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Best Answer

it has three cases 1st E1- two balls are white
E2- three balls are white
E4-four balls are white
E4/w= (p(E3)*4C2/4C2)/(P(E1)*2C2/4C2+P(E2)3C2/4C2+P(E3)*4C2/4C2)
WHERE-P(E1)=P(E2)=P(E3)=1/3
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2nd opinion

Let's say we have 3 boxes with 4 balls each.

Box A has 4 white balls.

Box B has 3 white balls.

Box C has 2 white balls.

The probability of drawing 2 W balls from;

Box A, P(2W│A)=

(4/4)∙(3/3)=

1

Box B, P(2W│B)=

(3/4)∙(2/3)=

1/2

Box C, P(2W│C)=

(2/4)∙(1/3)=

1/6

Say the probability of picking any of the 3 boxes is the same, we have;

P(A)=

1/3

P(B)=

1/3

P(C)=

1/3

Question is, given the event of drawing 2 W balls from a box taken blindly

from the 3 choices, what is the probability that the balls came from box A,

P(A│2W).

Recurring to Bayes Theorem:

P(A│2W)=

[P(A)P(2W│A)]/[P(A)P(2W│A)+P(B)P(2W│B)+P(C)P(2W│C)]=

[(1/3)(1)]/[(1/3)(1)+(1/3)(1/2)+(1/3)(1/6)]=

6/10=

0.60=

60%

P(A│2W)=

0.60=

60%



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Q: A bag contain 4 balls 2 balls are drawn at random and are found to be white What is the probability that all balls are white?
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