$A$ fair coin is tossed four times. $A$ person wins $Rs. 1$ for each head and loses $Rs. 1.50$ for each tail that turns up. From the sample space,calculate the different amounts of money one can have after four tosses and the probability of having each of these amounts.

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Since the coin is tossed four times,there are $2^4 = 16$ possible outcomes.
Let $H$ be the number of heads and $T$ be the number of tails,where $H + T = 4$.
The gain/loss $G$ is given by $G = 1(H) - 1.50(T)$.
$1$. If $H=4, T=0$: $G = 1(4) - 1.50(0) = Rs. 4.00$. Number of outcomes = $\binom{4}{4} = 1$. Probability = $\frac{1}{16}$.
$2$. If $H=3, T=1$: $G = 1(3) - 1.50(1) = Rs. 1.50$. Number of outcomes = $\binom{4}{3} = 4$. Probability = $\frac{4}{16} = \frac{1}{4}$.
$3$. If $H=2, T=2$: $G = 1(2) - 1.50(2) = 2 - 3 = -Rs. 1.00$. Number of outcomes = $\binom{4}{2} = 6$. Probability = $\frac{6}{16} = \frac{3}{8}$.
$4$. If $H=1, T=3$: $G = 1(1) - 1.50(3) = 1 - 4.50 = -Rs. 3.50$. Number of outcomes = $\binom{4}{1} = 4$. Probability = $\frac{4}{16} = \frac{1}{4}$.
$5$. If $H=0, T=4$: $G = 1(0) - 1.50(4) = -Rs. 6.00$. Number of outcomes = $\binom{4}{0} = 1$. Probability = $\frac{1}{16}$.

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