since the coin is tossed four time, there can be a maximum of $4$ heads and tails.

When $4$ heads turns up, $\mathrm {Rs.}$ $1+$ $\mathrm {Rs.}$ $1+$ $\mathrm {Rs.}$ $1+$ $\mathrm {Rs.}$ $1=$ $\mathrm {Rs.}$ $4$ is the gain.

When $3$ heads and $1$ tail turn up, $\mathrm {Rs.}$ $1+$ $\mathrm {Rs.}$ $1+$ $\mathrm {Rs.}$ $1-$ $\mathrm {Rs.}$ $1.50=$ $\mathrm {Rs.}$ $3-$ $\mathrm {Rs.}$ $1.50= $ $\mathrm {Rs.}$ $1.50$ is the gain.

When $2$ heads and $2$ tail turn up, $\mathrm {Rs.}$ $1+$ $\mathrm {Rs.}$ $1-$ $\mathrm {Rs.}$ $1.50-$ $\mathrm {Rs.}$ $1.50=-$ $\mathrm {Rs.}$ $1,$ ie., $\mathrm {Rs.}$ $1$ is the loss.

When $1$ heads and $3$ tail turn up, $\mathrm {Rs.}$ $1-$ $\mathrm {Rs.}$ $1.50-$ $\mathrm {Rs.}$ $1.50-$ $\mathrm {Rs.}$ $1.50=-$ $\mathrm {Rs.}$ $3.50,$  i.e., $\mathrm {Rs.}$ $3.50$ is the loss.

When $4 $ tails turn up, $-$ $\mathrm {Rs.}$ $1.50-$ $\mathrm {Rs.}$ $1.50-$ $\mathrm {Rs.}$ $1.50-$ $\mathrm {Rs.}$ $1.50=-$ $\mathrm {Rs.}$ $6.00,$ ie., $\mathrm {Rs.}$ $6.00$ 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, \, H T T T $,  $T H T T , \, T T H T ,\, T T H T $,  $T T T H , \,T T T T \}$

$\therefore n( S )=16$

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

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

The person wins $\mathrm {Rs.}$ $1.50$ when $3$ heads and one tail turns up, i.e., when the event $\{HHHT, \,H H T H , \,H T H H ,\, T H H H \}$ occurs. 

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

The person loses  $\mathrm {Rs.}$ $1.00$ when $2$ heads and $2$ tails turns up, ie., when the event $\{HHTT,\, HTTH , \,T T H H $, $H T H T ,\,T H T H , \, T H H T \}$ occurs.

$\therefore $  Probability $($ of loosing  $\mathrm {Rs.}$  $1.00)=\frac{6}{16}=\frac{3}{8}$

The person losses $\mathrm {Rs.}$ $3.50$ when $1$ head and $3$ tails turn up, ie.,  when the event $\{ HTTT,\, THTT , \,T T H T ,\,T T T H \}$ occurs.

$\therefore $ Probability $($ of loosing $\mathrm {Rs.}$ $3.50)=\frac{4}{16}=\frac{1}{4}$

The person losses $\mathrm {Rs.}$ $6.00$ when $4$ head and $3$ tails turn up, ie.,  when the event $\{TTTT\}$ occurs.

$\therefore $ Probability $($ of loosing $\mathrm {Rs.}$ $  6.00)=\frac{1}{16}$