An urn initially contains 5 white and 7 black balls. Each time a ball is selected, it’s color is noted and

Question

it is replaced in the urn along with two other balls of the same color. Compute the probability that
(a) one black and one white is selected.
(b) exactly two black balls are selected.

Answer 1

a) two cases: bw or wb
case 1:
first ball is b, prob(b) = 7/12
so 3 black balls are added,
now we have 5 w, 10 b
prob(w) = 5/15 = 1/3
prob(case1) = (7/12)(1/3) = 7/36

case 2:
first ball is w, prob(w) = 5/12
so 3 white balls are added,
now we have 8w, 7b
prob(b) = 7/15
prob(case2) = (5/12)(7/15) = 7/36

prob(black and white) = 7/36 + 7/36
= 7/18

b) using the same reasoning:
prob(2black) = (7/12)(10/15) = 7/18

check:
all cases:
ww ---> (5/12)(8/15) = 2/9
wb ---> (5/12)(7/15) = 7/36
bw ---> (7/12)(5/15) = 7/36
bb ---> (7/12)(10/15) = 7/18
add them up:
2/9 + 7/36+7/36+7/18 = 1

Answer 2

case 1