In a game, a person wins $10$ points if she gets all heads or all tails and loses $5$ points if she gets $1$ or $2$ heads when $3$ coins are tossed once. Find her expectation of number of points.
Let $X$ be the random variable denoting the number of points a person can win.
Then $X$ takes values $10, \; -5$.
P(getting $1$ head when $3$ coins are tossed) $= \dfrac{3}{8}$
P(getting $2$ heads when $3$ coins are tossed) $= \dfrac{3}{8}$
P(getting $3$ heads when $3$ coins are tossed) $= \dfrac{1}{8}$
P(getting $3$ tails when $3$ coins are tossed) $= \dfrac{1}{8}$
$\begin{aligned}
P \left(X = 10\right) & = P \left(\text{probability of getting 10 points}\right) \\\\
& = P \left(\text{getting all heads}\right) \; OR \;\; P \left(\text{getting all tails}\right) \\\\
& = P \left(\text{getting all heads}\right) + P \left(\text{getting all tails}\right) \\\\
& = \dfrac{1}{8} + \dfrac{1}{8} \\\\
& = \dfrac{2}{8} = \dfrac{1}{4}
\end{aligned}$
$\begin{aligned}
P \left(X = -5\right) & = P \left(\text{loosing 5 points}\right) \\\\
& = P \left(\text{getting 1 head}\right) \; OR \;\; P \left(\text{getting 2 heads}\right) \\\\
& = P \left(\text{getting 1 head}\right) + P \left(\text{getting 2 heads}\right) \\\\
& = \dfrac{3}{8} + \dfrac{3}{8} \\\\
& = \dfrac{6}{8} = \dfrac{3}{4}
\end{aligned}$
| $X$ | $10$ | $-5$ |
|---|---|---|
| $P \left(X = x\right)$ | $\dfrac{1}{4}$ | $\dfrac{3}{4}$ |
Mean value of $X = E \left(X\right) = \sum \limits_{i} x_i p_i$
$\therefore \;$ $E \left(X\right) = \left(10 \times \dfrac{1}{4}\right) + \left(-5 \times \dfrac{3}{4}\right) = - \dfrac{5}{4} = - 1.25$
i.e. $\;$ The person is expected to loose $1.25$ points.