Question

For \(\mathrm{k}=1,2,3\) the box \(\mathrm{B}_{\mathrm{k}}\) contains k red balls and \((k+1)\) white balls. Let \(P\left(B_1\right)=\frac{1}{2}, P\left(B_2\right)=\frac{1}{3}\) and \(\mathrm{P}\left(\mathrm{B}_3\right)=\frac{1}{6} . \mathrm{A}\) box is selected at random and a ball is drawn from it. If a red ball is drawn from it, then the probability that it comes from box \(\mathrm{B}_2\) is

• A. \(\frac{35}{78}\)
• B. \(\frac{14}{39}\)
• C. \(\frac{10}{13}\)
• D. \(\frac{12}{13}\)

Hint:

Apply Bayes' theorem using P(red given Bk) = k/(2k+1).

Answer 1

The Correct Option is B

Step 1: Joint probabilities
$P(B_1\cap R)=\frac16$, $P(B_2\cap R)=\frac{2}{15}$, $P(B_3\cap R)=\frac1{14}$.

Step 2: Common denominator 210
These are $\frac{35}{210},\frac{28}{210},\frac{15}{210}$.

Step 3: Ratio
$P(B_2|R)=\frac{28}{35+28+15}=\frac{28}{78}=\frac{14}{39}$.

Final Answer:
The required probability is $\frac{14}{39}$. \[ \boxed{\text{(B) }\dfrac{14}{39}} \]