You choose one of two identical looking bags at random. One bag has three black marbles and one white marble. The other has three white marbles and one black marble.
After choosing a bag you draw one marble out at random. You notice it is black. You then put it back and draw another marble out of the same bag at random.
What is the probability that the second marble drawn is black?
For this problem, I will refer to the first bag (the bag containing 3
black marbles and 1 white marble) as bag A and the second bag (the bag
containing 3 white marbles and 1 black marble) as bag B.
First, let's take into account the probability one picked bag A
(1/2). The probability that, on the first draw, one drew out a
black marble is 3/4. The probability that one drew out a black
marble on the second
draw would again be 3/4. Thus, the probability that one draws a
black marble on both tries is (3/4)*(3/4)=9/16. Do not forget the
probability they picked bag A, so (9/16)*(1/2)=9/32 of a chance that
one randomly drew two black marbles out of randomly picked bag A.
Second, we must take into account the probability of this also
happening with one randomly picking bag B (which has 1/2
probability). The probability of one drawing a black marble out
on both tries in bag B would be: (1/4)*(1/4)*(1/2)=1/32.
Thus, the probability of this event happening would be (9/32)+(1/32)=10/32 or 5/16.
a possible answer
