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  • If one ball is drawn at random from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls then the probability that 2 white and 1 black balls will b

If one ball is drawn at random from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls then the probability that 2 white and 1 black balls will be drawn is

Option: 1

\frac{13}{32}


Option: 2

\frac{1}{4}


Option: 3

\frac{1}{32}


Option: 4

\frac{3}{16}


Answers (1)

best_answer

Let \mathrm{E_{1}=} the event of drawing a white ball from the first box.

Similarly, \mathrm{E_{2}\; and\; E_{3}}.

Here , \mathrm{P\left(E_1\right)=\frac{3}{4}, \quad P\left(E_2\right)=\frac{1}{2}, \quad P\left(E_3\right)=\frac{1}{4} .}

The required probability \mathrm{=P\left(E_1 E_2 \bar{E}_3\right)+P\left(E_1 \bar{E}_2 E_3\right)+P\left(\bar{E}_1 E_2 E_3\right)}

              \begin{aligned} & =\mathrm{P\left(E_1\right) \cdot P\left(E_2\right) \cdot P\left(\bar{E}_3\right)+P\left(E_1\right) \cdot P\left(\bar{E}_2\right) \cdot P\left(E_3\right)+P\left(\bar{E}_1\right) \cdot P\left(E_2\right) \cdot P\left(E_3\right)} \\ \\& =\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{3}{4}+\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{1}{4}+\frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{4}=\frac{13}{32} . \end{aligned}

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shivangi.shekhar

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