(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A Bag contains 4 balls.two balls are drawn at random,and are found to be white.what is the probability that all balls are white?

2. Relevant equations

P(B/A) = P(A intersection B) / P (A)

3. The attempt at a solution

A = Two balls are white.

B = All four balls are white.

I wrote P(B/A) and P(A/B) according to Bayes' theorem and eliminated the intersection term, to arrive with P(B/A) = P(B)/P(A). Ofcourse, P(A|B) = 1.

How do i solve these two now?

I found an alternate solution on net which goes this way...

and another which says... P(E1)=p(E2)=p(E3)=1/3.

Calculate p(A1),p(A2),p(A3), you get 1, 1/2, 1/4

Apply Baye's theorem and you get the magical 3/5 as the answer

p(white)=1/2

p(nonwhite)=1/2

assuming equally likely outcomes

given balls 1&2 are white their respective proababilities of being white is 1.and is 1/2 for the other two.

hence,

P(all are white)=p(first ball being white)*p(2nd being white)*p(3rd being white)*p(4th being white)

=1*1*1/2*1/2

=1/4Can anyone, apart from helping me with my question, explain the first aliter and the mistake in assuming the probability = 1/2 in the second aliter?

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# Probability question - Bayes' Theorem - 4 Balls, 2 white, P(all white) = ?

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