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Question
A jewellery seller has precious gems in white and red colour which he has put in three boxes.
The distribution of these gems is shown in the table given below:
| Box | Number of Gems | |
| White | Red | |
| I | 1 | 2 |
| I | 2 | 3 |
| III | 3 | 1 |
He wants to gift two gems to his mother. So, he asks her to select one box at random and pick out any two gems one after the other without replacement from the selected box. The mother selects one white and one red gem.
Calculate the probability that the gems drawn are from Box II.
Solution
The probability of selecting each box is:
P(E1) = P(E2) = P(E3) = `1/3`
Two gems are chosen from the selected box.
Let A be the event where one white and one red gem are chosen.
P(A | E1) = Probability of drawing 1 white and 1 red gem when box I is chosen
= `(""^1C_1 xx ""^2C_1)/(""^3C_2)`
= `2/3`
P(A | E2) = Probability of drawing 1 white and 1 red gem when box II is chosen
= `(""^2C_1 xx ""^3C_1)/(""^5C_2)`
= `(2 xx 3)/10`
= `3/5`
P(A | E3) = Probability of drawing 1 white and 1 red gem when box III is chosen
= `(""^3C_1 xx ""^1C_1)/(""^4C_2)`
= `(3 xx 1)/6`
= `1/2`
According to Bayes theorem, we have
`P(E_2 | A) = (P(E_2)P(A | E_2))/(P(E_1)P(A | E_1) + P(E_2)P(A | E_2) + P(E_3)P(A | E_3)`
= `(1/3 xx 3/5)/(1/3 xx 2/3 + 1/3 xx 3/5 + 1/3 xx 1/2)`
= `(1/5)/(2/9 + 1/5 + 1/6)`
= `(1/5)/((20 + 18 + 15)/90)`
= `90/(5 xx 53)`
= `18/53`
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