Probability

Concept:The expected amount is the sum of (probability of each outcome × amount received).Explanation:Total balls = 10 (6 white, 4 red). Two balls are drawn without replacement.Let $X$ be the number of white balls drawn. Then red balls = $2 - X$.Amount received = $10X + 20(2 - X) = 40 - 10X$.Expected amount = $E[40 - 10X] = 40 - 10E[X]$.$X$ follows hypergeometric distribution: $E[X] = n \cdot \frac{N_w}{N} = 2 \cdot \frac{6}{10} = 1.2$.Thus expected amount = $40 - 10 \times 1.2 = 40 - 12 = 28$.Alternatively, compute directly:• $P(0\text{ white}, 2\text{ red}) = \frac{\binom{6}{0}\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45}$, amount = ₹40.• $P(1\text{ white}, 1\text{ red}) = \frac{\binom{6}{1}\binom{4}{1}}{\binom{10}{2}} = \frac{24}{45}$, amount = ₹30.• $P(2\text{ white}, 0\text{ red}) = \frac{\binom{6}{2}\binom{4}{0}}{\binom{10}{2}} = \frac{15}{45}$, amount = ₹20.Expected amount = $\frac{6}{45}\times 40 + \frac{24}{45}\times 30 + \frac{15}{45}\times 20 = \frac{1260}{45} = 28$.Answer:₹28