# 1 2-step and other basic conditional probability problems

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1 Name M362K Exam 2 Instructions: Show all of your work. You do not have to simplify your answers. No calculators allowed. 1 2-step and other basic conditional probability problems 1. Suppose A, B, C are three events and Find P (A B C). P (A B C) 0.1, P (A C) 0.3, P (B C) 0.3. Solution. By inclusion-exclusion P (A B C) P (A C)+P (B C) P (A B C) 0.5. (remember that P ( C) is a probability function; this is why inclusion-exclusion applies). 2. Suppose A, B are two events satisfying Find P (A B). P (A) 0.4, P (B) 0.6, P (A B) Three actors audition for parts in a forthcoming TV series. On the basis of their past experience and the caliber of the competition they face, A has a 40% chanice of being hired, B has a 50% chance and C a 30% chance. If exactly two of the three are cast, whar is the probability that A was rejected? (You can assume that whether A is hired or not is independent of what happens to B and C). 4. Roll 2 die. Let A be the event of getting at least one 6. Let B be the event of getting two 6 s. Find P (B A). 5. Consider a well-shuffled deck of cards. Assuming the first ace is before the 2 of clubs, what s the probability that the ace of hearts is before the 2 of clubs? Solution. Let E be the event that the first ace is before the 2 of clubs and let F be the event that the ace of hearts is before the 2 of clubs. It is important that E does not mean that there is only one ace before the 2 of clubs or that there is an ace immediately before the 2 of clubs. Instead, it could be that the first card is an ace the next could be the 10 of diamonds, the next could be another ace, then a 3 of spades and then the 2 of clubs. In this situation we would still say that the first ace is before the 2 of clubs so E occurs. The question asks for P (F E). By definition, P (F E) P (F E). P (E) Note that F E. So F E F. Now the ace of hearts is just as likely to before as after the 2 of clubs. So P (F ) 1/2. How do we find P (E)? E occurs if there is some ace before the 2 of clubs. We don t care which ace. The other cards in the deck (the non-ace cards and all cards other

3 (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/ John takes the bus with probability 0.2, the subway with probability 0.3, and catches a ride with probability 0.5. He is late 20% of the time when he takes the bus, 30% of the time when we takes the subway and 50% of the time when we gets a ride from his neighbor. What is the probability that he is late for work? Solution. (0.2)(0.2) + (0.3)(0.3) + (0.5)(0.5) 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, 4, or 5 then you roll the die an additional 2, 3, 4, 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? 2 Bayes formula 13. Three distinct methods A, B and C, are available for teaching a certain skill. The failure rates are 30%, 20% and 10%, respectively. However, due to costs, A is used twice as frequently as B which is used twice as frequently as C. (a) What is the overall failure rate in teaching the skill? (b) A worker is taught the skill, but fails to learn it correctly. What is the probability he was taught by method A? 14. A message is coded into the binary symbols 0 and 1 and the message is sent over a communication channel. The probability a 0 is sent is 0.4, the probability a 1 is sent is 0.6. The channel, however, has a random error that changes a 1 to a 0 with a probability 0.2 and a 0 to a 1 with probability 0.1. (a) What is the probabilty a 0 is received? (b) If a 1 is received, what is the probability that a 0 was sent? 15. Suppose that there is a test for a certain disease with the following properties. If the subject has the disease then test results are positive with probability 0.9 and negative with probability 0.1. If the subject does not have the disease then test results are positive with probability 0.1 and negative with probability 0.9. Assume 10% of the

4 population has the disease. What is the probability that a randomly chosen person has the disease given that he tests positive? Answer. Let D be the event that he has the disease and T the event that he tests positive. P (D T ) P (D T ) P (T ) P (T D)P (D) P (T D)P (D) + P (T D c )P (D c ) 16. 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 S)P (S) 1/2 1/2 1/2 1/2 + 1/3 1/2 1/2 1/2 + 1/ (b) What is the probability that it will land on heads on the next flip? Solution. P (HH H) P (HH F, H)P (F H) + P (HH S, H)P (S H) 1/2 3/5 + 1/3 2/ There are 3 urns. Urn #1 contains 3 red balls and 4 blue balls. Urn #2 contains 4 red balls and 5 blue balls. Urn #3 contains 5 red balls and 6 blue 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 + 4/9 + 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 1)P (E 1 ) P (R) (3/7)(1/3) (1/3) (3/7 + 4/9 + 5/11).

5 (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 i)p (E i ) 3 i1 P (R E i)p (E i ) (1/3) (3/7 2/6 + 4/9 3/8 + 5/11 4/10). (1/3) (3/7 + 4/9 + 5/11) 3 Joint distributions 18. An urn contains 100 balls: 10 red, 20 blue, 30 green and 40 yellow. 50 of these balls are selected at random. Let X be the number of red balls selected and Y be the number of blue balls selected. (a) Assuming the selection is done without replacement, find the joint distribution of (X, Y ). Answer. If 0 i 10 and 0 j 20 then ) P (X i, Y j) ( 10 )( 20 )( 70 i j 50 i j ( ). (b) Assuming the selection is done without replacement, find P (X 5 Y 10). (c) Assuming the selection is done with replacement, find the joint distribution of (X, Y ). Answer. If 0 i 10 and 0 j 20 then ( P (X i, Y j) 50 i, j, 50 i j ) (10/100) i (20/100) j (70/100) 50 i j. (d) Assuming the selection is done with replacement, find P (X 5 Y 10). (e) Suppose the selection is done one ball at a time with replacement. Let Z be the time the first red ball is chosen. Let W be the time the first blue ball is chosen. Find the joint distribution of (Z, W ). Answer. If 1 i < j then P (Z i, W j) (70/100) i 1 (10/100)(80/100) j i 1 (20/100). Similarly, if 1 j < i then P (Z i, W j) (70/100) j 1 (20/100)(90/100) i j 1 (10/100). In all other cases, the probability is zero. (f) Suppose the selection is done one ball at a time without replacement. Let Z be the time the first red ball is chosen. Let W be the time the first blue ball is chosen. Find the joint distribution of (Z, W ).

6 Answer. If 1 i < j then ) ( (i 1)! i P (Z i, W j) ( 70 i 1 j i 1) (j i 1)! 20 (100 j)!. 100! Reason: there are 100! ways to order all 100 balls. That s why 100! is in the denominator, it is the size of the sample space. Next we choose the first (i 1) balls. There are ( 70 i 1) (i 1)! choices: first we choose the balls (we have 70 balls to choose from since we can t choose red or blue) and we have to put them in order. Next we choose the i-th ball. There are 10 choices since there are 10 red balls. Next we choose the (i + 1)-st ball up to the (j 1)-th ball. We have 80 i balls to choose from (since we can t choose any of the i balls that we have already used and we can t choose any blue balls) and we have to choose (j i 1) balls and put them in order. The number of ways to do this is ( 80 i j i 1) (j i 1)!. Next we choose the j-th ball. There are 20 ways to choose this since there are 20 blue balls. Next we order the remaining (100 j) balls. There are (100 j)! ways to do that. If 1 j < i then P (Z i, W j) ( 70 ) ( j 1 (j 1)! j i j 1) (i j 1)! 10 (100 i)!. 100! The reason is similar to the previous case. In all other cases, the probability is zero. 19. Suppose we draw two tickets from a hat that contains tickets numbered 1,2,3,4. Let X be the first number drawn and Y be the second. Find the joint distribution of (X, Y ). 20. Consider the following joint distribution. (a) P (X 1)? (b) P (Y 1)? (c) P (X 1 Y 1)? (d) Are X and Y independent? Y X Flip a coin repeatedly. Let X be the time of the first heads and Y be the time of the second heads. For example, if the flips are T, T, H, T, H, T,... then X 3, Y 5. Find the joint distribution of (X, Y ). 4 Markov Chains 22. Physical experiments with real coins show that the probability that a coin will land on heads given that it starts on heads is about 51%. Similarly, if the coin starts on tails

7 then the probability that it will land on tails in the next flip is about 51%. Suppose that you have a coin which starts heads up. You flip it three times. What is the probability that it lands on heads exactly two of the three times? (This is not a binomial random variable). Answer. This is P (X 1 X 2 X 3 HHT X 0 H) + P (X 1 X 2 X 3 HT H X 0 H) + P (X 1 X 2 X 3 T HH X 0 H) The following transition matrix describes the migration patterns of birds between three habitats If there are 100 birds each habitat at the beginning of the first year, how many should we expect to be in the habitat at the end of the year? At the end of the second year? Answer. [ ] [ ] [ ]. [ ]. So at the end of the first year, there are 80 birds in #1, 120 in #2 and 100 in #3. At the end of the second year, there are 64 birds in #1, 132 birds in #2 and 104 birds in # A health study indicates that from one year to the next, 80% of smokers will continue to smoke while 20% will quit. 10% of nonsmokers will start smoking while 90% will not. (a) Find a transition matrix to describe this chain. Answer. S N p S N (b) Find p 2. (c) If P (X 0 S) 0.2, what is P (X 1 S)? P (X 2 S)? (d) Find P (X 2 S X 0 S). (e) If this trend continues, what percentage of the population will be smokers in long run? Answer

8 25. The graph K 3,3 has 6 vertices labeled v 1, v 2, v 3, w 1, w 2, w 3. There is an edge from v i to w j for all i, j. These are all of the edges. Consider simple random walk on this graph. This is the Markov chain with state space equal to {v 1, v 2, v 3, w 1, w 2, w 3 } and transition probabilities given by p(v i, w j ) 1/3 p(w j, v i ) for all i, j (all other transition probabilities are zero). Is this Markov chain irreducible? Does it have an aperiodic state? 26. Answer. It is irreducible but there are no aperiodic states.

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