Question

Suppose we have four boxes. A, B, C and D containing coloured marbles as given below:

Box	Marble colour
	Red	White	Black
A	1	6	3
B	6	2	2
C	8	1	1
D	0	6	4

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

Answer 1

The probability of selecting one box out of the given 4 boxes = `1/4`

i.e., P(E₁) = P(E₂) = P(E₃) = P(E₄) = `1/4`

Let the 4th event be drawing a red piece. There are a total of 10 pieces in box A, of which 1 is red.

∴ `P(A/E_1) = 1/10`

Similarly `P(A/E_2) = 6/10, P(A/E_3) = 8/10, P(A/E_4) = 0`

(i) Again, by Bayes' theorem,

`P((E_1)/A) = (P(E_1) xx P((A)/E_1))/(P(E_1) xxP(A/E_1) + P(E_2) xx P(A/E_2) + P(E_3) xx P(A/E_3) + P(E_4) xx P(A/E_4))`

= `(1/4 xx 1/10)/(1/4 xx 1/10 + 1/4 xx 6/10 + 1/4 xx 8/10 + 1/4 xx 0)`

= `1/(1 + 6 + 8)`

= `1/15`

(ii) Again, by Bayes' theorem,

= `P((E_2)/A) = (P(E_2) xx P((A)/E_2))/(P(E_1) xxP(A/E_1) + P(E_2) xx P(A/E_2) + P(E_3) xx P(A/E_3) + P(E_4) xx P(A/E_4))`

= `(1/4 xx 6/10)/(1/4 xx 1/10 + 1/4 xx 6/10 + 1/4 xx 8/10 + 1/4 xx 0)`

= `6/(1 + 6 + 8)`

= `6/15`

= `2/5`

(iii) and by Bayes' theorem,

= `P((E_3)/A) = (P(E_3) xx P((A)/E_3))/(P(E_1) xxP(A/E_1) + P(E_2) xx P(A/E_2) + P(E_3) xx P(A/E_3) + P(E_4) xx P(A/E_4))`

= `(1/4 xx 8/10)/(1/4 xx 1/10 + 1/4 xx 6/10 + 1/4 xx 8/10 + 1/4 xx 0)`

= `8/(1 + 6+ 8)`

= `8/15`

Hence, the probability of a red piece being selected from box A, box B, and box C is `1/5`, `2/5` and `8/15`, respectively.