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A fair coin is tossed four times, and a person win $\mathrm {Rs.}$ $1$ for each head and lose $\mathrm {Rs.}$ $1.50$ 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 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}$
