Problem:
There are 3 bags, each containing 6 counters; one contains 5 white and one black; another, 4 white and 2 black; the third, 3 white and 3 black. From two of the bags (it is not known which) 2 counters are drawn, and prove to be black and white. What is the chance of drawing a white counter from the remaining bag?
[4/3/80
Answer:
Seventeen-twentyfifths.
Solution:
Call the bags A, B, C.
if remaining bag be A, chance of observed event = ½ chance of drawing white from B and black from C + ½ chance of drawing black from B and white from C:
i.e. it = ½ · { × ½ + × ½} = ¼.
Similarly, if remaining bag be B, it is ½ · { · ½ + · ½} = ¼ and, if it be C, it is ½ · { · + · } = .
∴ chances of remaining bag being A, B, or C, are as ¼ to ¼ to ; i.e. as 9 to 9 to 7. ∴ they are, in value, .
Now, if remaining bag be A, chance of drawing white from it is ; ∴ chance, on this issue, is · = ; similarly, for B, it is · = ; and, for C, ½ · = . And either chance of drawing white from the remaining bag is the sum of these; i.e. = =
.