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Q)

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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A)
Given 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.
P(getting a six)$p = \frac{1}{6} \to q = 1 - p = \frac{5}{6}$
There are three possibilities here:
1. He gets a six in the first throw and he will receive Rs. 1. The required probability is $\frac{1}{6}$
2. He gets some other number on the first throw and throws again and gets a six on the second throw.
Since he loses a Rupee for not getting a six on the first throw, but wins back the Rupee on the second throw, his net gain/loss = 0.
Probability of this occuring is $\frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$
3. Does not get a six on the first two throws, but on the third throw.
The required probability is $\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{25}{216}$
The amount he will receive/lose is Rs. ( - 1 -1 + 1) = -Rs. 1.
Expected value of the amount =
\begin{align*}E(X)& = 1 \times \frac{1}{6} + 0 \times \frac{5}{36} + (-1) \times \frac{25}{216} \\ & = \frac{36 + 0 - 25}{216} \\ & = \frac{11}{54} \end{align*}