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?
P(B/A) = P(A intersection B) / P (A)
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
Can 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?p(white)=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.
P(all are white)=p(first ball being white)*p(2nd being white)*p(3rd being white)*p(4th being white)