Question

A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:
a. Let F = the event of getting the white ball twice.
b. Let G = the event of getting two balls of different colors.
c. Let H = the event of getting white on the first pick.
d. Are F and G mutually exclusive?
e. Are G and H mutually exclusive?

Answer 1

Answer:

See explanation

Step-by-step explanation:

Given

Represent the balls with the first letters

$W =1$

$R =1$

Solving (a): P(F) --- White balls twice

The event of F is:

$F = \{(W,W)\}$

So:

$P(F) = P(W) * P(W)$

$P(F) = \frac{n(W)}{n} * \frac{n(W)}{n}$

$P(F) = \frac{1}{2} * \frac{1}{2}$

$P(F) = \frac{1}{4}$

Solving (b): P(G) --- two different colors

The event of G is:

$G = \{(W,R),(R,W)\}$

So:

$P(G) = P(W) * P(R) + P(R) * P(W)$

$P(G) = \frac{n(W)}{n} * \frac{n(R)}{n} + \frac{n(R)}{n} * \frac{n(W)}{n}$

$P(G) = \frac{1}{2} * \frac{1}{2} + \frac{1}{2} * \frac{1}{2}$

$P(G) = \frac{1}{4} + \frac{1}{4}$

$P(G) = \frac{1}{2}$

Solving (c): P(H) --- White picked first

The event of H is:

$H = \{(W,R),(W,W)\}$

So:

$P(H) = P(W) * P(R) + P(W) * P(W)$

$P(H) = \frac{n(W)}{n} * \frac{n(R)}{n} + \frac{n(W)}{n} * \frac{n(W)}{n}$

$P(H) = \frac{1}{2} * \frac{1}{2} + \frac{1}{2} * \frac{1}{2}$

$P(H) = \frac{1}{4} + \frac{1}{4}$

$P(H) = \frac{1}{2}$

Solving (d): F and G, mutually exclusive?

We have:

$F = \{(W,W)\}$

$G = \{(W,R),(R,W)\}$

Check for common elements

$n(F\ n\ G) = 0$

Hence, F and G are mutually exclusive

Solving (e): G and G, mutually exclusive?

We have:

$G = \{(W,R),(R,W)\}$

$H = \{(W,R),(W,W)\}$

Check for common elements

$n(G\ n\ H) = 1$

Hence, F and G are not mutually exclusive