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.

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 $6.00$ ) $=\frac{1}{16}$