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In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/ loses.
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When a die is thrown, probability of getting a six $=\frac{1}{6}$ $=\mathrm{p}$ (say) $\therefore \mathrm{q}=1-\frac{1}{6}=\frac{5}{6}$
(i) If he gets a six in first throw then, Probability of getting a six $=\frac{1}{6}$
(ii) If he does not get a six in first throw but he gets a six in second throw. Its probability $=\frac{5}{6} \times \frac{1}{6}=\frac{5}{36}$ Probability that he does not get a six in any of the three throws $=\left(\frac{5}{6}\right)^3=\frac{125}{216}$ In first throw he gets a six, he will received $₹ 1$ If the gets a six in second throw, he will received $₹(1-1)=0$
If he gets a six in third throw, he will lose ₹ 1
Expected value
$$
\begin{aligned}
&=\frac{1}{6} \times 1+\left(\frac{5}{6}+\frac{1}{6}\right) \times 0+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right) \times(-1) \\
&=\frac{1}{6}-\frac{25}{216}=\frac{36-25}{216}=\frac{11}{216}
\end{aligned}
$$
Expected value is, he will loose $₹ \frac{11}{216}$
(i) If he gets a six in first throw then, Probability of getting a six $=\frac{1}{6}$
(ii) If he does not get a six in first throw but he gets a six in second throw. Its probability $=\frac{5}{6} \times \frac{1}{6}=\frac{5}{36}$ Probability that he does not get a six in any of the three throws $=\left(\frac{5}{6}\right)^3=\frac{125}{216}$ In first throw he gets a six, he will received $₹ 1$ If the gets a six in second throw, he will received $₹(1-1)=0$
If he gets a six in third throw, he will lose ₹ 1
Expected value
$$
\begin{aligned}
&=\frac{1}{6} \times 1+\left(\frac{5}{6}+\frac{1}{6}\right) \times 0+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right) \times(-1) \\
&=\frac{1}{6}-\frac{25}{216}=\frac{36-25}{216}=\frac{11}{216}
\end{aligned}
$$
Expected value is, he will loose $₹ \frac{11}{216}$
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