Q.Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.

Concept: Conditional Probability (Bayes' Theorem)

Let $E_1$ = choosing Bag I, $E_2$ = choosing Bag II, and $A$ = drawing a red ball.

• $P(E_1) = P(E_2) = \frac{1}{2}$
• $P(A|E_1) = \frac{3}{7}$, $P(A|E_2) = \frac{5}{11}$

We need $P(E_2|A)$.

Step 1: Apply Bayes' theorem:

Step 2: Substitute values:

Step 3: Simplify denominator:

Thus,

The problem asks: given that a red ball was drawn, what is the chance it came from Bag II? This is a classic case of inverse probability — we know the outcome (red ball) and want the probability of a particular cause (Bag II). The natural tool is Bayes’ theorem, which reverses the conditional probability.

Why Bayes’ theorem works here:

We have two mutually exclusive and exhaustive events (choosing Bag I or Bag II), each with a prior probability of $1/2$ (since the bag is chosen at random). For each bag, we know the probability of drawing a red ball (the “likelihood”). Bayes’ theorem combines these to give the posterior probability — the probability of the cause given the observed effect.

Let’s define the events clearly:

• $B_1$: the ball is drawn from Bag I
• $B_2$: the ball is drawn from Bag II
• $R$: the ball drawn is red

We need $P(B_2 \mid R)$.

1. Write down the prior probabilities. Since the bag is chosen at random,

2. Find the likelihoods — probability of drawing a red ball from each bag.

   • Bag I: 3 red + 4 black = 7 balls total. So $P(R \mid B_1) = \frac{3}{7}$.
   • Bag II: 5 red + 6 black = 11 balls total. So $P(R \mid B_2) = \frac{5}{11}$.

3. Compute the total probability of drawing a red ball (the denominator in Bayes’ theorem).

   By the law of total probability:

Find a common denominator (154):

So

4. Apply Bayes’ theorem.

Simplify step by step:

Numerator: $\frac{5}{11} \cdot \frac{1}{2} = \frac{5}{22}$.

So

Cancel common factors: $77 = 7 \times 11$, $22 = 2 \times 11$. The $11$ cancels:

5. Interpret the result. The probability $\frac{35}{68} \approx 0.5147$ is slightly more than $1/2$. This makes sense: Bag II has a higher proportion of red balls ($5/11 \approx 0.455$) than Bag I ($3/7 \approx 0.429$), so observing a red ball shifts the probability slightly in favour of Bag II.

Method: Bayes' Theorem (finding the cause from the observed outcome)

Use this whenever you observe an outcome and are asked which source most likely produced it — the conditioning is reversed from what the problem hands you (you know the chance of a red ball given each bag, but want the chance of a bag given a red ball).

Steps

Step 1: Set up the partition of possible causes.

List the mutually exclusive, exhaustive hypotheses (here: the ball came from Bag I or Bag II) and write their prior probabilities $P(E_i)$ (these must sum to $1$).

Step 2: Write each likelihood.

For each cause, state $P(A\mid E_i)$ — the probability of the observed outcome (here: drawing a red ball) under that cause.

Step 3: Get the total probability of the outcome (the denominator).

By the law of total probability,

Step 4: Apply Bayes' theorem for the cause you want.

The numerator is just the one term of the denominator that belongs to the cause you are asked about — so the answer is that cause's share of the total chance of the outcome.

Key subtlety for bag/urn problems: the likelihood of the observed colour from each source is computed against that source's own total (Bag I has $3+4=7$ balls, Bag II has $5+6=11$), so do not compare raw colour counts as if the totals were equal.

Common Mistakes

Mistake 1: Reporting $P(\text{red}\mid\text{Bag II})=\tfrac{5}{11}$ as the answer.

Why it's wrong: that is the likelihood (red given Bag II), not the required posterior $P(\text{Bag II}\mid\text{red})$. Correct approach: apply Bayes' theorem to reverse the conditioning.

Mistake 2: Leaving Bag I out of the denominator.

Why it's wrong: the total probability of drawing a red ball must include contributions from both bags. Correct approach: denominator $=P(B_1)P(R\mid B_1)+P(B_2)P(R\mid B_2)$.

Mistake 3: Comparing raw red-ball counts (5 vs 3).

Why it's wrong: the bags have different totals (7 vs 11), so counts are not comparable. Correct approach: use probabilities $\tfrac{3}{7}$ and $\tfrac{5}{11}$, each over its own bag's total.