From a bag containing 4 white and 5 red balls, if 3 balls are drawn at random, then the mean of the number of red balls among the balls drawn, is

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Aicharya · Accepted Answer

Let the random variable is $X$, then $\begin{array}{ccccc} \hline \boldsymbol{x} & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} \ \hline P(X) & \frac{{ }^4 C_3}{{ }^9 C_3} & \frac{{ }^4 C_2 \times{ }^5 C_1}{{ }^9 C_3} & \frac{{ }^4 C_1 \times{ }^5 C_2}{{ }^9 C_3} & \frac{{ }^4 C_0 \times{ }^5 C_3}{{ }^9 C_3} \end{array}$ $\begin{aligned} \therefore \quad \text { Mean }= & 0 \times\left(\frac{{ }^4 C_3}{{ }^9 C_3}\right)+1 \times\left(\frac{{ }^4 C_2 \times{ }^5 C_1}{{ }^9 C_3}\right) \ & +2 \times\left(\frac{{ }^4 C_1 \times{ }^5 C_2}{{ }^9 C_3}\right)+3 \times\left(\frac{{ }^4 C_0 \times{ }^5 C_3}{{ }^9 C_3}\right) \ & =\frac{(6 \times 5)+(2 \times 4 \times 10)+(3 \times 10)}{\frac{9 \times 8 \times 7}{3 \times 2}} \ = & \frac{30+80+30}{84}=\frac{140}{84}=\frac{5}{3} \end{aligned}$ Hence, option (a) is correct.

From a bag containing 4 white and 5 red balls, if 3 balls are drawn at random, then the mean of the number…

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