# 3 The multiplication rule/miscellaneous counting problems

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1 Practice for Exam 1 1 Axioms of probability, disjoint and independent events 1 Suppose P (A 0, P (B 05 (a If A and B are independent, what is P (A B? What is P (A B? (b If A and B are disjoint, what is P (A B? What is P (A B? 2 Permutations and Combinations 2 Roll a die ten times What the probability of getting exactly four 6 s? (For example, this occurs if you roll In tossing fair dice, what s the probability of tossing at most one 3? How many anagrams of MISSISSIPPI are there? Answer ( 11 1,,,2 5 The price of a certain stock changes each day at random: it goes up \$1 with probability 06 and down \$1 with probability 0 Assuming that changes on different days are independent, (a what s the probability that it will be back to its starting price after 10 days? Solution The stock must go up 5 times and down 5 times for this to happen The probability is ( 10 5 (06 5 (0 5 (b what s the probability that it will be up by at least \$1 after days? 6 What is the coefficient on x y 6 in the product (2x + 3y 10? Solution This uses the binomial formula The answer is ( The multiplication rule/miscellaneous counting problems 7 A hand of poker consists of 5 cards randomly drawn from a deck of 52 cards What s the probability of getting -of-a-kind? (There are kinds and suits Each card has a suit and a kind for a total of x52 cards A 5-card hand is dealt from a well-shuffled deck of 52 playing cards What is the probability that the hand contains at least one card from each of the four suits?

2 9 You are dealt 6 cards from a standard deck of 52 cards What is the probability that you get a four-of-a-kind and two cards that don t match For example, you could get Kings, a 10 and a 3 But you couldn t get Kings and two 10s 10 A licence plate consists of 3 letters followed by 3 numbers (for example XTY3 If no letter can be used more than once how many licence plates are there? The numbers can be repeated For example XTY00 is valid Answer In a bridge game, each of players gets cards drawn at random from an ordinary deck of 52 cards What is the probability that two players each have 2 aces? There are aces in the deck Answer The sample space has cardinality Ω 52!! The event can be described as an experiment in stages: (a First choose which 2 players get 2 aces There are ( 2 choices For example, suppose the players are called North, South, East, West We could have chosen North and South (b Next decide which aces each player gets There are ( 2 choices In the example, there are ( 2 ways to give two aces to North and South gets the other two (c Last, divide up the remaining cards There are ( 11,11,,! choices This 11! 2! 2 is because we only have to divide up the non-ace cards and two of the players are just getting 11 cards each (since they ve already received two cards each In the example, we are choosing 11 non-ace cards for North, 11 non-ace cards for South, non-ace cards for East and non-ace cards for West So the answer is ( 2! 2 11! 2! 2 52!! ( 2! 2 52!!2 11! 2 12 A certain group of 20 people consists of 7 doctors, 3 lawyers and 10 bankers They are all seated at random around a round table (a What is the probability that the 3 lawyers sit next to each other? Solution 19! is the number of ways to seat 20 people around a round table (we consider two seatings to be the same if they only differ by a rotation There are 16! ways to seat the non-lawyers (up to rotation, 17 ways to pick a spot for the three lawyers and 3! ways to order the lawyers So the answer is: 17! 3! 19! (b What is the probability that no doctor sits next to another doctor? Solution There are 12! ways to seat the non-doctors (up to rotation We choose 7 positions (out of to seat the doctors so that no two of them sit together (there

3 are ( 7 choices Then we have to decide which doctor sits where There are 7! choices So the answer is: 12! ( 7 7! 12!! 19! 6!19! A school play has distinct male roles and 5 distinct female roles If 7 men and women audition for the play, how many possible casts are there? What if Bob and Alice refuse to be in the play together? Solution The number of possible casts is ( ( 7! 5 5! (choose men, then assign them roles, choose 5 women, then assign them roles The number of ways that Alice and Bob can both be in the play is ( ( 6 3 3! 5 7! (choose a role for Bob, then roles for the other males, then a role for Alice, then a role for the other females So if Alice and Bob refuse to be in the play together the number of possible casts is ( ( 7 Urn problems! 5 5! ( 6 3 3! 5 1 Urn #1 contains 3 black and red balls Urn #2 contains 5 black and 2 red balls A ball is chosen at random from urn #1 A ball is also chosen at random from urn #2 What s the probability the two balls have the same color? 15 An urn contains 100 balls: 25 red, 25 blue, 50 green Select 12 balls at random from the urn (a Assume that the selection is done without replacement Compute the probability that 3 red, blue ball and 5 green balls are selected Solution ( 25 3 ( 25 ( 50 5 ( (b Assume the same question if the selection is done with replacement Solution ( ( 3 ( ( ,, (c What is the probability that all 12 balls have the same color? (assume the selection is done without replacement, then answer the same problem with replacement (d An urn contains 10 red balls and 5 blue balls You grab balls at random (without replacement What is the probability that you grabbed 2 red balls and 2 black balls? Solution ( 10 2 ( 5 2 ( 15 ( 7!

4 5 Inclusion/Exclusion and union problems 16 Suppose you draw cards out of a regular deck of 52 cards What is the probability that you will have at least 6 cards of the same suit? Solution There are four suits Let us number them 1,2,3, Let E i be the event that we get at least 6 cards of suit i These sets are disjoint, have the same probability, and the question asks for the probability of their union So P ( i1e i P (E 1 Now P (E 1 P (F 6 + P (F 7 + P (F where F i is the event of getting exactly i cards of suit 1 Also P (F i ( ( 39 i i So the answer is ( 39 ( ( ( 17 In a certain city, 50% of the people speak Spanish, 5% speak English, 0% speak French, 15% speak Spanish and English, 15% speak English and French, 10% speak Spanish and French If everyone speaks at least one of these three languages then what percentage of people speak all three? Solution By inclusion-exclusion x (where x is the percentage that speak all three Solving for x, we obtain x Suppose you are dealt 5 cards at random from an ordinary deck of 52 cards What s the probability of getting exactly four face cards? A face card is a king, queen or jack There are 12 face cards in the deck Solution (12 ( Suppose that E 1, E 2, E 3, E are four events such that P (E i 0 (for all i, P (E i E j 03 (for all i j, P (E i E j E k 01 if i, j, k are distinct and P (E 1 E 2 E 3 E 0 What is P (E 1 E 2 E 3 E? Answer (0 6( Suppose you draw 9 cards out of a regular deck of 52 cards What is the probability that you will have at least 7 cards of one suit? For example, you could get 7 hearts and one diamond or 6 spades, 1 club and 1 heart (There are suits: hearts, diamonds, clubs and spades There are cards of each suit Answer ( 7 ( ( ( ( 9 9

5 21 Suppose you draw cards out a regular deck of 52 cards What is the probability that you get at least one -of-a-kind? (For example, this event occurs if you draw Jacks and Kings or if you draw Jacks, a 2, a 3, a and a 5 Hint: Let E i be the event that you get cards of kind i (i {1,, } since there are kinds Compute P ( i1e i using inclusion-exclusion Solution Let E i be the event that you get cards of kind i (i {1,, } since there are kinds Then P ( i E i i P (E i i<j P (E i E j by the inclusion-exclusion formula (note that E i E j E k is empty if i, j, k are distinct Now ( P (E i and So the answer is P ( i E i P (E i E j 1 ( ( ( 1 52 ( 1 ( 2 52 Alternative solution You can answer this problem without using inclusion-exclusion To do so, let A be the event of getting exactly one -of-a-kind and B the event of getting exactly two -of-a-kinds Then P ( i E i P (A B P (A + P (B since A and B are disjoint Now ( 2 P (B To see this, observe that the sample space consists of all ways of selecting cards from 52 (this explains the ( 52 in the denominator and the number of ways to get 2 -of-a-kinds is the number of ways to choose 2 kinds amongst or ( 2 Also, P (A [( ] 12 Why? Well, there are ways to choose the kind that will be the -of-a-kind Then you have to choose more cards from amongst the that are left (and there are ( to do that However, you should not choose cards of a kind again; so we have to subtract the 12 ways that you can select another -of-a-kind So the final answer is P (A B [( ] (

6 6 Conditional expectation 22 There are 3 cards in a hat One is black on both sides, one is red on both sides and one is black on one side and red on the other side A card chosen at random and placed (at random on a table so that only one side is showing Given that the side showing is red, what is the probability that the other side is also red? Solution Let F denote the event that the side showing is red, RR the event of getting the red-red card, RB the event of getting the red-black card and BB the event of getting the black-black card Then P (RR F P (RR F P (F 1/3 P (RR + 1/2P (RB 1/3 1/3 + 1/ Consider the following game Roll a 6-sided die If it lands on six, you win outright If it lands on one, you lose If it lands on 2, 3,, or 5 then you roll the die an additional 2, 3,, or 5 times respectively If it lands on a 6 at least one time, then you win Otherwise you lose (a What are your chances of losing given that you rolled a 3 on the first roll? (b What are your chances of winning? 7 Recursive conditional probability problems 2 Alice and Bob are playing a game in which they take turns flipping a coin The coin lands on heads with probability 1/3 Alice goes first Alice wins if she flips heads before Bob flips tails What s the probability that Alice wins? Answer Bayes (2/3 k (1/3 k 1 (1/3 (1/3 1 2/9 (3/7 k0 25 There are 2 coins in a box The first lands on heads with probability 1/2 The second lands on heads with probability 1/3 You choose a coin at random (each possibility being equally likely and flip it once It comes up heads (a What is the probability that it was the first coin? Solution Let F denote the first coin, S the second coin, and H landing on heads We want to know P (F H P (F H P (F H P (H F P (F P (H P (H F P (F + P (H SP (S 1/2 1/2 1/2 1/2 + 1/3 1/2 1/2 1/2 + 1/3 3 5

8 2 Suppose that there is a test for a certain disease with the following properties If the subject has the disease the test results are positive with probability 09 and negative with probability 01 If the subject does not have the disease the rest results are positive with probability 01 and negative with probability 09 Assume 20% of the population has the disease What is the probability that a randomly chosen person has the disease given that she tests positive? Solution Let D be the event of having the disease and T the event of testing positive Then P (D T P (T DP (D P (T DP (D + P (T D c P (D c There are 3 urns Urn #1 contains 3 red balls and blue balls Urn #2 contains red balls and 5 blue balls Urn #3 contains 5 red balls and 6 red balls An urn is chosen at random and a ball is selected from the urn (a What is the probability that a red ball is selected? Solution Let E i be the event that the ith urn is selected and R the event that a red ball is selected Then P (R 3 P (R E i P (E i (1/3 (3/7 + /9 + 5/11 i1 (b Given that a red ball is selected, what is the probability that Urn #1 was chosen? Solution P (E 1 R P (R E 1P (E 1 P (R (3/7(1/3 (1/3 (3/7 + /9 + 5/11 (c Given that a red ball is selected, what is the probability that a second ball selected at random from the same urn (without replacement is red? Solution Let RR be the event that two red balls are selected Then P (RR R P (RR 3 P (R i1 P (RR E ip (E i 3 i1 P (R E ip (E i (1/3 (3/7 2/6 + /9 3/ + 5/11 /10 (1/3 (3/7 + /9 + 5/11 (d There are two coins in a box One is fair and the other is 2-headed (The fair coin lands on heads 50% of the time You choose a coin at random and flip it 10 times It comes up heads each time What s the probability that it is the 2-headed coin? Solution Let H denote the event of flipping heads 10 times, F of picking the fair coin P (F c H P (H F c P (F c P (H F c P (F c + P (H F P (F 1 (1/2 1/ (1/

9 9 Independence 30 A coin lands on heads with probability 01 It is flipped 100 times What is the probability that it lands on heads exactly 50 out of the 100 times? 31 A coin is tossed twice Consider the events: A heads on the first toss B heads on the second toss C the two tosses come out the same (a Are A, B independent? Solution yes (b Are B, C independent? Solution yes (c Are A, C independent? Solution yes (d Are A, B, C jointly independent? Justify your answer Solution no because P (C A B 1 1/2 P (C 32 Suppose P (A 02, P (B 03 (a If A and B are independent, what is P (A B? What is P (A B? (b If A and B are disjoint, what is P (A B? What is P (A B? 33 The price of a certain stock changes each day at random: it goes up \$1 with probability 06 and down \$1 with probability 0 Assuming that changes on different days are independent, what s the probability that it will be back to its starting price after 10 days? Solution The stock must go up 5 times and down 5 times for this to happen The probability is ( 10 5 (06 5 (0 5

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