Probability 5 Question 5
5. In a game, a man wins $₹ 100$ if he gets 5 or 6 on a throw of a fair die and loses ₹ 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is
(2019 Main, 12 Jan II)
(a) $\frac{400}{3}$ loss
(b) $\frac{400}{9} \operatorname{loss}$
(c) 0
(d) $\frac{400}{3}$ gain
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Answer:
Correct Answer: 5. (c)
Solution:
- Let $p$ and $q$ represents the probability of success and failure in a trial respectively. Then,
$$ p=P(5 \text { or } 6)=\frac{2}{6}=\frac{1}{3} \text { and } q=1-p=\frac{4}{6}=\frac{2}{3} \text {. } $$
Now, as the man decides to throw the die either till he gets a five or a six or to a maximum of three throws, so he can get the success in first, second and third throw or not get the success in any of the three throws.
So, the expected gain/loss (in ₹)
$$ \begin{aligned} & =(p \times 100)+q p(-50+100) \\ & \quad+q^{2} p(-50-50+100)+q^{3}(-50-50-50) \\ & =\frac{1}{3} \times 100+\frac{2}{3} \times \frac{1}{3}(50)+\frac{2}{3}^{2} \frac{1}{3}(0)+\frac{2}{3}^{3}(-150) \end{aligned} $$
$=\frac{100}{3}+\frac{100}{9}+0-\frac{1200}{27}$
$=\frac{900+300-1200}{27}=\frac{1200-1200}{27}=0$
