A leather bag contains two balls, about which nothing is known except that each is either black or white. We assert that the colours of the balls can be ascertained without removing them from the bag!
Observe that if a bag contains three balls, two being black and one white, then the probability of drawing a black one is 2/3, and no other state of things will give this probability.
With two balls, there are four equally likely cases: both balls can be black, or the first black and the second white, or the first white and the second black, or both white. Call these cases BB, BW, WB, WW, respectively. Each is equally likely, and since one of them must be the true situation, the probability of each being the true situation is 1/4.
Now add a black ball. Then the probabilities of BBB, BWB, WBB, and WWB are each 1/4 . In the case BBB, the probability of drawing a black ball is 1; in the cases BWB or WBB, 2/3, and in the case WWB, 1/3 . Since the four cases are mutually exclusive, by the law of addition of probabilities and Fermat’s Principle, the probability of drawing a black ball from the bag with three balls is (1/4)×1 + (1/4)×(2/3) + (1/4)×23 + (1/4)×(1/3)=8/12=2/3 .
But, as observed earlier, the probability of drawing a black ball is 2/3 only if the bag contains two black balls and one white ball. Hence, before the black ball was added, the bag must have contained one white and one black ball.
Discuss.
Observe that if a bag contains three balls, two being black and one white, then the probability of drawing a black one is 2/3, and no other state of things will give this probability.
With two balls, there are four equally likely cases: both balls can be black, or the first black and the second white, or the first white and the second black, or both white. Call these cases BB, BW, WB, WW, respectively. Each is equally likely, and since one of them must be the true situation, the probability of each being the true situation is 1/4.
Now add a black ball. Then the probabilities of BBB, BWB, WBB, and WWB are each 1/4 . In the case BBB, the probability of drawing a black ball is 1; in the cases BWB or WBB, 2/3, and in the case WWB, 1/3 . Since the four cases are mutually exclusive, by the law of addition of probabilities and Fermat’s Principle, the probability of drawing a black ball from the bag with three balls is (1/4)×1 + (1/4)×(2/3) + (1/4)×23 + (1/4)×(1/3)=8/12=2/3 .
But, as observed earlier, the probability of drawing a black ball is 2/3 only if the bag contains two black balls and one white ball. Hence, before the black ball was added, the bag must have contained one white and one black ball.
Discuss.
