Math 30530: Introduction to Probability, Spring 2012

Transcription

1 Name: Math 30530: Introduction to Probability, Spring 01 Midterm Exam I Monday, February 0, 01 This exam contains problems on 7 pages (including the front cover). Calculators may be used. Show all your work on the paper provided. Unsupported answers may not receive full credit. The honor code is in effect for this exam. Scores Question Score Out of Total 100 GOOD LUCK!!!

2 1. (15 points) A. A fair coin is tossed 3 times, the tosses being independent. What is the probability that there are exactly heads? The uniform sample space Ω = {hhh, hht, hth, htt, thh, tht, tth, ttt} is appropriate. (Recall that uniform means that all sample points have the same probability.) The event that there are exactly two heads is the event A = {hht, hth, thh}. Thus P (A) = 3/8. B. A fair die is rolled. What is the probability that an even face turns up? The uniform sample space Ω = {1,, 3, 4, 5, } is appropriate. The event that an even face turns up is the event A = {, 4, }. Thus, P (A) = 1/. C. An urn contains 3 green balls and red balls. What is the probabiliity that a randomly drawn ball is red? Let R denote the event that a red ball is drawn. There are red balls among the 9 balls in the urn, thus, P (R) = 3.

3 . (15 points) A fair die is rolled twice. Let A denote the event that an even face turns up on the first roll, B the event that an even face turns up on the second roll, and C the event that the sum of the faces is even. Assume that all outcomes of the experiment are equally probable. The uniform sample space Ω = {(i, j) : i, j = 1,..., } is appropriate, where i is the number showing on the first roll and j is the number showing on the second. A. Find P (A), P (B), and P (C). The event A consists of the sample points (i, j) such that i {, 4, } and j {1,, 3, 4, 5, }. Thus A contains 18 sample points, and so P (A) = 1/. The event B consists of the sample points (i, j) such that i {1,, 3, 4, 5, } and j {, 4, }. Thus B contains 18 sample points, and so P (B) = 1/. Arrange the sample points in a table with the sample point (i, j) in the i th row and j th column of the table. The table has rows and columns. In each row of the table, sample points with even and odd sums alternate. Thus each row contains three sample points belonging to C. Therefore the event C contains 18 sample points, and so P (C) = 1/. B. Show that A and B are independent. Also show that A and C are independent. (The argument showing that A and C are independent will also show that B and C are independent.) Note that AB = {(i, j) : i, j {, 4, }} Thus AB contains 9 sample points, and so P (AB) = 1/4. Since P (AB) = 1 4 = 1 1 the events A and B are independent. = P (A)P (B), 3

4 Now AB C, for if an even number shows on each roll, then the sum must be even. Since AB A, it now follows that AB AC. Conversely, if the first die shows an even number and the sum is even, then the second die must show an even number. Thus AC B. Since AC A, it now follows that AC AB. Therefore AC = AB, so P (AC) = P (AB) = 1/4. Since P (AC) = 1 4 = 1 1 the events A and C are independent. = P (A)P (C), C. Are A, B, and C mutually independent? Justify your answer. We have shown above that AB = AC, so ABC = ACC = AC. Thus P (ABC) = P (AB) = 1/4. However, P (A)P (B)P (C) = 1/8. Therefore the events A, B, and C are not mutually independent. 4

5 3. (18 points) After removing the pits from 4 of cherries, you were distracted from this real-world task by the necessity of studying for your probability exam. You left all cherries in a bowl on your backyard picnic table and went off to study for a while. While you were out, the neighbor s pet pig selected cherries from the bowl at random and ate them. A. What are the probabilities that k unpitted cherries, for k = 0, 1,, remain in the bowl after the pig has feasted? There are ( ) = 15 ways for the pig to select cherries from the bowl. For k = 0, 1,, let A k denote the event that k unpitted cherries remain in the bowl after the pig s visit. Now A 0 is also the event that the pig selected unpitted cherries from the bowl. There is only one way the pig could have done this, so P (A 0 ) = 1/15. Similarly, A 1 is the event that the pig selected 1 pitted and 1 unpitted cherry from the bowl. There are ( ( 1) 4 1) = 8 ways the pig could have done this, so P (A1 ) = 8/15. Finally, A ) is the event that the pig selected pitted cherries from the bowl. There are = ways the pig could have done this, so P (A ) = /15. ( 4 B. Let X be the number of pitted cherries remaining in the bowl. What is the probability mass function of X? Now, p X () = P ({X = }) = P (A ) = 15. Similar reasoning shows p X (3) = 8 15, p X (4) = 1 15 Thus p X (x) = 15 if x = 8 15 if x = if x = 4 0 otherwise 5

6 C. Assessing the damage, you randomly choose 1 cherry at random from the 4 in the bowl, intending to console yourself by eating it. If the cherry you chose was pitted, what is the probability that the pig ate exactly 1 pitted cherry? Let B denote the event that you choose a pitted cherry. P (A 1 B) = P (A 1B) P (B) P (A 1 B) = P (A 0 B) + P (A 1 B) + P (A B) P (A 1 )P (B A 1 ) = P (A 0 )P (B A 0 ) + P (A 1 )P (B A 1 ) + P (A )P (B A ) = 3 5 Remark. One could also use a tree diagram to analyze the problem.

7 4. (14 points) Let A and B be events in a probability space such that P (A) = 3/4 and P (B) = 1/3. What are the minimum and maximum possible values of P (A B)? What are the minimum and maximum possible values of P (AB)? The inclusion-exclusion principle P (A B) + P (AB) = P (A) + P (B) tells us that if P (A) + P (B) is fixed, the maximum value of P (A B) and the minimum value of P (AB) will occur together, as will the minimum and maximum values of P (A B) and P (AB). Since we must have P (A B) 1, it follows that P (AB) P (A) + P (B) 1 = 1 1. On the other hand, AB B, so P (AB) 1/3. Thus P (A B) P (A)+P (B) 1/3 = 3/4. We have shown that 1 1 P (AB) P (A B) 1 Strictly speaking we have given lower and upper bounds for the probabilities of the two events. We have not given examples showing these bounds can be attained. Although I did not require it for the exam, we really should give such examples. Let Ω = {1,,..., 1} be (the sample space of) a uniform probability space, and define the random variable X by X(ω) = ω for each ω Ω. (If you want a more colorful interpretation, think of rolling a dodecahedral die.) For the first example, let A = {X 9}, and let B 1 = {X 4}. Then B 1 A, so P (AB 1 ) = P (B 1 ) = 1/3. Also P (A B 1 ) = P (A) = 3/4. For the second example, let B = {X 9}. Then P (AB ) = P ({9}) = 1/1 and P (A B ) = P (Ω) = 1. 7

8 5. (18 points) Consider a sequence of independent rolls of a fair die. A. Suppose that the die is rolled until a appears. What is the probability that the experiment terminates (i.e., the first appears) on the n-th roll, where n is a positive integer? The first appearing on the n th trial means that the die showed some other face on the first n 1 trials. Since the trials are independent the probability of this event is ( ) n B. Again assume that the die is rolled until a appears. What is the probability that the experiment terminates after a finite number of rolls? (That is, what is the probability that the experiment does not go on forever?) We give two solutions and an example. First solution. Let A denote the event that the experiment terminates after a finite number of rolls and, for each n in N, let A n denote the event that the experiment terminates on the n th trial. Then A = n=1a n and the events A n, n N, are pairwise mutually exclusive. Thus P (A) = = P (A n ) n=1 n=1 = = 1 ( ) n 1 5 ( ) 1 8

9 Second solution. In the notation of the first solution, A c is the event that the experiment does not terminate. We show that P (A c ) = 0. Let X denote the number of the turn on which the first appears. For each n in N, the event {X > n} is just the event that a does not appear in the first n rolls of the die. Thus P (X > n) = (5/) n for each n N. Now A c {X > n} for each n N, so ( ) n 5 P (A c ) for each n N. Thus and so P (A c ) = 0. ( ) n 5 0 P (A c ) lim = 0, n Remark. Some people argued that the probability is zero that the experiment goes on forever because the probability that the experiment terminates on the n th trial tends to zero as n tends to infinity. The example below shows that this argument is not valid. Example. Suppose that we have an infinite sequence of coins such that the probability that the first coin comes up heads is 1/, the probability that the second comes up heads is 1/4, and in general, the probability that the k th coin comes up heads is 1/ k, k N. We toss these coins in turn until a head appears for the first time. (Note that we no longer have Bernoulli trials.) The probability that this experiment terminates on the n th trial is p n = 1 3 ( ) 1 n 1 n Note that p n 1/ n, so lim n p n = 0. We show next that the probability that the experiment terminates is strictly less than 1, or equivalently, the probability that a head never appears is positive. Indeed the probability that an unbroken infinite sequence of tails occurs is lim n n k=1 (1 1 n ). We shall show that each of these products is at least 1/4. Indeed, one can prove by induction that n (1 1 ) 1 n n+1 k=1 Thus the probability that a head never appears is at least 1/4. 9

10 C. Suppose that the die is rolled 10 times. What is the probability that the third appears on the 10-th roll. For the third to appear on the 10-th roll, there must have been two s in the first 9 rolls. This event occurs with probability ( ) 9 ( ) 1 ( ) 7 5. A occurs on the third roll with probability 1/. By the independence of the trials, the required probability is ( ) ( ) 7 ( )

11 . (0 points) Consider forming 4-letter words from the letters A, B, C, D, and E. (For our purposes, a 4-letter word is just a string of four letters. We require that each letter in the word be one of A, B, C, D, E.) A. How many such words are possible? We will call the 4-letter words described above words of type A. Each letter of a word of type A can be chosen in 5 ways. This choice is made 4 times in succession. Thus, by the generalized basic principle of counting (GBPC) there are 5 4 = 5 words of type A. B. How many such words are possible if no letter is repeated? There are 5 choices for the first letter, 4 for the second, 3 for the third, and for the fourth. By the GBPC there are = 10 words of type A in which no letter is repeated. C. How many such words are possible if exactly one of the letters A, B, C, D, E occurs twice? There are 5 choices for the letter to be repeated, and there are ( 4 ) choices for the positions in the word it will occupy. This leaves two vacant positions in the word. There are 4 choices for the letter to occupy the first of these and 3 choices for the letter to occupy the second. Thus there are ( ) = 30 words of type A in which exactly one of the letters is repeated. 11

12 D. How many such words are possible if none of the letters occurs more than twice? First solution. We begin by finding the number of words of type A in which two letters are each repeated twice. There are ( 5 ) choices for the two letters to be repeated. There are ( 4 ) choices for the positions that two instances of the first letter, alphabetically, is to occupy. Two instances of the second letter, alphabetically, will then occupy the remaining two positions. Thus there are ( ) ( ) 5 4 = 0 words of type A with two letters repeated twice. Now there are 10 words of type A with no repeated letters, 30 with exactly one letter repeated twice, and 0 with two letters repeated twice. Thus there are = 540 words of type A in which no letter is repeated more than twice. Second solution. We begin by finding the number of words of type A in which one letter is repeated three times. There are 5 choices for the letter to be repeated and 4 choices for the other letter. There are 4 choices of position for the letter that occurs only once. Thus there are = 80 words of type A in which one letter occurs three times. There are 5 words of type A in which one letter occurs 4 times. Thus there are 5 (80 + 5) = 540 words of type A in which no letter is repeated more than twice. 1