# 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.

Question:

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.

Solution:

Since the coin is tossed four times, there can be a maximum of 4 heads or tails.

When 4 heads turns up, Rel+Rel+Rel+Rel = Rs 4 is the gain.

When 3 heads and 1 tail turn up, Re 1 + Re 1 + Re 1 – Rs 1.50 = Rs 3 – Rs 1.50 = Rs 1.50 is the gain.

When 2 heads and 2 tails turns up, Re 1 + Re 1 – Rs 1.50 – Rs 1.50 = – Re 1, i.e., Re 1 is the loss.

When 1 head and 3 tails turn up, Re 1 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 3.50, i.e., Rs 3.50 is the loss.

When 4 tails turn up, – Rs 1.50 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 6.00, i.e., Rs 6.00 is the loss.

There are $2^{4}=16$ elements in the sample space $S$, which is given by:

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT, THTT, TTHT, TTTH, TTTT}

$\therefore n(S)=16$

The person wins Rs 4.00 when 4 heads turn up, i.e., when the event {HHHH} occurs.

$\therefore$ Probability (of winning Rs $4.00$ ) $=\frac{1}{16}$

The person wins Rs 1.50 when 3 heads and one tail turn up, i.e., when the event {HHHT, HHTH, HTHH, THHH} occurs.

$\therefore$ Probability (of winning Rs $1.50)=\frac{4}{16}=\frac{1}{4}$

The person loses Re 1.00 when 2 heads and 2 tails turn up, i.e., when the event {HHTT, HTTH, TTHH, HTHT, THTH, THHT} occurs.

$\therefore$ Probability (of losing Re $1.00)=\frac{6}{16}=\frac{3}{8}$

The person loses Rs 3.50 when 1 head and 3 tails turn up, i.e., when the event {HTTT, THTT, TTHT, TTTH} occurs.

Probability (of losing Rs $3.50$ ) $=\frac{4}{16}=\frac{1}{4}$

The person loses Rs 6.00 when 4 tails turn up, i.e., when the event {TTTT} occurs.

Probability (of losing Rs $6.00$ ) $=\frac{1}{16}$