Question

In a game, a man wins Rs. $1000$ if he gets an even number $\ge 4$ on a fair die and loses Rs. $200$ for getting any other number on the die. If he decides to throw the die until he wins or maximum of three times, then his expected gain/loss (in Rupees) is

• A. $\frac{3800}{9}\text{loss}$
• B. $\frac{3800}{9}\text{gain}$
• C. $0$
• D. $400\text{gain}$

Answer 1

The correct option is B $\frac{3800}{9}\text{gain}$

Even number on a die $\ge 4$ is $4,6$
Let $A$ be the event that either $4$ or $6$ appears on the die
and $B$ be the event that a number other than $4,6$ appears on the die.
Then $P\left(A\right)=\frac{1}{3}$ and $P\left(B\right)=\frac{2}{3}$

Possible sequence of events are
$A,BA,BBA,BBB$

Let $X$ be a random variable denoting the winning amount.
$\begin{array}{ccccc}X& +1000& +800& +600& -600\\ P\left(X\right)& \frac{1}{3}& \frac{2}{9}& \frac{4}{27}& \frac{8}{27}\end{array}$

$\therefore E\left(X\right)=1000\times \frac{1}{3}+800\times \frac{2}{9}+600\times \frac{4}{27}-600\times \frac{8}{27}=\frac{3800}{9}$