Statistics 5101, Fall 2000, Geyer

Homework Solutions #2

Note: Original done by Laura Pontiggia, Fall 1999. Additions by Yumin Huang, Fall 2000.

Problem 2-12

Since $(A \cup B^c)^c = A^c B$, and since $\mathop{\rm pr}\nolimits(B) = \mathop{\rm pr}\nolimits(A^c B) + \mathop{\rm pr}\nolimits(A B)$[using prop. (7)] then $\mathop{\rm pr}\nolimits(A^c B) = \mathop{\rm pr}\nolimits(B) - \mathop{\rm pr}\nolimits(A B) = .30 - .21 = .09$. Using prop. (1): $\mathop{\rm pr}\nolimits(A \cup B^c)= 1 - \mathop{\rm pr}\nolimits[(A \cup B^c)^c] = 1 - .09 = .91$.

Problem 2-20

The inequality on the right (the only one we were asked to do) $\mathop{\rm pr}\nolimits(E \cup F) \leq \mathop{\rm pr}\nolimits(E) + \mathop{\rm pr}\nolimits(F)$ follows from Property (3) and Axiom 1.

Problem 2-22

(b)

By Theorem 2, $\mathop{\rm pr}\nolimits(E_{1} \cup E_{2}) = \mathop{\rm pr}\nolimits(E_{1}) + \mathop{\rm pr}\nolimits(E_{2}) - \mathop{\rm pr}\nolimits(E_{1} E_{2})$. Since $\mathop{\rm pr}\nolimits(E_{1} E_{2})$ is nonnegative. The inequality holds for n = 2.

Now, assume the inequality holds when n = k - 1. Let $F = E_{1} \cup E_{2} \cup \cdots \cup E_{k-1}$. Applying Theorem 2(3) again, it follows that $\mathop{\rm pr}\nolimits(F \cup E_{k}) \leq \mathop{\rm pr}\nolimits(F) + \mathop{\rm pr}\nolimits(E_{k})$. Therefore the inequality holds by mathematical induction.

Problem 2-26

Since each characteristic two possible values, the number of outcomes in this experiment is 4²=16.

The probability distribution that reflects the proportions 9:3:3:1 is $(\frac{9}{16},\frac{3}{16},\frac{3}{16},\frac{1}{16})$. Hence the probability that a plant has yellow peas is given by

$$\mathop{\rm pr}\nolimits(\text{a plant has yellow peas}) = ... ... + pr(\text{RY}) = \frac{3}{16} + \frac{9}{16} = \frac{3}{4}.$$

Problem 2-27

(a)

$$\sum_{i=1}^\infty \sum_{j=1}^\infty \left(\frac{1}{2}\right)^... ...sum_{j=1}^\infty \left(\frac{1}{2}\right)^j = 1 \times 1 = 1.$$

(b)

Call the event in question A, so

$$A = \{\, (1,1), (1,2), (2,1), (1,3), (3,1), (2,2) \,\}.$$

Then

$$\mathop{\rm pr}\nolimits(A) = \frac{1}{2^{1+1}} + \frac{2}{... ... = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} = \frac{11}{16}.$$

Problem 3-3

(a)

	Y	1	2	3	4	f(x)
X
1		0	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{4}$
2		$\frac{1}{12}$	0	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{4}$
3		$\frac{1}{12}$	$\frac{1}{12}$	0	$\frac{1}{12}$	$\frac{1}{4}$
4		$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{12}$	0	$\frac{1}{4}$
f(y)		$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$	1

The distribution is uniform (equally likely outcomes) except for the outcomes on the diagonal (with X = Y), which are impossible because of the sampling ``without replacement.''

(b)

X	1	2	3	4
fX(x)	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$

Y	1	2	3	4
fY(y)	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$	$\frac{1}{4}$

(c)

Z	3	4	5	6	7
f(z)	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$

$$\mathop{\rm pr}\nolimits(Z = X + Y = 3) = \mathop{\rm pr}\nol... ...limits(X=2, Y=1) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}.$$

$$\mathop{\rm pr}\nolimits(Z = X + Y = 4) = \mathop{\rm pr}\nol... ...limits(X=3, Y=1) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}.$$

$$\begin{align*}\mathop{\rm pr}\nolimits(Z = X + Y = 5) & = \mathop{\rm pr}\noli... ...{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{1}{3}. \end{align*}$$
and so forth for $\mathop{\rm pr}\nolimits(Z=6)$ and $\mathop{\rm pr}\nolimits(Z=7)$.

Problem 3-4

There are six points in the sample space

	number
guess	correct
A B C	3
B C A	0
C A B	0
C B A	1
B A C	1
A C B	1

Thus

X	0	1	2	3
f(x)	$\frac{1}{3}$	$\frac{1}{6}$	0	$\frac{1}{2}$

(or the point 2 can be deleted from the sample space if you prefer).

Problem 3-6

There are $\binom{12}{3}$ points in the sample space (ways to choose 3 eggs from 12 eggs).

Similarly, there are are $\binom{2}{k}$ to choose k rotten eggs from the 2 rotten eggs in the carton and $\binom{10}{k}$ to choose k non-rotten eggs from the 10 non-rotten eggs in the carton. If there are k rotten eggs drawn, there are 3 - k non-rotten eggs drawn. Thus

$$f(y) = \frac{\binom{2}{y} \binom{10}{3-y}}{\binom{12}{3}}, \qquad y=0,1,2.$$