2. One box contains three white and six black balls. A second box contains five
ID: 3049202 • Letter: 2
Question
2. One box contains three white and six black balls. A second box contains five white and four black balls. One ball is chosen at random from the first box and put into the second box. Then a ball is randomly selected from the second box and put in the first box. (a) What is the probability that the first and second selected ball are both white? (b) At the conclusion of the experiment, what is the probability that the configua tion of white and black balls in the two boxes is the same as in the beginning?Explanation / Answer
(a) Probability that the first selected ball is white = 3/9=1/3
After the first ball is put into the second box, there could be:
either 6 white and 4 black balls (if white ball was selected from box 1, with probability 1/3)
or 5 white and 5 black balls in the second box (if black ball was selected from box 1, with probability 6/9 = 2/3).
Probability that the second selected ball is white = (6/10) x (1/3) + (5/10) x (2/3) = (6+10) / 30 = 16/30
The probability that both the first and second selected balls are white = (1/3) x (16/30) = 16/90
(b) The probability of satus quo at the end of the experiment =
probability that both the first and second selected balls are white + probability that both the first and selected balls are black.
We already have calculated the probability that both the first and second selected balls are white = 16/90
The probability that both the first and second selected balls are black:
(i) the probability that the first selected balls is black = 6/9 = 2/3
(ii) The probability that the second selected ball is black = probability that first selected ball was white x probability of selecting a black ball from 2nd box + probability that first selected ball was black x probability of selecting a black ball from 2nd box = (1/3) x (4/10) + (2/3) x (5/10) = 14/30
The probability that both the first and second selected balls are black = (2/3) x (14/30) = 28/90
The probability that the configuration is the same as at the beginning = 16/90 + 28/90 = 44/90 =22/45
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