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.

Question:

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.

Solution:

When a die is thrown, then probability of getting a six $=\frac{1}{6}$ then, probability of not getting a six $=1-\frac{1}{6}=\frac{5}{6}$

If the man gets a six in the first throw, then

probability of getting a six $=\frac{1}{6}$

If he does not get a six in first throw, but gets a six in second throw, then

probability of getting a six in the second throw $=\frac{5}{6} \times \frac{1}{6}=\frac{5}{36}$

If he does not get a six in the first two throws, but gets in the third throw, then

probability of getting a six in the third throw $=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}=\frac{25}{216}$

probability that he does not get a six in any of the three throws $=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{125}{216}$

In the first throw he gets a six, then he will receive Re 1.

If he gets a six in the second throw, then he will receive Re (1 - 1) = 0

If he gets a six in the third throw, then he will receive Rs(-1 - 1 + 1) = Rs (-1), 
that means he will lose Re 1 in this case.

Expected value $=\frac{1}{6} \times 1+\left(\frac{5}{6} \times \frac{1}{6}\right) \times 0+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right) \times(-1)=\frac{11}{216}$

So, he will loose Rs $\frac{11}{216}$.

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