditioning on the event F does not change the probability of E. In other words, knowing that the first die came up 5 provides no additional information about the likelihood that the two dice will have the same number. Two events are independent if conditioning on one event does not change the probability of the other event.

What is probability distribution? What is the formula?

Consider a dishonest dice player who shows up to a game with a loaded die. The player's die is biased so that an outcome of 6 is twice as likely to occur as the other numbers. The probability that you roll 1-5 or roll a 6 is as follows: p(1) = p(2) = p(3) = p(4) = p(5) = 1 / 7 chance p(6) = 2 / 7 chance

What does it mean when a set is countably infinity or uncountably infinite?

Discrete probability is concerned with experiments in which the sample space is a finite or countably infinite set. A set is countably infinite if there is a one-to-one correspondence between the elements of the set and the integers. A set that is not countably infinite is said to be uncountably infinite. Examples of countably infinite sets include the set of all binary strings (of any length), the set of ordered pairs of integers (Z × Z), the set of all rational numbers. The set of real numbers is an example of an uncountably infinite set. In fact the set of real numbers in a finite interval (for example the set of real numbers from 0 to 1) is also uncountably infinite.

What is the inclusion-exclusion principle?

If two events are not mutually exclusive, the probability of the union of events can be determined by a version of the Inclusion-Exclusion principle

What is the conditional probability formula?

Let's roll our blue and red dice again and look at the event E that the two numbers on the dice sum to at least 11. Now suppose we start the experiment by rolling the blue die first and the blue die comes up 5. How does the information that the blue die has come up 5 change the probability that the event E happens? In a sense, the sample space has shrunk from the set of all 36 outcomes to the set of outcomes (5, *) in which the blue die comes up 5 There are two events needed in the analysis. The first is the original event E that the sum of the two numbers is at least 11. The second event F is that the blue die comes up 5. If the event F happens, the new probability of E is the conditional probability of E given F, denoted by p(E|F).

Using the blue and red spinners and formula from the previous question, calculate the expected value of the sum of the spinners ( E[M] )

List out all of the potential outcomes of the spinners and add each set together M(s) = 2 , 3 , 4 , 3 , 4 , 5 , 4 , 5 , 6 Multiply each number by the probability of each outcome, 1/9 M = 4

Are events A and B mutually exclusive? A coin is tossed five times. A: The first three flips come up heads B: The last three flips come up tails A coin is tossed six times. A: The first three flips come up heads B: The last three flips come up tails A 5-card hand is dealt from a standard playing deck. A: The hand has three aces B: The hand has a pair of kings A 5-card hand is dealt from a standard playing deck. A: The hand has three aces B: The hand has three kings

Mutually exclusive If the first three flips come up heads, then the outcome looks like HHH**. If the last three flips come up tails then the outcome looks like **TTT. In the first event, the third flip comes up heads. In the second event, the third flip comes up tails. The third flip can not both be heads and tails, so there is no outcome in which the first three flips come up heads and the last three flips come up tails. Not mutually exclusive HHHTTT is an outcome in which the first three flips come up heads and the last three flips come up tails. Therefore the intersection of the two events is not empty. Not mutually exclusive Below is a 5-card hand with a pair of kings and three aces: {K♠, K♣, A♥, A♦, A♣} Mutually exclusive Since there are only five cards in the hand, the hand can not have three aces and three kings.

What is the complement of an event?

Sometimes it is easier to determine the probability that an event doesn't happen than to determine that the event does happen. The complement of an event E is S − E

Using the previous "Alternative way to calculate the expectation of a random variable" formula and the red and blue spinner example, calculate the expectation of a random variable again.

Sort the outcomes by the value of M 2 = 1/9 because it appears once 3 = 2/9 because it appears twice 4 = 3/9 5 = 2/9 6 = 1/9 Multiply each number by their fraction M = 4

What is uniform distribution? What is the formula for it?

The probability distribution in which every outcome has the same probability is called the uniform distribution. Since there are |S| outcomes in sample space S and their probabilities sum to 1, under the uniform distribution, for each s ∈ S, p(s) = 1/|S|. Returning to the experiment with the red and blue die, if the dice are fair, then each of the 36 outcomes is equally likely. The event that the sum of the two numbers is 8 is the set E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}. The probability of the event E is |E|/|S| = 5/36.

A dealer deals a 5-card hand. If the deck is perfectly shuffled, then each 5-card hand is equally likely and the distribution over the sample space of 5-card hands is uniform. What is the probability that the hand has two pairs? The definition of having two pairs is that there are two pairs of cards, each pair has the same rank. The pairs have different rank from each other and the fifth card (that is not part of a pair), has different rank from the pairs Examples: {2♥ , 2♦ , 8♠ , 8♣ , A♣} A hand with 2 pairs {2♥ , 2♦ , 2♠ , 2♣ , A♣} Not 2 pairs, must be different ranks {2♥ , 2♦ , 8♠ , 8♣ , 2♣} Not 2 pairs, last card must have different rank

There are (13 / 2) ways to select the ranks for the two pairs from {2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A} There are (4 / 2) ways to select the suits for one pair from the set {Diamonds, Hearts, Clubs, Spades}. There are (4 / 2) ways to select the suits for the other pair. The last card can be any card whose rank is not the same as the pairs. There are 52 − 4 − 4 = 44 choices. The probability that a random hand is a 2-pair hand is (13 / 2) * (4 / 2) * (4 / 2) * 44 / (52 / 5) ≈ .04754

What does it mean when two events are mutually exclusive?

Two events are mutually exclusive if the two events are disjoint (i.e., the intersection of the two events is empty).

The following questions pertain to an experiment in which a fair coin is flipped three times. Define the following three events: A: The first two flips come up heads. B: The last two flips come up heads. C: The last two flips come up tails. Enter your answer as a simplified fraction Are the events A and C mutually exclusive? What is the probability that the first two flips come up heads and the last two flips come up tails? (What is p(A ∩ C)?) What is the probability that the first two flips are heads or the last two flips are tails. (What is p(A ∪ C)?) Are the events A and B mutually exclusive? What is the probability that the first two flips come up heads and the last two flips come up heads? (What is p(A ∩ B)?) What is the probability that the first two flips are heads or the last two flips are heads. (What is p(A ∪ B)?)

Yes A = {HHH, HHT} and C = {HTT, TTT}. Since A ∩ C = ∅, A and C are mutually exclusive. 0 |A ∩ C| = 0. p(A ∩ C) = |A ∩ C|/8 = 0/8 = 0. 1/2 p(A) = 2/8 = 1/4. p(C) = 2/8 = 1/4. Since A and C are mutually exclusive, p(A ∪ C) = 1/4 + 1/4 = 1/2. No A = {HHH, HHT} and B = {HHH, THH}. The outcome HHH is in both A and B, so A and B are not mutually exclusive. 1/8 A ∩ B = {HHH}, so |A ∩ B| = 1 and p(A ∩ B) = |A ∩ B|/8 = 1/8. 3/8 p(A) = 1/4, p(B) = 1/4, and p(A ∩ B) = 1/8. Therefore, P(A ∪ B) = 1/4 + 1/4 - 1/8 = 3/8.

In the experiment of a 5-card hand, is the following an outcome or an event? {4♣, 6♠, 7♥, Q♦, K♦} outcome event In the experiment of a 5-card hand, is the following an outcome or an event? {{4♣, 6♠, 7♥, Q♦, K♦}, {4♠, 9♥, 9♣, 9♦, K♦}} outcome event In the example of a red and blue die that are thrown, define the event E to be that the number on both dice are multiples of 3. Which set corresponds to E? (3, 6) {(3, 3), (6, 6)} {(3, 3), (3, 6),(6, 3), (6, 6)} Suppose a coin is flipped three times. The outcome of the experiment is the sequence of outcomes from each flip. For example, HHH denotes the outcome in which the coin comes up heads in each flip. How many distinct outcomes are there? 8 2 4 In the experiment where the coin is flipped three times, which set corresponds to the event that at least two of the three flips come up heads? {HHH, HHT} {HHH, HHT, HTH, THH} {HHH, TTH, THT, HTT

outcome Every outcome is a subset of 5 cards. Therefore the set of 5 cards given is an outcome. event The given set has two elements, each of which is a set of 5 cards. Therefore, the set is an event with two outcomes. {(3, 3), (3, 6),(6, 3), (6, 6)} The set includes exactly those outcomes in which both dice are a multiple of 3. 8 There are two possibilities for each coin flip. Therefore the number of outcomes is 2^3 = 8. {HHH, HHT, HTH, THH} The set corresponds to exactly those outcomes in which the coin comes up heads at least two times.