Question 1183791
i believe the probability that A draws a white ball first is .7.


my thinking:


A gets the white ball on the first draw.
that probability is 3/5 = .6


A gets the white ball on the second draw.
in order for this to happen, both A and B must have drawn a black ball on their first try..
if they do, then there are only white balls left and A is sure to draw a white ball on his second try.
the probability of this happening is 2/5 * 1/4 * 3/3 = 6/60 = .1.


the probability that A is first to get a white ball is therefore .6 + .1 = .7.


since the total probability must be equal to 1, then the probability that B gets the white ball first must be .3.


the ways that this could happen are:


B draws a white ball on the first try.
in order for this to happen, A must have drawn a black ball on the first try.
the probability that B is the first to draw a white ball on the first try is therefore 2/5 * 3/4 = 6/20 = .3


B draws a white ball first on the second try.
in order for this to happen, both A and B must have drawn a black ball on their first try.
sine there are only white balls left, then A is assured to get a white ball on the second try.
the probability that B gets a white ball on the second try is therefore 0.


the probabilities seem to be true.


the probability that A gets a white ball first is .7
the probability that B gets a white ball first is .3
the total probabilities are equal to 1.


this seems reasonable, which is why i think the probability that A gets a white ball first is .7.