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

Answer: (A)

$\frac{1}{3}$

Answer 2

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: $\frac{1}{3}.$