MHT CET · Maths · Probability
Three boxes contain respectively 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, from each of the boxes one ball is drawn at random. The probability that 2 white and 1 black balls will be drawn, is
- A \(13 / 12\)
- B \(1 / 4\)
- C \(1 / 32\)
- D \(3 / 16\)
Answer & Solution
Correct Answer
(A) \(13 / 12\)
Step-by-step Solution
Detailed explanation
Given, Box \((\mathrm{I})=3 W, 1 B ;\) Box \((\mathrm{II})=2 W, 2 B ;\) Box
\((\mathrm{III})=1 W, 3 B .\)
\(\therefore\) Required probability
\(\begin{array}{l}=P(B W W)+P(W B W)+P(W W B) \\=P(B) P(W) P(W)+P(W) P(B) P(W) \\\quad+P(W) P(W) P(B) \\
=\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4}+\frac{3}{4} \times \frac{2}{4} \times \frac{1}{4}+\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4} \\=\frac{2+6+18}{64}=\frac{26}{64} \\=\frac{13}{32}\end{array}\)
\((\mathrm{III})=1 W, 3 B .\)
\(\therefore\) Required probability
\(\begin{array}{l}=P(B W W)+P(W B W)+P(W W B) \\=P(B) P(W) P(W)+P(W) P(B) P(W) \\\quad+P(W) P(W) P(B) \\
=\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4}+\frac{3}{4} \times \frac{2}{4} \times \frac{1}{4}+\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4} \\=\frac{2+6+18}{64}=\frac{26}{64} \\=\frac{13}{32}\end{array}\)
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