One bag contains 4 white balls and 6 black balls. Another bag contains 8 white balls and 2 black balls. A coin tossed to select a bad, then a ball is randomly selected from that bag.

Question

Suppose white ball was drawn. What is the probability that it came from the first bag?

Answer 1

Since a coin toss is used to select either bag1 or bag2
the prob that bag1 is choses is 1/2

prob that a white ball is choses from bag1 = 4/10 = 2/5

so prob of your 'event' = (1/2)(2/5) = 1/5

Answer 2

but the answer in my book says 1/3... Maybe its wrong?

Answer 3

PR=BROWN

Answer 4

Box X contains 8 oranges of which 3 are Bad and Box Y contains 5 pears of which 2 are bad. A fruit is drawn from each box.
a) What is the probability that both are not bad?
b) What is the probability that one item is bad and one is not?
c) If one fruit is a bad and one is not what is the probability that the bad fruit came from box X

Answer 5

@ the man - yes your book is wrong when it says 1/3, which is just totaling white balls in bag 1 (4) and bag 2 (8) to get 12, of which the 1st 4 represent 1/3.

this can can easily be shown to be wrong, if you make the 1st bag 4 white and 600 black instead, your book would still say 1/3, but the probability a white ball from bag 1 would in fact be much, much lower.

incidentally, 1/5 is close, but also wrong. the correct answer is 1 / 5.5.