In a game, a person wins 5 rupees for getting a number greater than 4 and loses 1 rupee otherwise, when a…
In a game, a person wins 5 rupees for getting a number greater than 4 and loses 1 rupee otherwise, when a fair die is thrown. A man participated in the game, but decided to quit as and when he gets a number greater than 4. Then the expected value (mean value) of the amount he wins/loses is
\(\frac{9}{19}\)
\(\frac{8}{19}\)
\(\frac{19}{9}\)
\(\frac{19}{8}\)
Solution
In a game die is thrown. A man participated in the game, he decided to throw a die thrice but to quit as and when the amount he win/loses is
The probability to win in a throw is \(\frac{2}{6}\) and loss a throw is \(\frac{4}{6}\).
Now following cases are possible
(i) \(\mathrm{W}\)
(ii) \(\mathrm{L} \mathrm{W}\)
(iii) L L W
(iv) L L L
So, the expected value (mean value) of the amount he wins/loses is
\(\begin{aligned}
& =5\left(\frac{2}{6}\right)+4\left(\frac{4}{6} \times \frac{2}{6}\right)+3\left(\frac{4}{6} \times \frac{4}{6} \times \frac{2}{6}\right) +(-3)\left(\frac{4}{6} \times \frac{4}{6} \times \frac{4}{6}\right) \\
& =\frac{10}{6}+\frac{32}{36}+\frac{96}{216}-\frac{192}{216} \\
& =\frac{360+192+96-192}{216}=\frac{456}{216}=\frac{19}{9}
\end{aligned}\)