Mathematics 375 – Probability and Statistics I Problem Set 2 – Solutions September 19, 2003

2.19/ a) Let A = {R, L, S} be the set of possible behaviors of a single car. The sample space for two cars is the Cartesian product

S = A × A = {(R, R), (R, L), (R, S), (L, R), (L, L), (L, S), (S, R), (S, L), (S, S)}

By the mn rule, |S| = 3 · 3 = 9.

b) Since the behaviors are equally likely for each car, each sample point should be assigned P (E_{i}) = \frac{1}{9}, and hence the probability that at least one car turns left is

P (\{(R, L), (L, R), (S, L), (L, S), (L, L)\}) = \frac{5}{9}.

c) Similiarly the probability that at most one car turns is

P (\{(S, L), (S, R), (R, S), (L, S), (S, S)\}) = \frac{5}{9}.

2.24/ a) Call the two types of consoles 1, 2. The sample space is {1, 2}^{4}, so there are 16 points in the sample space:

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

b) Since we are given that the two types of consoles are in equal demand, each sample point should have same probability, P (E_{i}) = \frac{1}{16}.

c) A = \{(1, 1, 1, 1), (2, 2, 2, 2)\} so P (A) = \frac{2}{16} = \frac{1}{8}.

2.25/ a) For each family, let outcome 1 represent income ≤ $35353, and let outcome 2 represent income > $35353. Then the sample space is represented by the same set as given in 2.24 above. b)

\begin{matrix}A = & \{(1, 1, 2, 2), (1, 2, 1, 2), (1, 2, 2, 1), (2, 1, 1, 2), (2, 1, 2, 1), (2, 2, 1, 1), \\ (1, 2, 2, 2), (2, 1, 2, 2), (2, 2, 1, 2), (2, 2, 2, 1), (2, 2, 2, 2)\} \end{matrix}

B = {(1, 1, 2, 2), (1, 2, 1, 2), (1, 2, 2, 1), (2, 1, 1, 2), (2, 1, 2, 1), (2, 2, 1, 1)}

C = {(1, 2, 2, 2), (2, 1, 2, 2), (2, 2, 1, 2), (2, 2, 2, 1)}

c) By the definition of the median, half the families in the population would have 1 and the other half 2. Hence P (E_{i}) = \frac{1}{16} for each of the 16 simple events. Hence P (A) = \frac{11}{16}, P (B) = \frac{3}{8}, and P (C) = \frac{1}{4}.

2.27/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.

2.28/This is the number of permutations of three objects P_{3}^{3} = 3! = 6.

2.30/Using the mn rule repeatedly, we have 4 · 3 · 4 · 5 = 240 possible meals (each appetizer can be ordered with every combination of salad, entree, and dessert).

2.33/Using the mn rule, there are 9 × 10^{6} = 9000000 such numbers.

2.35/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 the multinomial coefficient

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

2.36/ a) We want to count the sample points in the event “taxi needing repair goes to C”.

There is no choice for the taxi needing repair – it goes to C. But there are \binom{8}{3 5} = 56

ways the other 8 taxis can be divided into two groups of sizes 3 and 5. Hence the probability of this event is \frac{56}{504} = \frac{1}{9}.

b) The three taxis needing repair each go to a separate airport, so there are 6 = P_{3}^{3} ways in which those three taxis can be assigned to the three airports A, B, C. There are

\binom{6}{2 4} = 15 ways to assign the other taxis to airports (note the one taxi for airport C

is already accounted for). The probability of this event is \frac{6·15}{504} = \frac{5}{28}.

2.41 For all four parts, the sample space is the set of all possible ways to assign the three prizes among the 50 lottery tickets. Hence the sample space has size \binom{50}{3}. For the events given, we want to think about choosing the three prizewinners in the specified ways.

a) There are \binom{4}{3}\binom{46}{0} ways to select the prizewinners if the officers win them all. The probability of this event is \binom{4}{3}\binom{46}{0}/\binom{50}{3}.

b) If exactly two officers win prizes there are \binom{4}{2} possibilities for the officers who win, and \binom{46}{1} possibilities for the other prizewinner. Hence the probability of this event is \binom{4}{2}\binom{46}{1}/\binom{50}{3}.

c) Similar to b): \binom{4}{1}\binom{46}{2}/\binom{50}{3}

d) Similar to b) and c): \binom{4}{0}\binom{46}{3}/\binom{50}{3}. Comment: It should be clear that the four events in the four parts of the problem account for the whole sample space. Do our formulas agree with that? That is, do the answers from parts a), b), c), d) sum to 1? Or, in other words, is

(1)  \binom{4}{3}\binom{46}{0} + \binom{4}{2}\binom{46}{1} + \binom{4}{1}\binom{46}{2} + \binom{4}{0}\binom{46}{3} = \binom{50}{3}?

The answer is yes(!), and we can see it like this. By the polynomial interpretation of the binomial coefficients, \binom{50}{3} is the coefficient of x^{3} in the expansion of (1 + x)^{50}. But

(1 + x)^{50} = (1 + x)^{4}(1 + x)^{46}

= (\binom{4}{0} + \binom{4}{1}x + \binom{4}{2}x^{2} + \binom{4}{3}x^{3} + ⋯)·

(\binom{46}{0} + \binom{46}{1}x + \binom{46}{2}x^{2} + \binom{46}{3}x^{3} + ⋯)

Collecting all x^{3} terms in the product gives exactly the left side of (1) above. This is a special case of a general binomial coefficient identity sometimes called the Vandermonde convolution formula.

2.54/ a) \binom{n}{n} = \frac{n!}{n!0!} = 1. This says there is only one way to choose a subset of size n from a set with n elements – you have to take all of the elements!

b) \binom{n}{0} = \frac{n!}{0!n!} = 1. This says there is only one way to choose a subset of size 0 from a set with n elements – you have to take the empty subset!

c) \binom{n}{r} = \frac{n!}{r!(n−r)!} = \binom{n}{n−r}, since n − (n − r) = r. This says that the number of ways to choose a subset of size r from a set with n elements is the same as the number of ways to choose a subset of size n − r. This makes sense because for each subset A of size r, we have the complement \overline{A} which contains exactly n − r elements. There are as many ways to choose A as there are to choose \overline{A}.

d) We have (1 + x)^{n} = ∑_{k=0}^{n} \binom{n}{k}x^{k} by a result proved in class. This is an equality of polynomials, so we get another true equality by substituting any value for x. With x = 1, we get 2^{n} = ∑_{k=0}^{n} \binom{n}{k} which is what we had to show.

2.56/By Theorem 2.3, the multinomial expansion is:

(x_{1} + ⋯ + x_{k})^{n} = ∑_{n_{1}+···+n_{k}=n} \binom{n}{n_{1} ⋯ n_{k}} x_{1}^{n_{1}} ⋯ x_{k}^{n_{k}}.

Similarly to 2.54d, set x_{1} = ⋯ = x_{k} = 1, and we get

k^{n} = ∑_{n_{1}+···+n_{k}=n} \binom{n}{n_{1} ⋯ n_{k}}

as claimed.

2.57/ For all parts we use the formula P (X|Y) = \frac{P (X∩Y)}{P (Y)} from the definition of the conditional probability.

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

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

c) For P (A|A∪B), we need to know P (A∪B). By the sum rule, since we know P (A) = .5, P (B) = .3, and P (A ∩ B) = .1, we have P (A ∪ B) = .5 + .3 − .1 = .7. Moreover, by usual “set algebra”, A ∩ (A ∪ B) = A. Hence P (A|A ∪ B) = \frac{.5}{.7} = \frac{5}{7}. d) P (A|A ∩ B) = \frac{P (A∩(A∩B))}{P (A∩B)} = \frac{P (A∩B)}{A∩B} = 1. Comment: The same reasoning shows P (X|Y) = 1 whenever Y ⊂ X, which is certainly what we would hope!

e) P (A ∩ B|A ∪ B) = \frac{P (A∩B)}{P (A∪B)} = \frac{1}{7}. (Note: Here we used the set algebra fact that (A ∩ B) ∩ (A ∪ B) = A ∩ B.

2.58/ A and M are independent since P (A ∩ M) = \frac{24}{100} = \frac{6}{25} = P (A)P (M). Similiarly \overline{A} and F are independent since P (\overline{A} ∩ F) = \frac{24}{100} = P (\overline{A})P (F). Comment: Intuitively, these independences are evident from the table, since the fraction of the total males who passed equals the fraction of the total females who passed (24/40 = 3/5 = 36/60) and similarly for the fractions who failed.