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 ?

 Two balls are certainly white. So the remaining balls may be one of the following

{WW, WN, NW, NN}

where W means a white ball and N means non-white ball

Let the events be:

E- all four balls are white(WW)

F- 3 balls are white(WN, NW)

G- only two balls are white(NN)

Now, P(E) = P(G) = 1/4, and P(F) = 1/2

Let A be the event that the two balls drawn are white

P(A|E) = ⁴C₂/⁴C₂ = 1

P(A|F) = ³C₂/⁴C₂ = 1/2

P(A|G) = ²C₂/⁴C₂ = 1/6

Now By Bayes' Theorem,

required probability = P(E|A) = (1*1/4) / ( 1*1/4 + 1/2*1/2 + 1/6*1/4 )

  = (1/4) / ( 13/24 )

  = 6/13

it is not corrent answer

To solve the problem carefully, we need to set up some machinery. Let A be the event they are all white, and F be the event the first two balls removed are white. We want Pr(W|F).

By using the defining formula Pr(W|F)=Pr(W∩FPr(F for conditional probabilities, we can calculate a formal answer. If you wish, I can later complete that approach.

But let's do it more informally. We put balls into an urn 600 times, where 120 times we put in all white, 120 times we put in 1 white and 3 black, and so on.

How many times roughly did we draw 2 white? If we were drawing from a 2 white 2 black, the probability of 2 white is 12⋅13, so we got 2 white about 120/6=20 times. If we are drawing from a 3 white 1 black, the probability of 2 white is 34⋅23, so we got 2 white about 60 times. Finally, if we drew from an all white, we got 2 white for sure, so 120 times.

The total number of times we got 2 white was 200. Of these times, 120 came from an all white situation. So the probability is 120200=35.