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Home  >>  CBSE XII  >>  Math  >>  Probability
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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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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= \large\frac{1}{6}$$ \rightarrow q = 1 - p = $$\large\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 $\large\frac{1}{6}$
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 $\large\frac{5}{6}$$ \times\large \frac{1}{6} = \frac{5}{36}$
3. Does not get a six on the first two throws, but on the third throw.
The required probability is $\large\frac{5}{6}$$\times \large\frac{5}{6}$ $\times\large\frac{1}{6} = \frac{25}{216}$
The amount he will receive/lose is Rs. ( - 1 -1 + 1) = -Rs. 1.
Expected value of the amount =
\(\;E(X)=1 \times \large \frac{1}{6}\)\(+\;0\; \times \large\frac{5}{36}\)\(+(-1)\times\large\frac{25}{216}\;\)
=\(\large\;\frac{36+0-25}{216}=\;\frac{11}{54}\)

 

answered Jun 22, 2013 by balaji.thirumalai
 

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