Mathematics 375 – Probability and Statistics 1 Solutions for Problem Set 5 October 22, 2009

3.39. (a) Y = the number of components that last longer than 1000 hours has a binomial distribution. P (Y = 2) = \binom{4}{2}(.8)^{2}(.2)^{2} ≐ .1536.

(b) The subsystem will operate if 2, 3, or 4 of the components last longer than 1000 hours. So this probability is P (Y ≥ 2) = \binom{4}{2}(.8)^{2}(.2)^{2} + \binom{4}{3}(.8)^{3}(.2) + \binom{4}{4}(.8)^{4} ≐ .973.

3.40. These are all binomial probabilities. Let Y be the number who recover out of the 20. (a) P (Y = 14) = \binom{14}{20}(.8)^{14}(.2)^{6} ≐ .11.

(b) P (Y ≥ 10) = ∑_{y=10}^{20} \binom{20}{y}(.8)^{y}(.2)^{20−y} ≐ .9994.

(c) P (14 ≤ Y ≤ 18) = ∑_{y=14}^{18} \binom{20}{y}(.8)^{y}(.2)^{20−y} ≐ .844.

(d) P (0 ≤ Y ≤ 16) = ∑_{y=0}^{16} \binom{20}{y}(.8)^{y}(.2)^{20−y} ≐ .589.

3.43. Let Y be the number who qualify for the favorable rates (out of the 5), a binomial random variable. (a) P (Y = 5) = (.7)^{5} ≐ .168.

(b) P (Y ≥ 4) = ∑_{y=4}^{5} \binom{5}{y}(.7)^{y}(.3)^{5−y} ≐ .528.

3.44. Let Y be the number of successful operations (binomial). (a) P (Y = 5) = (.8)^{5} ≐ .328. (b) P (Y = 4) = \binom{5}{4}(.6)^{4}(.4) ≐ .259.

(c) P (Y < 2) = ∑_{y=0}^{1} \binom{5}{y}(.3)^{y}(.7)^{5−y} ≐ .528. (Note that this is the same as the answer for 3.43

(b). Do you see why? For a formal statement of the pattern here, see Exercise 3.54.)

3.53. Y = the number of children (out of 3) that develop Tay-Sachs has a binomial distribution.

(a) P (Y = 3) = (.25)^{3} ≐ .0156.

(b) P (Y = 1) = \binom{3}{1}(.25)(.75)^{2} ≐ .422.

(c) This conditional probability is \frac{(.75)^{2}(.25)}{(.75)^{2}} = .25. (This follows because of the independence.)

3.59. Let Y be the number of good motors, so 10 − Y is the number of defectives. The problem is asking for the expected value of G = 1000 − 200(10 − Y) = 200Y − 1000. Since Y is binomial with n = 10 and p = .92, we have E(G) = E(200Y −10000) = 200E(Y)−1000 = (200)(9.2)−1000 = 840 (dollars).

3.62. (a) To justify multiplying p_{1}p_{2}p_{3} to get the probability of detecting a wing crack, you need to assume the individual events are independent.

(b) p = (.9)(.8)(.5) = .36. Then the probability desired is P (Y ≥ 1) = ∑_{y=1}^{3} \binom{3}{y}(.36)^{y}(.64)^{3−y} ≐

.738.

3.67. This is geometric with p = .3. So P (Y = 5) = (.7)^{4}(.3) ≐ .072.

3.69. Y is a geometric random variable with p = .59, so P (Y = y) = (.41)^{y−1}(.59).

3.71. (a) For a geometric random variable Y,

P (Y > a) = ∑_{y=a+1}^{∞} q^{y−1}p = \frac{pq^{a}}{1 − q} = q^{a}

(where we used the geometric series sum formula – this is a geometric series with first term q^{a}p and ratio q).

(b) Using the definition of conditional probability and part a),

P (Y > a + b|Y > a) = \frac{P (Y > a + b)}{P (Y > a)} = \frac{q^{a+b}}{q^{a}} = q^{b}.

(Note Y > a + b implies Y > a, so the event (Y > a + b) ∩ (Y > a) is the same as Y > a + b.)

(c) This is the same as P (Y > b). In other words, for geometric random variables, the probability that Y > a + b given that Y > a (i.e. the probability that the first success occurs after trial a + b, given that the first success occurs after trial a) is the same as the probability that the first success occurs after trial b. Geometric random variables thus are “memoryless” in the sense that starting from the a-th trial (with no previous successes), the probability that the first success occurs more than b trials later is the same as if we started from the beginning with the 0th trial. The previous non-successes don’t affect the probability of later successes – no “memory”.

3.105. (a) Y is hypergeometric because the teachers are selected without replacement from the finite pool of 8 candidates.

(b) The number of internship candidates chosen is hypergeometric with N = 8, r = 5 (the internship candidates) n = 3 (the chosen ones). So the desired probability is

P (Y = 2) + P (Y = 3) = \begin{matrix}\binom{5}{2} & \binom{3}{1} \\ ( & {8 \atop 3} & ) \end{matrix} + \begin{matrix}\binom{5}{3} & \binom{3}{0} \\ ( & {8 \atop 3} & ) \end{matrix} ≐ .536 + .179 ≐ .714.

(c) Using the formulas from Theorem 3.10, μ = \frac{nr}{N} = \frac{15}{8} = 1.875, and

σ^{2} = n · \frac{r}{N} · \frac{N − r}{N} · \frac{N − n}{N − 1} = 3 · \frac{5}{8} · \frac{3}{8} · \frac{5}{7} = \frac{225}{448}.

So the standard deviation is σ = \sqrt{\frac{225}{448}} ≐ .709.

3.122. (a) P (Y ≤ 3) = ∑_{y=0}^{3} \frac{7^{y}e^{−7}}{y!} ≐ .0818.

(b) P (Y ≥ 2) = 1 − P (Y = 0) − P (Y = 1) ≐ .993.

(c) P (Y = 5) = \frac{7^{5}e^{−7}}{5!} ≐ .128.

3.123. For a Poisson random variable, P (Y = y) = \frac{λ^{y}e^{−λ}}{y!} . So if P (Y = 0) = P (Y = 1), then we have the equation

e^{−λ} = λe^{−λ}

This implies λ = 1. Hence P (Y = 2) = \frac{e^{−1}}{2} ≐ .184.

3.125. If Y is the number of customers who arrive, the total service time is T = 10Y. Then E(T) = E(10Y) = 10E(Y) = 10 · 7 = 70 and V (T) = V (10Y) = 100V (Y) = 100 · 7 = 700. To answer the last part of the question, note that σ = \sqrt{700} ≐ 26.46. 2.5 hours is 150 minutes, which is more than 2 standard deviations above μ. So it is not likely that the total service time will exceed 2.5 hours.

3.126. (a) The arrivals over the two-hour period are a Poisson process with λ = 14. So the

probability that there are exactly 2 arrivals is \frac{14^{2}e^{−14}}{2!} .

(b) Method 1 – “direct approach”: Let I be the number of customers arriving between 1 and 2 and II the number between 3 and 4. Let T = I + II be the total number. We can get T = 2 in three different ways (I = 2 and II = 0, or I = 1 and II = 1, or I = 0 and II = 2.) Then using independence:

P (T = 2) = ∑_{k=0}^{2} P (I = k)P (II = 2 − k)

= ∑_{k=0}^{2} \frac{7^{k}e^{−7}}{k!} · \frac{7^{2−k}e^{−7}}{(2 − k)!}

= e^{−14} (\frac{7^{2}}{2} + 7^{2} + \frac{7^{2}}{2} )

= e^{−14} \frac{4 · 7^{2}}{2!}

= \frac{14^{2}e^{−14}}{2!}

(same as in part (a)!)

Method 2 – “clever”: The fact that there are two noncontiguous hours is really irrelevant for the total number T of customers arriving if the numbers from the two hours are independent. It should be the same as a situation where you have a Poisson distribution with λ = 7 + 7 = 14 customers arriving per hour. Then P (T = 2) = \frac{14^{2}e^{−14}}{2!} . (Same as above!)

Comment: In fact, this problem illustrates a general fact about Poisson random variables. If you have X Poisson with parameter λ and Y Poisson with parameter μ and X, Y are independent, then Z = X + Y also has a Poisson distribution with parameter λ + μ. One proof comes from an argument like the “direct approach” above:

P (Z = z) = ∑_{x=0}^{z} P (X = x)P (Y = z − x)

= ∑_{x=0}^{z} P (X = x)P (Y = z − x)

= ∑_{x=0}^{z} \frac{λ^{x}e^{−λ}}{x!} · \frac{μ^{z−x}e^{−μ}}{(z − x)!}

= e^{−(λ+μ)} ∑_{x=0}^{z} \frac{1}{x!(z − x)!}λ^{x}μ^{z−x}

= \frac{e^{−(λ+μ)}}{z!} ∑_{x=0}^{z} \binom{z}{x}λ^{x}μ^{z−x}

= \frac{e^{−(λ+μ)}}{z!} (λ + μ)^{z}

(using the binomial theorem). This is exactly the right probability mass function for a Poisson random variable with parameter λ + μ.

There’s another proof via the moment generating functions too(!) If X, Y are independent, and Z = X + Y, then

m_{Z}(t) = E(e^{tX+tY}) = E(e^{tX})E(e^{tY}) = e^{λ(e^{t}−1)}e^{μ(e^{t}−1)} = e^{(λ+μ)(e^{t}−1)}

By the uniqueness theorem, Z has a Poisson distribution with parameter λ + μ.

3.139. By the linearity of the expected value, we have E(X) = E(50 − 2Y − Y^{2}) = 50 − 2E(Y) − E(Y^{2}). By Theorem 3.11, E(Y) = λ = 2. Also, E(Y^{2}) = V (Y) + E(Y)^{2} = 2 + 4 = 6. So E(X) = 50 − 4 − 6 = 40.