Question

A bag contains 4 balls. Two balls are drawn at random (without replacement) and are found to be white. What is the probability that all balls in the bag are white?

Answer 1

The events are defined as follows:

E: Two ball drawn are white

A: There are 2 white balls in the bag

B: There are 3 white balls in the bag

C: There are 4 white balls in the bag

Then, P(A) = P(B) = P(C) = 1/3

Also,

`P(E/A)=(""^2C_2)/("^4C_2)=1/6`

`P(E/B)=(""^3C_2)/("^4C_2)=3/6 = 1/2`

`P(E/C)=(""^4C_2)/("^4C_2)=1`

∴ Required probability `= P(C/E)`

Apply Baye’s theorem:

`P(C/E) = (P(C).P(E/C))/(P(A).P(E/A) + P(B).P(E/B) + P(C).P(E/C))`

`=(1/3 × 1)/(1/3 × 1/6 + 1/3 × 1/2 + 1/3 × 1) = 3/5`

Thus, the required probability is 3/5.