## Problem:

There are 3 bags; one containing a white counter and a black one, another two white and a black, and the third 3 white and a black. It is not known in what order the bags are placed. A white counter is drawn from one of them, and a black from another. What is the chance of drawing a white counter from the remaining bag?

## Answer:

Eleven-seventeenths.

## Solution:

Call the bags `A`, `B`, `C`; so that `A` contains a white counter and a black one; &c.

The chances of the orders `ABC`, `ACB`, `BAC`, `BCA`, `CAB`, `CBA`, are, a priori,

each. Since they are equal, we may, instead of multiplying each by the probability it gives to the observed event, simply assume those probabilities as being proportional to the chances *after* the observed event.
These probabilities are:—

## Solution:

Hence the chances are proportional to 4, 3, 8, 4, 9, 6; i.e. they are these Nos. divided by 34.

Hence the chance, of drawing a white counter from the remaining bag; is

· {4 × ¾ + 3 ×

+ 8 × ¾ + 4 × ½ + 9 ×

+ 6 × ½};

i.e.

× {3 + 2 + 6 + 2 + 6 + 3}; i.e.

; i.e.

.