Solutions

1.1

a.

Toss coin 4 times:

$$\{(H,H,H,H), \ldots\} = \{ (x_{1},x_{2},x_{3},x_{4}): x_{i} \in \{H,T\}\}$$

or

$$\{0, 1, 2, 3, 4\}$$

b.

Count number of insect-damaged leaves:

$$\{0, 1, \ldots, N\} N$$

$$\{0, 1, 2, \ldots\}$$    if no upper bound is available

c.

Measure lifetime in hours:

$$\{0, 1, 2, \ldots\}$$    if rounded (can put in upper limit)

$$[0,\infty)$$    if fractional hours are allowed

d.

Weight of ten-day-old rats:

$$(0,\infty)$$

$$(0,1000)$$    in grams

e.

Proportion of defectives:

$$[0,1]$$

$$\{r \in \mathbb{Q}: 0 \le r \le 1\}$$

$$\{0/N, 1/N, \ldots, N/N\} N$$

1.2

a.

$A \backslash B$ is defined as $A \cap B^{c}$ . To see that $A \backslash B = A \backslash (A \cap B)$ :

$$A \backslash (A \cap B) = A \cap (A \cap B)^{c}$$

$$= A \cap (A^{c} \cup B^{c})$$ De Morgan's law

$$= (A \cap A^{c}) \cup (A \cap B^{c})$$ distributive law

$$= \emptyset \cup (A \cap B^{c})$$

$$= A \backslash B$$

b.

$B = (B \cap A) \cup (B \cap A^{c})$ :

$$(B \cap A) \cup (B \cap A^{c}) = B \cap (A \cup A^{c})$$ distributive law

$$= B \cap S$$

$$= B$$

c.

By definition.

d.

$A \cup B = A \cup (B \cap A^{c})$ :

$$A \cup B = A \cup ((B \cap A) \cup (B \cap A^{c}))$$ by part (b)

$$= (A \cup (B \cap A)) \cup (B \cap A^{c})$$ associative law

$$= A \cup (B \cap A^{c})$$

1.4

a.

$A$ or $B$ or both:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

b.

$A$ or $B$ but not both:

$$A \Delta B = (A \cup B) \backslash (A \cap B)$$

$$= (A \cap B^{c}) \cup (B \cap A^{c})$$

and the sets $A \cap B^{c}$ and $B \cap A^{c}$ are disjoint. Since $A \cap B$ and $A \cap B^c$ are disjoint and their union is $A$ ,

$$P(A) = P(A \cap B) + P(A \cap B^c)$$

and therefore $P(A \cap B^c) = P(A) - P(A \cap B)$ . Similarly $P(B \cap A^c) = P(B) - P(A \cap B)$ and therefore

$$P(A \Delta B) = P(A) + P(B) - 2 P(A \cap B)$$

c.

At least one of $A$ or $B$ : $A \cup B$ .

d.

At most one of $A$ or $B$ = not $A \cap B$ = $(A \cap B)^{c}$ :

$$P((A \cap B)^{c}) = 1 - P(A \cap B)$$

1.5

The event $A \cap B \cap C$ is the event that the birth results in identical twins who are female. The proportion of births satisfying this description is

$$P(A \cap B \cap C) = \frac{1}{90} \times \frac{1}{3} \times{1}{2} = \frac{1}{540} = 0.001852.$$

1.13

If $A$ and $B$ are disjoint then

$$P(A \cup B) = P(A) + P(B) \le 1,$$

but

$$P(A) + P(B) = P(A) + 1 - P(B^c) = \frac{1}{3} + 1 - \frac{1}{4} = \frac{13}{12} > 1,$$

so $A$ and $B$ cannot be disjoint if $P(A) = \frac{1}{3}$ and $P(B^c) = \frac{1}{4}$ . Another way to see that this is not possible: If $A$ and $B$ are disjoint then $A \subset B^c$ and therefore $P(A) \le P(B)$ , but the opposite is true.