Bag A contains 3 red and 4 black balls, and Bag B contains 5 red and \'n\' black balls. One ball is drawn at random from one of the bags and it is found to be red. If the probability that it was from bag B is 35/68, then the value of \'n\' is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 6
Each bag is equally likely to be chosen, so \( P(A)=P(B)=\frac{1}{2} \). Let R be the event that the ball is red:
\( \displaystyle P(R|A)=\frac{3}{7} \)
\( \displaystyle P(R|B)=\frac{5}{5+n} \)
By Bayes’ theorem (the factors \( \frac{1}{2} \) cancel):
\( \displaystyle P(B|R)=\frac{\frac{5}{5+n}}{\frac{3}{7}+\frac{5}{5+n}} \)
\( \displaystyle =\frac{35}{68} \)
Write \( u=\frac{5}{5+n} \):
\( \displaystyle 68u=35\left(\frac{3}{7}+u\right) \)
\( \displaystyle 68u=15+35u \)
\( \displaystyle 33u=15\ \Rightarrow\ u=\frac{5}{11} \)
\( \displaystyle \frac{5}{5+n}=\frac{5}{11}\ \Rightarrow\ 5+n=11 \)
\( \displaystyle n=6 \)
Shortcut: Use the ratio form of Bayes’ theorem. With equal priors, \( \frac{P(B|R)}{P(A|R)}=\frac{P(R|B)}{P(R|A)} \), and \( P(A|R)=1-\frac{35}{68}=\frac{33}{68} \):
\( \displaystyle \frac{5/(5+n)}{3/7}=\frac{35}{33} \)
\( \displaystyle \frac{5}{5+n}=\frac{35}{33}\times\frac{3}{7}=\frac{5}{11} \)
\( \displaystyle n=6 \)
NCERT (2026-27 edition):
Bayes' Theorem
