Mathematics 375 – Probability Theory Solutions for Problem Set 2 September 16, 2011

2.25. This is clearly nonsense. While it is true that every car is either a Volkswagen or not, the numbers of Volkswagens and other cases are certainly not the same. The probability should be

\frac{#Volkswagens}{#all cars}.

2.26. (a) Number the cans 1 - 4 in such a way that cans 1 and 2 contain water and 3 and 4 do not. (Note: There are other, more complicated, ways to solve this too, in which each possible assignment of the cans actually containing water is also taken into account, but this approach is better since the numbering can be done in one particular way to start.) Each sample point will consist of a 4-tuple where the entries indicate whether the “expert” found water in the corresponding can using the divining rod. Say W means the “expert” found water in that can, and NW means the “expert” found no water in that can. The sample space consists of all 4-tuples containing 2 W ’s and 2 NW ’s:

S = {(W, W, NW, NW), (W, NW, W, NW), (W, NW, NW, W), (NW, W, W, NW), (NW, W, NW, W),

(NW, NW, W, W)}.

(b) If the “expert” is just guessing, that is equivalent to selecting one of the sample points at random with each possibility equally likely. The probability that the correct one (W, W, NW, NW) is chosen is \frac{1}{6}.

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.

2.33. (a) The sample points are ordered 4-tuples where the entries indicate whether the corresponding family’s income was above (H) or below or equal to the median (L). (b) We have

A = {(H, H, L, L), (H, L, H, L), (H, L, L, H), (L, H, H, L), (L, H, L, H), (L, L, H, H),
(H, H, H, L), (H, H, L, H), (H, L, H, H), (L, H, H, H), (H, H, H, H)}
B = {(H, H, L, L), (H, L, H, L), (H, L, L, H), (L, H, H, L), (L, H, L, H), (L, L, H, H)}
C = {(H, H, H, L), (H, H, L, H), (H, L, H, H), (L, H, H, H)}.

(c) Since each sample point is equally likely, we assign probability \frac{1}{16} to each one. Then P (A) = \frac{11}{16}, P (B) = \frac{6}{16} = \frac{3}{8} and P (C) = \frac{4}{16} = \frac{1}{4}.

2.34. (a) The sample points are triples representing the stations chosen by each of the three patients. There are 27 of these:

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

(b) The event A = {(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)}. (c) If the selection of the station is completely random, we would expect each sample point in S to be equally likely. Therefore P (A) = \frac{6}{27} = \frac{2}{9}.

2.36. The number is the same as the number of permutations of the set {1, 2, 3}: 3 · 2 · 1 = 3! = 6.

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, there are several different possible approaches.

“Brute Force:” 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 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}.

“Slick:” For each ordering in which Denver is visited before San Francisco, there is a corresponding ordering in which San Francisco is visited before Denver – just switch the order of those two cities, leaving the others fixed. For instance

DxSyzw ↦ SxDyzw.

Since we can always switch back too:

SxDyzw ↦ DxSyzw,

this mapping is invertible, hence one-to-one and onto. Therefore, the number of orderings in which D comes before S is the same as the number in which S comes before D. Since every ordering of the 6 cities falls into one of these classes or the other, this means that each of them is exactly half of the total.

2.38. By the m × n rule applied repeatedly, there are 4 × 3 × 4 × 5 = 240 possible dinners.

2.40. (a) This is just like 2.38 above: 5 × 4 × 2 = 40 different combinations. (Note that the color is not included in this part.) (b) Now by the m × n rule again taking colors into account too, 40 × 8 = 320 different combinations of all 4 features (style, engine, transmission, color).

2.45. The coefficient is the multinomial coefficient

\binom{17}{2 5 10} = \frac{17!}{2!5!10!} = 408408.

2.46. Note that by numbering the games, we are thinking of them as distinguishable. But the two teams playing in game 1 do not have any particular order – just a set of two teams. So there are \binom{10}{2} possible assignments of teams for game 1, then \binom{8}{2} assignments for game 2 from the remaining 8 teams, etc. The total number is

\binom{10}{2} · \binom{8}{2} · \binom{6}{2} · \binom{4}{2} · \binom{2}{2}.

This is the same as the multinomial coefficient

\binom{10}{2 2 2 2 2} = 113400.

2.50. The potential winning numbers were the ones containing only 4 and 6 digits:

444, 446, 464, 644, 466, 646, 664, 666.

(So if one of these numbers was chosen at random and played in the lottery the chance of winning would have been \frac{1}{8} rather than the probability \frac{1}{1000} without the “doctoring” of the Ping-Pong balls.)

2.55. (a) The sample is an unordered set of size 10 chosen from the 90 total nurses so the number of possible samples is \binom{90}{10}. (b) The possible samples described here consist of 4 out of the 20 men, and 6 out of the 70 women. The number of such samples is found by the m × n rule: \binom{20}{4}\binom{70}{6}. So the probability of getting a sample of this form is:

\begin{matrix}\binom{20}{4} & \binom{70}{6} \\ ( & {90 \atop 10} & ) \end{matrix} ≐ .111

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} = \frac{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}.

2.70. The multinomial coefficients are the coefficients of the various terms x_{1}^{n_{1}} ⋯ x_{k}^{n_{k}} in the expanded and collected form of

(x_{1} + ⋯ + x_{k})^{n}

(note n_{1} +⋯+n_{k} = n in each term). If we substitute x_{1} = ⋯ x_{k} = 1 in both sides of the equation, we get

k^{n} = sum of all multinomial coefficients with upper index n.

This establishes the fact to be proved.

Extra Credit: The formula

(y_{1} + ⋯ + y_{k})^{n} = ∑ \binom{n}{n_{1} n_{2} ⋯ n_{k−1} n_{k}} y_{1}^{n_{1}} ⋯ y_{k}^{n_{k}}

can be established by induction on k. The base case is k = 2, where this formula is the ordinary binomial theorem, since when n = n_{1} + n_{2},

\binom{n}{n_{1} n_{2}} = \frac{n!}{n_{1}!n_{2}!} = \frac{n!}{n_{1}!(n − n_{1})!} = \binom{n}{n_{1}}.

For the induction step we want to assume the formula holds for the expansion of the nth power of a sum of k terms, and consider the expansion of the nth power of a sum of k + 1 terms. We have by the induction hypothesis for powers of sums of k terms, then the base case for the expansion of the (y_{k} + y_{k+1})^{n_{k}}:

(y_{1} + ⋯ + y_{k} + y_{k+1})^{n} = ((y_{1} + ⋯ + y_{k−1} + (y_{k} + y_{k+1}))^{n}
= ∑ \binom{n}{n_{1} n_{2} ⋯ n_{k−1} n_{k}} y_{1}^{n_{1}} ⋯ y_{k−1}^{n_{k−1}}(y_{k} + y_{k+1})^{n_{k}}
= ∑ \binom{n}{n_{1} n_{2} ⋯ n_{k−1} n_{k}} y_{1}^{n_{1}} ⋯ y_{k−1}^{n_{k−1}} · ∑_{ℓ+m=n_{k}} \binom{n_{k}}{ℓ}y_{k}^{ℓ}y_{k+1}^{m}.

For any particular ℓ and m, since ℓ + m = n_{k}, the coefficient of

y_{1}^{n_{1}} ⋯ y_{k−1}^{n_{k−1}}y_{k}^{ℓ}y_{k+1}^{m}

when we expand this out will be:

\binom{n}{n_{1} n_{2} ⋯ n_{k−1} n_{k}} \binom{n_{k}}{ℓ} = \frac{n!}{n_{1}! ⋯ n_{k−1}!n_{k}!} · \frac{n_{k}!}{ℓ! m!}
= \frac{n!}{n_{1}! ⋯ n_{k−1}! ℓ! m!}
= \binom{n}{n_{1} n_{2} ⋯ n_{k−1} ℓ m} .

This shows what we want since (n_{1}, …, n_{k−1}, ℓ, m) now runs over all collections of k+1 nonnegative integers satisfying n_{1} + ⋯ + n_{k−1} + ℓ + m = n.

2.71. (a) P (A|B) = \frac{P (A∩B)}{P (B)} = \frac{1}{3}.

(b) P (B|A) = \frac{P (A∩B)}{P (A)} = \frac{1}{5}

(c) We have P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = .7. Therefore, P (A|A ∪ B) = \frac{P (A)}{P (A∪B)} = \frac{5}{7}.

(d) P (A|A ∩ B) = 1, since (A ∩ B) ∩ A = A ∩ B.

(e) P (A ∩ B|A ∪ B) = \frac{1}{7}.

2.72. (a) P (A ∩ M) = .24 while P (A) = .6 and P (M) = .4. Since P (A ∩ M) = P (A)P (M), the events A, M are independent (see Definition 2.10).

(b) P (\overline{A} ∩ F) = .24 while P (\overline{A}) = .4 and P (F) = .6. Since P (\overline{A} ∩ F) = P (\overline{A})P (F), these events are also independent.

2.74. (a) P (A ∩ D) = .2, while P (A) = .61 and P (D) = .3 since .2 ≠ (.61)(.3), A and D are not independent.

(b) P (B ∩ D) = .09 = P (B)P (D) = (.3)(.3). So B and D are independent.

(c) P (C ∩ D) = .01, but P (C)P (D) = (.09)(.3) = .027. So C and D are not independent.

2.77. (a) P (A) = .4 (b) P (B) = .37 (c) P (A ∩ B) = .1 (d) P (A ∪ B) = .67 (e) P (\overline{A}) = .6 (f) P (\overline{A ∪ B}) = .33 (g) P (\overline{A ∩ B}) = .9 (h) P (A|B) = \frac{P (A∩B)}{P (B)} ≐ .27 (i) P (B|A) = \frac{P (A∩B)}{P (A)} = .25.