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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?
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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 = 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.
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