## 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`, &rac12; · = . And either chance of drawing white from the remaining bag is the sum of these; i.e. = =
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