QJEEM24Ma0d67: JEE Main 2024 Mathematics - Class 12 JEEM24Ma0d67 Solution: There are three bags and . Bag contains 5 one-rupee coins and 4 …

To solve this problem, we use Bayes' theorem. Let's define the events:

• $X$ : Selecting bag $X$
• $Y$ : Selecting bag $Y$
• $Z$ : Selecting bag $Z$
• $A$ : Drawing a one-rupee coin

We are given that a bag is selected at random, so the probabilities for choosing any of the bags are:

$P(X) = P(Y) = P(Z) = \frac{1}{3}$

Next, we need the probability of drawing a one-rupee coin from each bag:

• From bag $X$ : $P(A|X) = \frac{5}{5+4} = \frac{5}{9}$
• From bag $Y$ : $P(A|Y) = \frac{4}{4+5} = \frac{4}{9}$
• From bag $Z$ : $P(A|Z) = \frac{3}{3+6} = \frac{3}{9} = \frac{1}{3}$

We need to find the probability that the coin came from bag $Y$ given that a one-rupee coin was drawn, i.e., we need $P(Y|A)$ . Using Bayes' theorem:

$P(Y|A) = \frac{P(A|Y) \cdot P(Y)}{P(A)}$

To find $P(A)$ , the total probability of drawing a one-rupee coin can be calculated as follows:

$P(A) = P(A|X) \cdot P(X) + P(A|Y) \cdot P(Y) + P(A|Z) \cdot P(Z)$

Substituting the values:

$P(A) = \left(\frac{5}{9} \cdot \frac{1}{3}\right) + \left(\frac{4}{9} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right)$

Calculating the above, we get:

$P(A) = \frac{5}{27} + \frac{4}{27} + \frac{1}{9}$

Note that $\frac{1}{9} = \frac{3}{27}$ , so:

$P(A) = \frac{5}{27} + \frac{4}{27} + \frac{3}{27} = \frac{12}{27} = \frac{4}{9}$

Now, substituting back into Bayes' theorem:

$P(Y|A) = \frac{\left(\frac{4}{9}\right) \cdot \left(\frac{1}{3}\right)}{\frac{4}{9}} = \frac{4}{9} \cdot \frac{1}{3} \cdot \frac{9}{4} = \frac{1}{3}$

Hence, the probability that the coin came from bag $Y$ is:

Option B: $\frac{1}{3}$