Bag I contains 3 red and 4 black balls while 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. then the probability that it was drawn from Bag I is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{33}{68} \)
This is a Bayes' theorem question, and each bag is chosen with probability \( \frac{1}{2} \). Let \( E_1,E_2 \) be "Bag I, Bag II is chosen" and A be "the ball is red".
\( \displaystyle P(A|E_1)=\frac{3}{7} \)
\( \displaystyle P(A|E_2)=\frac{5}{11} \)
\( \displaystyle P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)} \)
\( \displaystyle =\frac{\frac{1}{2}\cdot\frac{3}{7}}{\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{2}\cdot\frac{5}{11}} \)
\( \displaystyle =\frac{\frac{3}{7}}{\frac{3}{7}+\frac{5}{11}} \)
\( \displaystyle =\frac{\frac{33}{77}}{\frac{33}{77}+\frac{35}{77}} \)
\( \displaystyle =\frac{33}{68} \)
Shortcut: With equal priors, the posterior probabilities are in the ratio of the likelihoods: \( \frac{3}{7}:\frac{5}{11}=33:35 \), so \( P(E_1|A)=\frac{33}{33+35}=\frac{33}{68} \).
Why the other options are incorrect
Option (1) \( \displaystyle\frac{34}{77} \) — Incorrect. \( \frac{34}{77}=\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{2}\cdot\frac{5}{11} \) is P(A), the total probability of drawing a red ball, which is only the denominator of Bayes' formula.
Option (2) \( \displaystyle\frac{3}{14} \) — Incorrect. \( \frac{3}{14}=\frac{1}{2}\cdot\frac{3}{7} \) is \( P(E_1\cap A) \), the numerator only; it has not been divided by P(A).
Option (4) \( \displaystyle\frac{35}{68} \) — Incorrect. \( \frac{35}{68} \) is the probability that the red ball came from Bag II.