Q.A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Concept: Conditional Probability (Bayes' Theorem)

Let $E_1$ = choosing first bag, $E_2$ = choosing second bag, and $A$ = drawing a red ball.

Step 1: Prior probabilities are equal:

$P(E_1) = P(E_2) = \frac12$

Step 2: Probability of drawing a red ball from each bag:

$P(A|E_1) = \frac{4}{8} = \frac12$

$P(A|E_2) = \frac{2}{8} = \frac14$

Step 3: By Bayes' Theorem:

$P(E_1|A) = \frac{P(E_1) \cdot P(A|E_1)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$

$= \frac{\frac12 \cdot \frac12}{\frac12 \cdot \frac12 + \frac12 \cdot \frac14} = \frac{\frac14}{\frac14 + \frac18} = \frac{\frac14}{\frac38} = \frac14 \times \frac83 = \frac{2}{3}$

We have two bags, each with a different mix of red and black balls. A bag is chosen at random, then a ball is drawn and turns out to be red. The question asks: given that we saw a red ball, what is the chance it came from the first bag?

This is a classic case of inverse probability — we know the outcome (red ball) and want to trace back to which bag it likely came from. The tool for this is Bayes’ theorem, which flips conditional probabilities using the law of total probability.

Let’s define the events clearly:

• $B_1$: first bag is chosen
• $B_2$: second bag is chosen
• $R$: a red ball is drawn

We are asked for $P(B_1 \mid R)$.

1. Prior probabilities Since the bag is chosen at random, each bag is equally likely:

2. Likelihoods — probability of drawing a red from each bag

   • First bag: 4 red out of 8 total balls → $P(R \mid B_1) = \frac{4}{8} = \frac{1}{2}$
   • Second bag: 2 red out of 8 total balls → $P(R \mid B_2) = \frac{2}{8} = \frac{1}{4}$

3. Total probability of drawing a red ball

   By the law of total probability:

4. Apply Bayes’ theorem

Method: Bayes' Theorem — reversing a conditional probability

Reach for Bayes' theorem whenever you are told an outcome (a red ball, a positive test, a defect) and asked which underlying cause or category produced it — the "given the effect, find the cause" pattern.

Steps

Step 1: Name the competing causes (a partition) and their priors.

List the mutually exclusive, exhaustive hypotheses $H_1, H_2, \dots$ — which bag/box/machine was chosen — with prior probabilities $P(H_i)$ (equal, if the choice is "at random").

Step 2: Write the likelihood of the observed evidence under each cause.

For each hypothesis find $P(E \mid H_i)$: the chance of the observed evidence $E$ if that cause were true (e.g. the fraction of red balls in that bag).

Step 3: Get the total probability of the evidence.

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

The numerator is just one branch of the denominator's sum — so the posteriors across all causes always add to $1$, a handy self-check.

Common Mistakes

Mistake 1: Answering $P(\text{red}\mid \text{bag 1})$ instead of $P(\text{bag 1}\mid \text{red})$.

Why it's wrong: the question asks the reverse conditional — which bag, given a red ball. Correct approach: invert with Bayes' theorem, $P(B_1\mid R)=\dfrac{P(B_1)P(R\mid B_1)}{P(R)}$.

Mistake 2: Forgetting the second bag in the denominator.

Why it's wrong: $P(R)$ must count red draws from both bags, not just bag 1. Correct approach: $P(R)=\tfrac12\cdot\tfrac12+\tfrac12\cdot\tfrac14=\tfrac38$.

Mistake 3: Miscounting a bag's total.

Why it's wrong: each bag holds $8$ balls, so $P(R\mid B_1)=\tfrac48=\tfrac12$ and $P(R\mid B_2)=\tfrac28=\tfrac14$; using $4$ or $6$ as the total corrupts every later step.