Question

A box has three balls, one white and two 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

See explaation

Explanation:

Given

Represent the balls with the first letters

`W =1`

`R =2`

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

The event of F is:

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

So:

`P(F) = P(W) * P(W)`

`P(F) = (n(W))/(n) * (n(W))/(n)`

`P(F) = (1)/(3) * (1)/(3)`

`P(F) = (1)/(9)`

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) = (n(W))/(n) * (n(R))/(n) + (n(R))/(n) * (n(W))/(n)`

`P(G) = (1)/(3) * (2)/(3) + (2)/(3) * (1)/(3)`

`P(G) = (2)/(9) + (2)/(9)`

`P(G) = (4)/(9)`

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) = (n(W))/(n) * (n(R))/(n) + (n(W))/(n) * (n(W))/(n)`

`P(H) = (1)/(3) * (2)/(3) + (1)/(3) * (1)/(3)`

`P(H) = (2)/(9) + (1)/(9)`

`P(H) = (3)/(9)`

`P(H) = (1)/(3)`

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