An urn contains 5 white and 5 black balls. 4 balls are drawn from this urn and put into another urn. From this second urn a ball is drawn and is found to be white. What's the probability of drawing a white ball again at the next draw. (The first white ball drawn is not replaced.)
What I did was:-
E0= Drawing 0 white and 4 black balls
E1= Drawing 1 white and 3 black balls
E2= Drawing 2 white and 2 black balls
E3= Drawing 3 white and 1 black balls
E4= Drawing 4 white and 0 black balls
We ignore E0 because if there were 0 white balls transferred, the first one drawn couldn't be white. Out of the 4 remaining events, each of them have the probability of occurring at 1/4.
In E1, if the first ball drawn is white, there are 0 white balls left.(Total left 3)
In E2, if the first ball drawn is white, there is 1 white ball left. (Total left 3)
In E3, if the first ball drawn is white, there are 2 white balls left. (Total left 3.)
In E4, if the first ball drawn is white, there are still 3 more white balls left. (Total left 3)
So, the final answer should be [(1/4*0/3) + (1/4*1/3) + (1/4*2/3) + (1/4*3/3)]
Which equals to 1/12 + 2/12 + 3/12 = 1/2
I believe this to be the correct answer however the book has got 4/9 for the answer. Where exactly did I go wrong?
(Thanks for your time and sorry if there are any formatting errors in question.)