Bag I contains 2 red and 3 blue balls and bag II contains \( \alpha \) red and 5 blue balls. One ball is drawn at random from one of the bags and is found to be blue. If the probability that it was drawn from bag II is \( \displaystyle\frac{25}{52} \), then \( \alpha \) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
The bag is chosen at random, so each bag has probability \( \frac{1}{2} \). Let \( E_1,E_2 \) be the events that bag I, bag II is chosen, and B that the ball is blue.
\( \displaystyle P(B|E_1)=\frac{3}{5} \)
\( \displaystyle P(B|E_2)=\frac{5}{\alpha+5} \)
By Bayes’ theorem,
\( \displaystyle P(E_2|B)=\frac{\frac{1}{2}\cdot\frac{5}{\alpha+5}}{\frac{1}{2}\cdot\frac{3}{5}+\frac{1}{2}\cdot\frac{5}{\alpha+5}} \)
\( \displaystyle =\frac{25}{52} \)
Writing \( p=\frac{5}{\alpha+5} \) and cancelling \( \frac{1}{2} \):
\( \displaystyle \frac{p}{\frac{3}{5}+p}=\frac{25}{52} \)
\( 52p=15+25p \)
\( 27p=15 \)
\( \displaystyle p=\frac{5}{9} \)
\( \displaystyle \frac{5}{\alpha+5}=\frac{5}{9} \)
\( \alpha=4 \)
Shortcut: Ratio method: with equal priors, \( P(E_2|B):P(E_1|B)=25:27 \) must equal \( \frac{5}{\alpha+5}:\frac{3}{5} \). Try the options: \( \alpha=4 \) gives \( \frac{5}{9}:\frac{3}{5}=25:27 \).