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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?
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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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