There are three bags X, Y, and Z. Bag X contains 5 one-rupee coins and 4 five-rupee coins; Bag Y contains 4 one-rupee coins and 5 five-rupee coins, and Bag Z contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn f

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Solution

Q.
There are three bags X, Y, and Z. Bag X contains 5 one-rupee coins and 4 five-rupee coins; Bag Y contains 4 one-rupee coins and 5 five-rupee coins, and Bag Z contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability that it came from bag Y is [2024]

1 $\frac{1}{2}$

2 $\frac{5}{12}$

3 $\frac{1}{4}$

4 $\frac{1}{3}$

Ans.
(4)

Let $E_{1}, E_{2}, E_{3}$ be the event of selecting bags X, Y and Z respectively and A be the event that coin drawn is one-rupee coin.

$∴$ $P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}$

$P (A | E_{1}) = \frac{5}{9}, P (A | E_{2}) = \frac{4}{9}, P (A | E_{3}) = \frac{3}{9}$

By Bayes' theorem,

Required Probability = $P (E_{2} | A) = \frac{\frac{1}{3} \times \frac{4}{9}}{\frac{1}{3} \times \frac{5}{9} + \frac{1}{3} \times \frac{4}{9} + \frac{1}{3} \times \frac{3}{9}}$

$= \frac{4}{5 + 4 + 3} = \frac{4}{12} = \frac{1}{3}$

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