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: Understanding the Concept
Box \(B_k\) has \(k\) red and \(k+1\) white balls, so \(P(R|B_k)=\frac{k}{2k+1}\): \(\frac13,\frac25,\frac37\).

Step 2: Total probability of red
\(P(R)=\frac12\cdot\frac13+\frac13\cdot\frac25+\frac16\cdot\frac37=\frac{35+28+15}{210}=\frac{78}{210}\).

Step 3: Bayes' theorem
\(P(B_2|R)=\frac{\frac13\cdot\frac25}{P(R)}=\frac{28/210}{78/210}=\frac{28}{78}\).

Step 4: Simplify
\(\frac{28}{78}=\frac{14}{39}\).

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