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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?
Solution
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.
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