# What is his expectation?

A man draws 3 balls from a jug containing 5 white balls & 7 black balls. He gets Rs. 20 for each white ball & Rs. 10 for each black ball. What is his expectation?

$$\\$$ a) $$Rs. 21.25\ \quad$$ b) $$Rs. 42.50 \quad$$ c) $$Rs. 31.25\ \quad$$ d) $$Rs. 45.21\$$

My attempt:

Probability of drawing 2 white balls & 1 black ball out of 12 balls $$=\frac{\binom{5}{2}\cdot \binom{7}{1}}{\binom{12}{3}}=\frac{7}{22}$$

Probability of drawing 1 white ball & 2 black balls out of 12 balls $$=\frac{\binom{5}{1}\cdot \binom{7}{2}}{\binom{12}{3}}=\frac{21}{44}$$ as there are two different ways of drawing three balls hence the total expectation $$=(2\times20+10)\frac{7}{22}+(20+2\times10)\frac{21}{44}=Rs. 35$$

• You have not taken into account the possibilities that the man draws three black balls and no white balls or three white balls and no black balls. Dec 18, 2017 at 11:41

You missed the options of taking three white balls with probability: $$\frac{\binom{5}{3}}{\binom{12}{3}}= \frac{10}{220}=\frac{1}{22}$$ and taking three black balls with probability: $$\frac{\binom{7}{3}}{\binom{12}{3}}= \frac{35}{220}=\frac{7}{44}$$
• answer comes out to $466.5/11=42.41$ So the answer must be 42.5 am I right? Dec 18, 2017 at 11:51
• @jeanneclement The answer is $35+\frac{1}{22}[60]+\frac{7}{44}[30]=42.5$