[Math Lair] Pillow-Problems: Problem #27

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For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

Problem:

27. (22, 50)

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:

27. (6, 50)

Seventeen-twentyfifths.

Solution:

27. (6, 22)

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 · {
(2)/
(3)
× ½ +
(1)/
(3)
× ½} = ¼.

Similarly, if remaining bag be B, it is ½ · {
(5)/
(6)
· ½ +
(1)/
(6)
· ½} = ¼ and, if it be C, it is ½ · {
(5)/
(6)
·
(1)/
(3)
+
(1)/
(6)
·
(2)/
(3)
} =
(7)/
(36)
.

∴ chances of remaining bag being A, B, or C, are as ¼ to ¼ to
(7)/
(36)
; i.e. as 9 to 9 to 7. ∴ they are, in value,
(9, 9, 7)/
(25)
.

Now, if remaining bag be A, chance of drawing white from it is
(5)/
(6)
; ∴ chance, on this issue, is
(5)/
(6)
·
(9)/
(25)
=
(3)/
(10)
; similarly, for B, it is
(2)/
(3)
·
(9)/
(25)
=
(6)/
(25)
; and, for C, &rac12; ·
(7)/
(25)
=
(7)/
(50)
. And either chance of drawing white from the remaining bag is the sum of these; i.e.
(15 + 12 + 7)/
(50)
=
(34)/
(50)
=
(17)/
(25)
.