AP EAMCET · Maths · Probability
Three boxes \(\mathrm{B}_1, \mathrm{~B}_2\) and \(\mathrm{B}_3\) contain balls with different colors as follows:
\begin{array}{cccc}
& White & Black & Red \\
\mathrm{B}_1 & 2 & 1 & 2 \\
\mathrm{~B}_2 & 3 & 2 & 4 \\
\mathrm{~B}_3 & 4 & 3 & 2
\end{array}
A die is thrown. Box \(B_1\) is chosen if either 1 or 2 turns up. Box \(B_2\) is chosen if 3 or 4 turns up and box \(B_3\) is chosen if 5 or 6 turns up. Having chosen a box in this way, a ball is drawn at random from that box. If the ball drawn is found to be Red, then the probability that it is drawn from box \(\mathrm{B}_2\) is
- A \(\frac{7}{12}\)
- B \(\frac{5}{12}\)
- C \(\frac{1}{12}\)
- D \(\frac{3}{26}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{12}\)
Step-by-step Solution
Detailed explanation
If 1 or 2 turns up, box \(B_1\) is selected Hence, \(P\left(B_1\right)=\frac{2}{6}\) Similary, \(\mathrm{P}\left(\mathrm{B}_2\right)=\frac{2}{6}\) and \(\mathrm{P}\left(\mathrm{B}_3\right)=\frac{2}{6}\) For box \(\mathrm{B}_1\) : Total number of balls \(=5\) No. of red balls…
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