Q.Bag I contains $3$ black and $2$ white balls, Bag II contains $2$ black and $4$ white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball.

Concept: Conditional Probability (Law of Total Probability)

Step 1 – Define events

Let $B_1$ = selecting Bag I, $B_2$ = selecting Bag II.

$P(B_1) = P(B_2) = \frac12$.

Let $A$ = event of drawing a black ball.

Step 2 – Find conditional probabilities

Bag I: 3 black, 2 white → $P(A \mid B_1) = \frac{3}{5}$.

Bag II: 2 black, 4 white → $P(A \mid B_2) = \frac{2}{6} = \frac13$.

Step 3 – Apply Law of Total Probability …

The problem asks: you pick one of the two bags at random (each equally likely), then from that bag you pick one ball at random. What is the probability that the ball you end up with is black?

This is a classic setup for conditional probability and the law of total probability. The key idea: the event "black ball" can happen in two mutually exclusive ways — either you picked Bag I and then a black ball from it, OR you picked Bag II and then a black ball from it. Since the bag is chosen first, the probability of drawing black depends on which bag you're in. The law of total probability tells us to average the conditional probabilities, weighted by the probability of each condition.

Let's work it through.

1. Define the events clearly. Let $B_1$ be the event that Bag I is selected, and $B_2$ the event that Bag II is selected. Since a bag is chosen at random, each is equally likely:

Let $E$ be the event that the selected ball is black.

2. Find the conditional probabilities. In Bag I: 3 black, 2 white → total 5 balls. So

In Bag II: 2 black, 4 white → total 6 balls. So

3. Apply the law of total probability. The law states:

Substitute the numbers:

4. Compute. First term: $\frac{3}{5} \times \frac{1}{2} = \frac{3}{10}$. Second term: $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$. Add them: …

Method: Law of Total Probability (choose a source, then draw)

Use this two-stage template whenever you first pick one of several containers (each with its own chance) and then draw from the chosen one.

Steps

Step 1: Name the first-stage events and their probabilities.

Let the containers be $B_1, B_2, \dots$ with selection probabilities $P(B_i)$ (here each bag is equally likely, $P(B_i)=\tfrac12$).

Step 2: Find the conditional probability of the target inside each container. …

Common Mistakes

Mistake 1: Averaging the two conditional probabilities without weighting.

Why it's wrong: writing $\tfrac12\left(\tfrac35+\tfrac13\right)$ only works here because the bags are equally likely; if the bag-selection probabilities differed, this would give the wrong answer. Correct approach: always apply $P(A)=\sum P(B_i)P(A\mid B_i)$ with each bag's own weight.

Mistake 2: Using a common denominator across both bags (e.g. treating it as $5$ black out of $11$). …