$A$ 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.
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 = $\frac{1}{16}$.
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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