MHT CET · Maths · Probability
For \(\mathrm{k}=1,2,3\) the box \(\mathrm{B}_{\mathrm{k}}\) contains k red balls and \((\mathrm{k}+1)\) white balls. Let \(\mathrm{P}\left(\mathrm{B}_1\right)=\frac{1}{2}, \mathrm{P}\left(\mathrm{B}_2\right)=\frac{1}{3}\) and \(\mathrm{P}\left(\mathrm{B}_3\right)=\frac{1}{6}\). 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 \(B_2\) is
- A \(\frac{35}{78}\)
- B \(\frac{14}{39}\)
- C \(\frac{10}{13}\)
- D \(\frac{12}{13}\)
Answer & Solution
Correct Answer
(B) \(\frac{14}{39}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{P}\left(\mathrm{R} \mid \mathrm{B}_{\mathrm{k}}\right)=\frac{\mathrm{k}}{2 \mathrm{k}+1}\) \(\mathrm{P}\left(\mathrm{R} \mid \mathrm{B}_1\right)=\frac{1}{3}, \mathrm{P}\left(\mathrm{R} \mid \mathrm{B}_2\right)=\frac{2}{5}, \mathrm{P}\left(\mathrm{R} \mid \mathrm{B}_3\right)=\frac{3}{7}\)
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