Question

A bag contain four balls, two balls are drawn at random and they are found to be white. Find the probability that all balls are white.

Answer 1

3/5

E1 = ALL 4 R WHITE(WWWW) 

E2= 3 WHITE 1 BLACK             

E3 = 2 W AND 2 Black
each of these events hve equal prob.=1/3
let the event of withdrawing 2 white balls be A

now P(A/E1)=2C2/4C2
P(A/E2)=3C2/4C2
P(A/E3)=4C2/4C2
now apply bayes theorem
P(E3/A)= (P(E3)*P(A/E3))    /    P(E3)P(A/E3) + P(E2)P(A/E2)  + P(E1)P(A/E1)
THE ANS IS 3/5

Answer 2

This is a simple sounding question that hides a lot of complexity and some assumptions.
We need to compute P(all balls white | drew 2 white balls).
This is a conditional probability so we use Bayes Law.
P(A | B) = P(B | A) * P(A)/ P(B).

Where A = all balls white
B = draw 2 white balls

P(B) is not given. I will assume there are 3 possible worlds: 4 whites, 3 white and 1 non white,
and 2 whites and 2 non-white. I need to also know the probabilities for each of these worlds. I will assume they are equal, the standard assumption if no information is given.

P(B) = P(B| 4W)*P(4W) +P(B| 3W, 1non-white)*P(3W, 1 non-white) + P(B | 2white,2 non-white).

P(B|4W) = 1
P(4W) = 1/3 = P(3W and 2 non) = P(2W and 2 non)

P(B | 3W and 1 non) = 3/4*2/3 = 1/2

P(B | 2W and 1 non) = 2/4*1/3 = 1/6

Putting this all together with Bayes:
P(4W in bag | drew 2 white) = 1* 1/3/ ( 1*1/3 + 1/2*1/3 + 1/6*1/3) = 3/5 assuming I haven't made any arithmetic mistakes.
Read my solution, and then do it yourself to see if you understand the method.
good luck