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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?
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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