Mathematics 375 – Probability Theory Solutions for Problem Set 5 October 21, 2011

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

3.68. The expected number is \frac{1}{.3} ≐ 3.3 by Theorem 3.8.

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”.

3.72. This is the situation of 3.71 with a = 10 and b = 1. The probability is the same as P (Y > 1) for a geometric random variable with p = q = 1/2. Hence P (Y > 1) = 1 − (1/2) = 1/2.

3.75. (a) Geometric with p = .1, q = .9. P (Y = 3) = (.9)^{2}(.1) = .081. (b) P (Y ≥ 3) = 1 − P (Y ≤ 1) = 1 − .1 − (.9)(.1) = .81.

3.87. We have

E(1/Y) = ∑_{y=1}^{∞} \frac{1}{y}q^{y−1}p

= \frac{p}{q} ∑_{y=1}^{∞} \frac{q^{y}}{y}

= −\frac{p}{q} ln(1 − q)

= \frac{p}{q} ln (\frac{1}{p}) .

3.102. This is a hypergeometric probability. Lumping the 5 non-green marbles together, the probability is

\begin{matrix}\binom{5}{3} & \binom{5}{0} \\ ( & {10 \atop 3} & ) \end{matrix} = \frac{5 · 4 · 3}{10 · 9 · 8} = \frac{1}{12}.

3.103. Hypergeometric:

\begin{matrix}\binom{6}{5} & \binom{4}{0} \\ ( & {10 \atop 5} & ) \end{matrix} = \frac{1}{42}.

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.

3.106. The cost is determined by the number Y of defectives found among the 5 chosen machines. This is hypergeometric with N = 10, n = 5, r = 4. So the mean of the total repair cost is

E(50Y) = 50 · \frac{5 · 4}{10} = 100,

and the variance is

V (50Y) = 50^{2} · 5\frac{4}{10} \frac{6}{10} \frac{5}{9} = 1666.67.

3.108. The number of defectives found is hypergeometric with N = 20, r = 3. We want to choose n so that

P (Y = 0) = \begin{matrix}\binom{3}{0} & \binom{17}{n} \\ ( & {20 \atop n} & ) \end{matrix} ≤ .2.

The smallest n for which this is true is n = 8 (found by trial and error).

3.121. We have P (Y = y) = \frac{2^{y}e^{−2}}{y!} . So (a) P (Y = 4) ≐ .09. (b) P (Y ≥ 4) = 1 − P (Y ≤ 3) ≐ .143. (c) P (Y < 4) ≐ .857. (d) P (Y ≥ 4 | Y ≥ 2) ≐ \frac{.143}{.594} = .241.

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.

3.126. (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 contiguous hours in part (a) is really irrelevant for the total number T of customers arriving if the numbers from the two hours are independent. The same method used in (b) works in (a) as well.

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 function for a Poisson random variable with parameter λ + μ.

4.1. (a) We have

F (y) = P (Y ≤ y) = \begin{cases}0 & y < 1 \\ .4 & 1 ≤ y < 2 \\ .7 & 2 ≤ y < 3 \\ .9 & 3 ≤ y < 4 \\ 1 & y ≥ 4. \end{cases}

(b) The graph is a step function with jump discontinuities at y = 1, 2, 3, 4.

4.2. (a) We have P (Y = 1) = .2, P (Y = 2) = (.8)(.25) = .2, P (Y = 3) ≐ (.8)(.75)(.3333) = .2, P (Y = 4) ≐ (.8)(.75)(.6667)(.5) = .200, P (Y = 5) = (.8)(.75)(.6667)(.5) = .200. (b) The distribution function is

F (y) = P (Y ≤ y) = \begin{cases}0 & y < 1 \\ .2 & 1 ≤ y < 2 \\ .4 & 2 ≤ y < 3 \\ .6 & 3 ≤ y < 4 \\ .8 & 4 ≤ y < 5 \\ 1 & y ≥ 5. \end{cases}

(c) P (Y < 3) = .4, P (Y ≤ 3) = .6, and P (Y = 3) = .2. (d) There is no contradiction. Y is not a continuous random variable so the equation P (Y = a) = 0 does not apply. Note that F (y) has jump discontinuities at y = 1, 2, 3, 4, 5.

4.6. (a) The probability function for the (discrete) geometric random variable is p(y) = q^{y−1}p. Hence if y = i ≥ 1,

F (i) = P (Y ≤ i) = P (Y = 1) + ⋯ + P (Y = i)

= p(1 + q + q^{2} + ⋯ + q^{i−1})

= p · \frac{1 − q^{i}}{1 − q}

= 1 − q^{i}

(where we used the formula for the sum of a finite geometric series to get the next to last line, then simplified using q = 1 − p). Now, if y > 0 is not an integer, we get F (y) = F (i) where i is the largest integer ≤ y. Hence

F (y) = \begin{cases}0 & if y < 0 \\ 1 − q^{i} & if i ≤ y < i + 1 for i = 0, 1, 2, …. \end{cases}

b) lim_{y→−∞} F (y) = 0 by the definition. Then,

lim_{y→+∞} F (y) = lim_{i→∞} 1 − q^{i} = 1

since q < 1. Finally, we want to show F (y) is non-decreasing. If y_{1} ≤ y_{2} and y_{1} or y_{2} is negative then F (y_{1}) = 0 and the inequality 0 = F (y_{1}) ≤ F (y_{2}) is satisfied. So we can assume 0 < y_{1} ≤ y_{2}. If this is true, then it must be true that i_{1} = [y_{1}] ≤ i_{2} = [y_{2}] (I’m using the shorthand notation [x] = greatest integer ≤ x here). Hence since 0 < q < 1,

0 < y_{1} ≤ y_{2} ⇒ q^{i_{2}} ≤ q^{i_{1}} ⇒ −q^{i_{1}} ≤ −q^{i_{2}} ⇒ 1 − q^{i_{1}} ≤ 1 − q^{i_{2}} ⇒ F (y_{1}) ≤ F (y_{2}).

4.8. (a) We must have ∫_{−∞}^{+∞} f(y) dy = ∫_{0}^{1} ky(1−y) dy = 1. This gives the equation k (\frac{y^{2}}{2} − \frac{y^{3}}{3} )|_{0}^{1} =

1, so \frac{k}{6} = 1, and k = 6.

(b)

P (.4 ≤ Y ≤ 1) = ∫_{.4}^{1} 6y(1 − y) dy
= 3y^{2} − 2y^{3}|_{.4}^{1}
= .648.

(c) This is the same as (b) because Y is a continuous random variable, so including the endpoints or not makes no difference.

(d) By the definition of conditional probability,

P (Y ≤ .4|Y ≤ .8) = \frac{P (Y ≤ .4)}{P (Y ≤ .8)} = \frac{∫_{0}^{.4} 6y(1 − y) dy}{∫_{0}^{.8} 6y(1 − y) dy} ≐ .393.

(e) This is the same as (d) for the same reason as in (c). 4.10. (a) and (b) We have ∫_{0}^{y_{0}} 6y(1 − y) dy = −2b^{3} + 3b^{2}. This equals .95 when b ≐ .8646. (There are also two other real roots, but they lie outside the interval [0, 1]. (c) They are the same. This is a continuous random variable. 4.15. (a) The function values f(y) ≥ 0 for all y by definition. We also have

∫_{−∞}^{∞} f(y) dy = ∫_{b}^{∞} \frac{b}{y^{2}} dy

= \frac{−b}{y}|_{b}^{∞}

= 1.

Hence f has the required properties. (b) We have

F (y) = \begin{cases}0 & y ≤ b \\ 1 − \frac{b}{y} & b ≤ y < ∞. \end{cases}

(c) If c > 0, then

P (Y > b + c) = 1 − P (Y ≤ b + c) = 1 − (1 − \frac{b}{b + c}) = \frac{b}{b + c}.

(d) Using part (c) we have for d > c,

P (Y > b + d | Y > b + c) = \frac{\frac{b}{b+d}}{\frac{b}{b+c}} = \frac{b + c}{b + d}.