JEE Main 2024 — Probability Question with Solution

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