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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?

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#### Solution

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`

`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.

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