Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 2 September 17, 2009

2.10. (a) The sample space is the set {A, B, AB, O}.

(b) The probabilities of the simple events are P (A) = .41, P (B) = .1, P (AB) = .04, P (O) = .45.

(c) P ({A, AB}) = P (A) + P (AB) = .41 + .04 = .45.

2.11 (a) Since E_{i}, i = 1, 2, 3, 4, 5 make up the whole sample space, 1 = P (E_{1}) + P (E_{2}) + P (E_{3}) + P (E_{4}) + P (E_{5}) = .15 + .15 + .4 + 3P (E_{5}). Then P (E_{5}) = .1 and P (E_{4}) = .2.

(b) Similarly, P (E_{1}) = .3 and P (E_{2}) = .1, so 1 = .3 + .1 + 3P (E_{3}). Hence P (E_{3}) = P (E_{4}) = P (E_{5}) = .2.

2.13. (a) The simple events are the four possible responses, say S = {V L, SL, U, O}. (b) Not equally likely, P (V L) = .24, P (SL) = .24, P (U) = .4, P (O) = .12.

(c) P (SL, V L) = P (SL) + P (V L) = .24 + .24 = .48.

2.14. (a) The probability that a random adult needs glasses is .58 (add the numbers in the first row).

(b) The probability that a random adult needs glasses but does not use them is .14.

(c) The probability that a random adult uses glasses for reading (whether or not they are needed) is .46 (add numbers in first column).

2.19. (a) There are 9 points in the sample space:

S = {(V_{1}, V_{1}), (V_{1}, V_{2}), (V_{1}, V_{3}), (V_{2}, V_{1}), (V_{2}, V_{2}), (V_{2}, V_{3}), (V_{3}, V_{1}), (V_{3}, V_{2}), (V_{3}, V_{3})}.

(b) Assuming the selections are all random, it makes sense to assign the probability 1/9 to each simple event.

(c) There are three simple events in A, so P (A) = 3/9 = 1/3. There are 5 simple events in B (B = {(V_{2}, x), (x, V_{2}) | x ∈ {V_{1}, V_{2}, V_{3}}). So P (B) = 5/9. Then P (A∩B) = 1/9 since A∩B includes only (V_{2}, V_{2}). Finally A ∪ B includes all the pairs except (V_{1}, V_{3}) and (V_{3}, V_{1}), so P (A ∪ B) = 7/9.

2.20. (a) With no information, and with the choice of curtain made randomly, P (G) = P (D_{1}) = P (D_{2}) = 1/3.

(b) (i) The probability is P (G) = 1/3. (ii) She will get the dud prize in this case, since the first choice was the good prize. (iii) If she initially chooses one of the duds, the other dud is shown, and she switches, then she wins the good prize. (iv) This probability is 2/3. To see this, think of the three possible ways this can go – if she chooses the good prize (probability 1/3), the host shows one dud, and she chooses the other, which is also a dud. If she chooses one dud (probability 2/3), the host shows the other dud, and she switches to the good prize! (v) The strategy that maximizes the contestant’s chance of winning the good prize is to always switch to the other curtain.

2.28. Let the majority candidates be M1, M2, M3, and the minority candidate m. (a) S = {M1M2, M1M3, M2M3, M1m, M2m, M3m}.

(b) If the selection is really random each of the six simple events in S should be equally probable so each has probability 1/6.

(c) The event A where m is included in the two candidates chosen is A = {M1m, M2m, M3m}, so P (A) = 3/6 = 1/2.

2.29. Let the men be m1, m2, m3, m4 and the women w1, w2. (a) The experiment consists of selecting two individuals independently and at random out of the 6. A sample point will consist of a set (unordered pair) like m3w1.

(b) The sample space consists of

S = {m1m2, m1m3, m1m4, m1w1, m1w2, m2m3, m2m4, m2w1, m2mw2, m3m4,
m3w1, m3w2, m4w1, m4w2, w1w2}.

(c) The probability is 1/15.

2.30. (a) A sample point is an ordered listing of the 3 wines, like 312 (i.e. wine 3 is best, followed by wine 1 next, and wine 2 worst).

(b) The sample space is S = {123, 132, 213, 231, 312, 321}.

(c) Say the superior wine is 1 and the second best is 2. The cases where the “expert” ranks 1 no worse than 2 are 123, 132, 312. So the probability is 3/6 = 1/2.

2.35. Since the flights are on successive days, there are no time conflicts and all pairs are possible. By the mn rule there are 6 × 7 = 42 pairs.

2.37. (a) There are 6! = 720 different orderings of the six cities.

(b) Intuitively, the answer should be 1/2 because there should be an equal chance of visiting Denver first or of visiting San Francisco first if an itinerary is chosen randomly. To show this rigorously, note that the orderings in which Denver (D) is visited before San Francisco (S) are of \binom{6}{2} types: DSxyzw, DxSyzw, ⋯, xyzwDS (corresponding to the different possible placements of D, S among the 6 cities. For each of these types, the other four cities can be visited in any of the 4! possible orders. So there are \binom{6}{2} · 4! such itineraries. Then P (D before S) = \frac{\binom{6}{2}·4!}{6!} = \frac{1}{2}.

2.43. The set of 9 taxis is partitioned into three disjoint groups of sizes 3, 5, 1 depending on which airport they go to. This is exactly the situation of Theorem 2.3, and the number of ways the groups can be selected is \binom{9}{3} · \binom{6}{5}, which equals the multinomial coefficient

\binom{9}{3 5 1} = \frac{9!}{3!5!1!} = 504.

2.44. (a) If the one taxi that needs repairs is dispatched to airport C, there are still \binom{8}{3} = \frac{8!}{3!5!} = 56 ways the other taxis can be dispatched to the other two airports. So the probability of this case is

\frac{56}{504} = \frac{1}{9}.

(This can also be seen intuitively by thinking of picking the one taxi to go to airport C and asking what is the probability that that one is the bad one.)

(b) Similar to part (a). Take out the three bad ones, so there are 6 taxis left; 2 going to airport A and 4 to airport B. The probability is

\frac{\binom{6}{2}}{504} = \frac{5}{168}.

2.54. There are \binom{8}{4} ways to pick four people out of the whole group. There are \binom{3}{2} · \binom{5}{2} ways to pick groups that have exactly 2 undergraduates. So the probability of this case is

\frac{\binom{3}{2} · \binom{5}{2}}{\binom{8}{4}} = \frac{3}{7}.

2.64. There are 6! = 720 possible orders of the 6 distinct numbers 1, 2, 3, 4, 5, 6, and 6^{6} = 46656 possible outcomes for the six tosses of the dice, all equally likely. Hence the probability is

6! · (\frac{1}{6})^{6} = \frac{720}{46656} = 5/324.

2.65. Now there are 5! = 120 different orders of the 5 distinct numbers 1, 2, 3, 4, 5. The difference between this situation and the one in 2.64 is that it is now twice as likely that a 5 will be rolled. So the probability is

5! · (\frac{1}{6})^{4} \frac{2}{6} = \frac{5}{162}.