Solutions

1.24

Let $E_i$ be the event that the first head appears on toss $i$ and let $A$ be the event that player A wins. Then

$$A = E_1 \cup E_3 \cup E_5 \cup \dots$$

The $E_i$ are pairwise disjoint, and $P(E_i) = (1-p)^{i-1}p$ , so

$$P(A) = p + (1-p)^2 p + (1-p)^4 p + \dots$$

$$= p \sum_{k = 1}^\infty (1-p)^{k-1}$$

$$= \frac{p}{1-(1-p)^2}$$

$$= \frac{p}{1-(1-2p+p^2)} = \frac{p}{2p-p^2} = \frac{1}{2-p}$$

So if $p = \frac{1}{2}$ then $P(A) = \frac{2}{3}$ , and for all $p$

$$P(A) = \frac{1}{2-p} \ge \frac{1}{2}.$$

An alternative approach to deriving the formula for $P(A)$ is to condition on the results of the first two tosses. For the first toss,

$$P(A) = p P(A\vert E_1) + (1-p) P(A\vert E_1^c)$$

$$= p + (1-p) P(A\vert E_1^c)$$

For the second toss

$$P(A\vert E_1^c) = p P(A\vert E_2 \cap E_1^c) + (1-p) P(A\vert E_2^c \cap E_1^c)$$

$$= (1-p) P(A\vert E_2^c \cap E_1^c)$$

Since the tosses are independent, the game starts over if the first two tosses are tails, i.e.

$$P(A\vert E_2^c \cap E_1^c) = P(A)$$

So we have an equation in $P(A)$ :

$$P(A) = p + (1-p)^2 P(A)$$

and the solution is

$$P(A) = \frac{1}{2-p} \ge \frac{1}{2}.$$

1.26

The die must be cast more than five times if and only if none of the first five tosses results in a 6:

$$P( ) = \left(\frac{5}{6}\right)^5 = 0.4019$$

1.34

$$I: B,B,G P(B\vert I) = 2/3$$

$$II: B,B,B,G,G P(B\vert II) = 3/5$$

$P(I) = P(II) = 1/2$ (choose litter at random).

a.

$P(B)= P(B\vert I)P(I) + P(B\vert II)P(II) = \frac{2}{3} \times \frac{1}{2} + \frac{3}{5} \times \frac{1}{2} = \frac{1}{3} + \frac{3}{10} = \frac{19}{30}$

b.

$P(I\vert B) = \frac{P(I \cap B)}{P(B)} = \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{19}{30}} = \frac{\frac{1}{3}}{\frac{19}{30}} = \frac{10}{19}$

1.36

The probabilities of no hits and exactly one hits are:

$$P( ) = \left(\frac{4}{5}\right)^{10}$$

$$P( ) = 10 \times \frac{1}{5}\times \left(\frac{4}{5}\right)^{9} 0.6242$$

Therefore

$$P( ) = 1-P( )-P( )$$

$$= 1-\left(\frac{4}{5}\right)^{10}-10 \times \frac{1}{5} \times \left(\frac{4}{5}\right)^{9}$$

$$P( \vert ) = \frac{P(\text{at least twice})}{P(\text{at least once})} \approx 0.6242$$

$$= \frac{ 1-\left(\frac{4}{5}\right)^{10}-10 \times \frac{1}{5} \times \left(\frac{4}{5}\right)^{9}}{1-\left(\frac{4}{5}\right)^{10}} \approx 0.6993$$

1.45

(i)

$P_{X}(A) = P(X \in A) \ge 0$ since $P$ is a probability.

(ii)

$P_{X}(\mathbb{R})=P(X \in \mathbb{R}) = P(S) = 1$ since $P$ is a probability.

(iii)

For Borel sets $A_{1}, A_{2}, \ldots$ that are pairwise disjoint,

$$P_{X}\left(\bigcup A_{i}\right) = P\left(X \in \bigcup A_{i}\right)$$

$$= P\left(\bigcup \{X \in A_{i}\}\right)$$

$$= \sum P(X \in A_{i})$$

$$= \sum P_{X}(A_{i})$$

since $B_{i}=\{X \in A_{i}\}$ are pairwise disjoint.

1.46

$\mathcal{X} = \{0, 1, 2\}$ . For $X_3 = 2$ to occur there must be three balls in one cell, three in a second, and one in a third:

$$P(X_{3}=2) = \frac{\binom{7}{2}\binom{7}{3}\binom{4}{3}5}{7^{7}} = \frac{14,700}{7^7} \approx .0178.$$

For $X_{3}=1$ there are four patterns:

Pattern	Ways
34	$7\binom{7}{3} 6 =$	1,470
322	$7\binom{7}{3} \binom{6}{2}\binom{4}{2} =$	22,050
3211	$7\binom{7}{3} 6 \binom{4}{2} \binom{5}{2} 2 =$	176.400
31111	$7\binom{7}{3} \binom{6}{4} 4! =$	88,200
		288,120

So

$$P(X_3 = 1) = \frac{288,120}{7^7} \approx .3498.$$

The remaining case can be handled by complementarity:

$$P(X_{3}=0)=1 - P(X_{3}=2) - P(X_{3}=1) \approx 1 - .0178 - .3498 = 0.6324$$

1.47

All functions in (a)-(d) are continuous and therefore right continuous. We therefore only need to check that they are nondecreasing and have the right limits at $\pm\infty$ .

(a)

For all $x \in \mathbb{R}$

$$\frac{d}{dx}\left(\frac{1}{2} + \frac{1}{\pi}\tan^{-1}(x)\right) = \frac{1}{\pi}\frac{1}{1+x^2} > 0$$

so $F(x)$ is increasing, and

$$\lim_{x \to -\infty} \frac{1}{2} + \frac{1}{\pi}\tan^{-1}(x) = \frac{1}{2} + \frac{1}{\pi}\left(-\frac{\pi}{2}\right) = 0$$

$$\lim_{x \to \infty} \frac{1}{2} + \frac{1}{\pi}\tan^{-1}(x) = \frac{1}{2} + \frac{1}{\pi}\frac{\pi}{2} = 1.$$

So $F(x)$ is a CDF.

(b)

For all $x \in \mathbb{R}$

$$\frac{d}{dx} (1 + e^{-x})^{-1} = \frac{e^{-x}}{(1 - e^{-x})^2} > 0$$

so $F(x)$ is increasing, and

$$\lim_{x \to -\infty} (1 + e^{-x})^{-1} = (1 + \infty)^{-1} = 0$$

$$\lim_{x \to \infty} (1 + e^{-x})^{-1} = (1 + 0)^{-1} = 1.$$

So $F(x)$ is a CDF.

(c)

For all $x \in \mathbb{R}$

$$\frac{d}{dx} e^{-e^{-x}} = e^{-x}e^{-e^{-x}} > 0$$

so $F(x)$ is increasing, and

$$\lim_{x \to -\infty} e^{-e^{-x}} = e^{-\infty} = 0$$

$$\lim_{x \to \infty} e^{-e^{-x}} = e^0 = 1.$$

So $F(x)$ is a CDF.

(d)

$F(x) = 0$ for $x \le 0$ . For $x > 0$

$$\frac{d}{dx} (1 - e^{-x}) = e^{-x} > 0$$

so $F(x)$ is nondecreasing, and

$$\lim_{x \to \infty} (1 - e^{-x}) = 1 - 0 = 1.$$

So $F(x)$ is a CDF.

(e)

The function is continuous everywhere except possible at the origin, and at the origin it is right continuous because of the placement of the equality sign in the definition. The function is increasing for $y < 0$ and for $y > 0$ from part (b), and using the limit results from part (b)

$$\lim_{y \to -\infty} F(y) = \epsilon \times 0 = 0$$

$$\lim_{x \to \infty} F(y) = \epsilon + (1 - \epsilon) * 1 = 1.$$

So $F(x)$ is a CDF.