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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 $$=(2\times20+10)\frac{7}{22}+(20+2\times10)\frac{21}{44}=Rs. 35$$

My answer is not matching with any option. Can somebody please help me where I am wrong? Thanks.

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    $\begingroup$ 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. $\endgroup$ Dec 18, 2017 at 11:41

1 Answer 1

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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: $$\frac{\binom{7}{3}}{\binom{12}{3}}= \frac{35}{220}=\frac{7}{44}$$

So, you missed up a total of: ??

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  • $\begingroup$ answer comes out to $466.5/11=42.41$ So the answer must be 42.5 am I right? $\endgroup$ Dec 18, 2017 at 11:51
  • $\begingroup$ Yes you are right $\endgroup$ Dec 18, 2017 at 11:52
  • $\begingroup$ thank you for helping me $\endgroup$ Dec 18, 2017 at 11:53
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    $\begingroup$ @jeanneclement The answer is $35+\frac{1}{22}[60]+\frac{7}{44}[30]=42.5$ $\endgroup$
    – user371838
    Dec 18, 2017 at 11:58
  • $\begingroup$ thank you. your answer is correct $\endgroup$ Dec 18, 2017 at 12:08

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