# Pillow-Problems: Problem #19

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## Problem:

19. (21, 42)

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?

19. ()

Eleven-seventeenths.

## Solution:

19. (5, 21)

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,
 (1)/ (6)
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:—

for ABC, ½ ×
 (1)/ (3)
; i.e.
 (1)/ (6)
.

## Solution:

ACB, ½ × ¼; i.e.
 (1)/ (8)
BAC,
 (2)/ (3)
× ½; i.e.
 (1)/ (3)
BCA,
 (2)/ (3)
× ¼; i.e.
 (1)/ (6)
CAB, ¾ × ½ i.e.
 (3)/ (8)
CBA, ¼ ×
 (1)/ (3)
; i.e. ¼

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

 (1)/ (34)
· {4 × ¾ + 3 ×
 (2)/ (3)
+ 8 × ¾ + 4 × ½ + 9 ×
 (2)/ (3)
+ 6 × ½};
i.e.
 (1)/ (34)
× {3 + 2 + 6 + 2 + 6 + 3}; i.e.
 (22)/ (34)
; i.e.
 (11)/ (17)
.