Q.A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

The key idea is that the net amount depends only on the number of heads (or tails), not the order. Let $H$ be the number of heads in 4 tosses. Then $H$ can be $0,1,2,3,4$.

Step 1: Net amount formula

Each head gives +1, each tail gives –1.50. With $H$ heads and $4-H$ tails:

Step 2: Possible amounts

• $H=0$: $2.5(0)-6 = -6$
• $H=1$: $2.5(1)-6 = -3.50$
• $H=2$: $2.5(2)-6 = -1$
• $H=3$: $2.5(3)-6 = 1.50$
• $H=4$: $2.5(4)-6 = 4$

So there are 5 different amounts.

Step 3: Probabilities

Total outcomes $= 2^4 = 16$. Number of ways for $H$ heads is $\binom{4}{H}$.

• $P(-6) = \binom{4}{0}/16 = 1/16$ …

Why this approach works

When you toss a fair coin four times, each sequence of heads (H) and tails (T) is equally likely — there are $2^4 = 16$ equally probable outcomes. The net winnings depend only on the number of heads, not on the order of heads and tails. That’s because each head adds Re 1 and each tail subtracts Rs 1.50, and the total is just:

Since the number of tails = $4 - \text{number of heads}$, we can rewrite everything in terms of $h$, the number of heads.

This is a linear function of $h$, so each distinct $h$ gives a distinct net amount. The possible values of $h$ are 0, 1, 2, 3, 4 — that’s 5 different amounts.

Now, the probability of getting exactly $h$ heads in 4 tosses of a fair coin is given by the binomial distribution:

We’ll compute the net amount for each $h$, then list the probability.

Step-by-step solution

1. List the possible numbers of heads

   $h$ can be 0, 1, 2, 3, or 4. Each corresponds to a unique net amount because $2.5h - 6$ is strictly increasing in $h$.

2. Compute the net amount for each $h$

   • $h = 0$: $2.5(0) - 6 = -6$ (lose Rs 6)
   • $h = 1$: $2.5(1) - 6 = -3.5$ (lose Rs 3.50)
   • $h = 2$: $2.5(2) - 6 = -1$ (lose Re 1)
   • $h = 3$: $2.5(3) - 6 = 1.5$ (win Rs 1.50)
   • $h = 4$: $2.5(4) - 6 = 4$ (win Rs 4)

   So the five distinct amounts are: -6, -3.5, -1, 1.5, 4 (all in rupees).

3. Find the probability of each $h$

   Use $\binom{4}{h}$:

   • $h=0$: $\binom{4}{0}=1$, probability $= \frac{1}{16}$
   • $h=1$: $\binom{4}{1}=4$, probability $= \frac{4}{16} = \frac{1}{4}$
   • $h=2$: $\binom{4}{2}=6$, probability $= \frac{6}{16} = \frac{3}{8}$
   • $h=3$: $\binom{4}{3}=4$, probability $= \frac{4}{16} = \frac{1}{4}$
   • $h=4$: $\binom{4}{4}=1$, probability $= \frac{1}{16}$

4. Map these probabilities to the net amounts

   Since each $h$ gives exactly one net amount, the probability of a net amount equals the probability of the corresponding $h$.

| Net amount (Rs) | Number of heads $h$ | Probability | …