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. Let $p$ be the probability of person losing Rs $3.50$ after $4$ tosses.Find $4p$ ?

Answer 1

Number of elements in sample space S is $2^4=16$
$S=HHHH,HHHT,HHTH,HTHH,THHH,HHTT,HTTH,TTHH,HTHT,THTH,THHT,HTTT,THTT,TTHT,TTTH,TTTT$
$\therefore n\left(S\right)=16$

Since, the coin is tossed four times there can be a maximum of $4$ heads .
When $4$ heads turns up, $Re1+Re1+Re1+Re1=Rs4$ is the gain.
When $3$ heads and $1$ tail turn up, $Re1+Re1+Re1-Rs1.50=Rs3-Rs1.50=Rs1.50$ is the gain.
When $2$ heads and $2$ tails turns up, $Re1+Re1-Rs1.50-Rs1.50=-Re1$, i.e., Re $1$ is the loss.
When $1$ head and $3$ tails turn up,$Re1-Rs1.50-Rs1.50-Rs.1.50=-Rs3.50$, i.e., Rs $3.50$ is the loss.
When $4$ tails turn up, $-Rs1.50-Rs1.50-Rs1.50-Rs1.50=-Rs6.00$ i.e., Rs $6.00$ is the loss.
The person wins Rs $4.00$ when $4$ heads turn up i.e., when the event $\left\{HHHH\right\}$ 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 $\left\{TTTT\right\}$ occurs
Probability (of losing Rs $6.00$) = $\frac{1}{16}$