Question

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:

• A. \(\frac{1}{3}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{5}{12}\)

Answer 1

The Correct Option is A

Let the events \( E_X, E_Y, \) and \( E_Z \) denote the selection of bags \( X, Y, \) and \( Z \) respectively. Let the event \( A \) denote drawing a one-rupee coin. We are required to find the conditional probability: \[ P(E_Y|A) = \frac{P(E_Y) \times P(A|E_Y)}{P(A)}. \]

The probabilities of selecting each bag are: \[ P(E_X) = P(E_Y) = P(E_Z) = \frac{1}{3}. \] 

The probability of drawing a one-rupee coin from each bag is given by: \[ P(A|E_X) = \frac{5}{9}, \quad P(A|E_Y) = \frac{4}{9}, \quad P(A|E_Z) = \frac{3}{9}. \] 

The total probability of drawing a one-rupee coin, using the law of total probability: \[ P(A) = P(E_X) \times P(A|E_X) + P(E_Y) \times P(A|E_Y) + P(E_Z) \times P(A|E_Z). \] 

Substituting the values: \[ P(A) = \frac{1}{3} \times \frac{5}{9} + \frac{1}{3} \times \frac{4}{9} + \frac{1}{3} \times \frac{3}{9}, \] \[ P(A) = \frac{5}{27} + \frac{4}{27} + \frac{3}{27} = \frac{12}{27} = \frac{4}{9}. \] 

Now, the conditional probability that the coin came from bag \( Y \) given that it is a one-rupee coin is: \[ P(E_Y|A) = \frac{P(E_Y) \times P(A|E_Y)}{P(A)}, \] \[ P(E_Y|A) = \frac{\frac{1}{3} \times \frac{4}{9}}{\frac{4}{9}} = \frac{\frac{4}{27}}{\frac{4}{9}} = \frac{1}{3}. \] 

Therefore: \[ \frac{1}{3}. \]