Solutions to Homework 7 SOSC 30100, Sept. 190, 2013

1. (a) the probability that “3” does not come up in four tosses of a fair die is (1 − \frac{1}{6})^{4} = (\frac{5}{6})^{4} (since the tosses are independent. Hence the probability that a “3” does come up is 1 − (\frac{5}{6})^{4} = \frac{671}{1296} = .51774 …

(b) the probability that a ‘snake’s eye’ does not come up in twenty-four tosses of a pair of fair dice is (1 − \frac{1}{36})^{2}4 = (\frac{35}{36})^{2}4 (since the tosses are independent. Hence the probability that a snake’s eyes does come up is 1 − (\frac{35}{36})^{2}4 = .4914 . . . . . . .

I would rather bet on (a).

2. (a) If there is no limit to the number of committees that a person can chair, then for each committee that are 50 (independent) choices of chair. Hence the total number is 50^{10}.

If a person can chair at most one committee, then the number is

50 · 49 · 48 · · · ...41 = \frac{50!}{40!} .

(There are 50 possibilities for the chair of the first committee, 49 for the chair of the second committee, etc.).

(b) If you don’t care who the chair is, then the number of ways is the multinomial coefficient

\frac{50!}{5!^{6}10!^{2}} .

If in addition, you have to pick a chair for each committee, then the number of possibilities is

\frac{50!}{5!^{6}10!^{2}} · (5^{6} · 10^{2}) = \frac{50!}{4!^{6}9!^{2}} .

3.

(a) Using Bayes’ law, the probability that starting a negative campaign is indeed a bad idea, if the focus group decided that it is

\frac{(0.4)(0.8)}{(0.6)(0.3) + (0.4)(0.8)} = \frac{0.32}{0.5} = 0.64 .

Hence the probability that it is a good idea is 1 − 0.64 = 0.36.

(b) Now the modified prior that starting a negative campaign is a good idea is 0.36 and that it is a bad idea is 0.64. If the second focus group decided that it is a good idea after all, then the probability that it is indeed a good idea is

\frac{(0.36)(0.7)}{(0.36)(0.7) + (0.64)(0.2)} = \frac{0.32}{0.5} = 0.6631 ….

(c) If things are reversed, then the probability that starting a negative campaign, after the first focus group is

\frac{(0.6)(0.7)}{(0.6)(0.7) + (0.4)(0.2)} = \frac{0.42}{0.5} = 0.84 …,

and after the second focus group it is

\frac{(0.84)(0.3)}{(0.84)(0.3) + (0.16)(0.8)} = 0.6631 ….

You get the same probability.

4. (a)

∫_{b}^{∞} \frac{1}{b}exp(−\frac{x}{b})] dx = 1 − F (b) = 1 − (1 − exp(−b/b)) = e^{−1} = \frac{1}{e} .

(b)

∫_{0}^{2b} \frac{1}{b}exp(−\frac{x}{b}) dx = F (2b) = (1 − exp(−(2b)/b)) = 1 − e^{−2} = 1 − \frac{1}{e^{2}} .

5. (a)

P r(X = 1) = \frac{3}{28} + \frac{3}{14} + \frac{1}{28} = \frac{10}{28} = \frac{5}{14} ,

P r(X = 2) = \frac{9}{28} + \frac{3}{14} = \frac{15}{28} ,

P r(X = 3) = \frac{3}{28} .

P r(Y = 0) = \frac{3}{28} + \frac{9}{28} + \frac{3}{28} = \frac{15}{28} ,

P r(Y = 1) = \frac{3}{14} + \frac{3}{14} = \frac{3}{7} ,

P r(Y = 2) = \frac{1}{28} .

(b) The expectation of a random variable is ∑_{s∈S} X(s)P r(X = s), so:

E(X) = 1·P r(X = 1)+2·P r(X = 2)+3·P r(X = 3) = \frac{10}{28} +2· \frac{15}{28} +3· \frac{3}{28} = \frac{10 + 30 + 9}{28} = \frac{49}{28} = \frac{7}{4} .

Similarly,

E(Y) = 0 · P r(Y = 0) + 1 · P r(Y = 1) + 2 · P r(Y = 2) = 0 · \frac{15}{28} + 1 · \frac{3}{7} + 2 · \frac{1}{28} = \frac{1}{2} .

(c) The variance of a random variable is ∑_{s∈S}(X(s) − μ)^{2}P r(X = s), where μ is the expectation. So, since for E(X) = \frac{7}{4},

V ar(X) = (1−\frac{7}{4})^{2}·P r(X = 1)+(2−\frac{7}{4})^{2}·P r(X = 2)+(3−\frac{7}{4})^{2}·P r(X = 3) = (1−\frac{7}{4})^{2}·\frac{10}{28}+(2−\frac{7}{4})^{2}·\frac{15}{28}+(3−\frac{7}{4})^{2}·\frac{3}{28}

Also, since E(Y) = \frac{1}{2},

V ar(Y) = (0−\frac{1}{2})^{2}·P r(Y = 0)+(1−\frac{1}{2})^{2}·P r(Y = 1)+(2−\frac{1}{2})^{2}·P r(Y = 2) = \frac{1}{4}·\frac{15}{28}+\frac{1}{4}·\frac{3}{7}+\frac{9}{4}cdot\frac{1}{28} .

The Covarianace of two random variables X and Y is the expectation of (X − E(X))(Y − E(Y)) So

Cov(X, Y) = \frac{3}{28}(1 − \frac{7}{4})(0 − \frac{1}{2}) + ⋯

(d) They are not independent, since for example, the probability that X = 2 and Y = 1, \frac{3}{14} is not the product of the P r(X = 2) = \frac{5}{28} and P r(Y = 1) = \frac{3}{17}. 6.

(a) The expectation of a continuous random variable whose probability density function is f(x) is ∫_{−∞}^{∞} xf(x) dx.

So, in this case

E(X) = ∫_{x=0}^{∞} xe^{−x} = 1 .

(By integration by parts, or using Maple). (b)

E(X^{2}) = ∫_{0}^{∞} x^{2}e^{−x} = 2 dx.

(using Maple). (c)

V ar(X) = E(X^{2}) − E(X)^{2} = 2 − 1^{2} = 2 − 1 = 1 .

(d)

E(e^{2X/3}) = ∫_{0}^{∞} e^{2x/3}e^{−x} = ∫_{0}^{∞} e^{−x/3} = \frac{e^{−x/3}}{−1/3} |_{0}^{∞}= −3e^{−x/3}|_{0}^{∞}= −3(0 − 1) = 3 .

7.

(a) E(X + Y) = E(X) + E(Y) = −2 + 1 = −1

(b) V ar(X +Y) = V ar(X)+V ar(Y) = 9+16 = 25, V ar(X −Y) = V ar(X)+V ar(Y) = 9+16 = 25 (c) and (d) too advanced!