Q.
Statement I A fair coin is tossed four times and a person win $₹ 1$ for each head and lose $₹ 1.50$ for each tail that turns up. From the sample space, how many different amounts of money you can have after four tosses and then the probability of having each of these amounts is $\frac{1}{16}, \frac{1}{4}, \frac{3}{8}, \frac{1}{4}, \frac{1}{16}$ respectively?
Statement II If $\frac{2}{11}$ is the probability of an event, then the probability of the event 'not $A$ ' is $\frac{9}{11}$.
Probability
Solution:
I. If a coin is tossed four times, then total number of possible outcomes
$=2^4=16$
For these cases, sample space can be written as
Sample space
Amount
HHHH
$1 + 1 + 1 + 1 =4$
HH HT
$1+ 1+1+-1.50 = 3 -1.50=1.50$
HH TH
$1 + 1-1.50 + 1 = 3 - 1.50 = 1.50$
HHTT
$1+1-1.50-1.50 = 2-3 = -1.00$
HTHH
$1-1.50+1+1=3-1.50=1.50$
HTHT
$1-1.50+1-1.50=2-3=-1.00$
HTTH
$1-1.50-1.50+1=2-3=-1.00$
HTTT
$1-1.50-1.50-1.50=1-4.50=-3.50$
THHH
$-1.50+1+1+1=-1.50+3=1.50$
THHT
$-1.501+1-1.50=2-3.00=-1.00$
THTH
$-1.50+1-1.50+1=2-3.00=-1.00$
THTT
$-1.50+1-1.50-1.50=1-4.50=-3.50$
TTHH
$-1.50-1.50+1+1=2-3.00=-1.00$
TTHT
$-1.50-1.50+1-1.50=1-4.50=-3.50$
TTTH
$-1.50-1.50-1.50+1=-4.50+1=-3.50$
TTTT
$-1.50-1.50-1.50-1.50=-6.00$
Hence, from above sample space, we get five types of different amounts
i.e., $ 4,1.50,-1.00,-3.50,-6.00$ i.e.,
Amounts
Number of times occurance
4.00
1
1.50
4
-1.00
6
-3.50
4
-6.00
1
Total
16
(negative sign indicates the losing value of amount)
$\Rightarrow P \text { (winning ₹ 4.00) }$
$=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{1}{16}$
$\Rightarrow P(\text { having } ₹ 1.50)=\frac{4}{16}=\frac{1}{4} $
$\Rightarrow P(\text { having } ₹-1.00)=\frac{6}{16}=\frac{3}{8} $
$\Rightarrow P \text { (having } ₹-3.50)=\frac{4}{16}=\frac{1}{4} $
$\Rightarrow P(\text { having } ₹-6.00)=\frac{1}{16}$
II. If $A$ be any event, then given
$P(A) =\frac{2}{11}$
$\therefore P(\operatorname{not} A) =P\left(A^{\prime}\right)=1-P(A) $
$ =1-\frac{2}{11}=\frac{9}{11}$
$[\because$ sum of probabilities of an event $(A)$ and its complementary event $\left(A^{\prime}\right)$ is always 1]
| Sample space | Amount |
|---|---|
| HHHH | $1 + 1 + 1 + 1 =4$ |
| HH HT | $1+ 1+1+-1.50 = 3 -1.50=1.50$ |
| HH TH | $1 + 1-1.50 + 1 = 3 - 1.50 = 1.50$ |
| HHTT | $1+1-1.50-1.50 = 2-3 = -1.00$ |
| HTHH | $1-1.50+1+1=3-1.50=1.50$ |
| HTHT | $1-1.50+1-1.50=2-3=-1.00$ |
| HTTH | $1-1.50-1.50+1=2-3=-1.00$ |
| HTTT | $1-1.50-1.50-1.50=1-4.50=-3.50$ |
| THHH | $-1.50+1+1+1=-1.50+3=1.50$ |
| THHT | $-1.501+1-1.50=2-3.00=-1.00$ |
| THTH | $-1.50+1-1.50+1=2-3.00=-1.00$ |
| THTT | $-1.50+1-1.50-1.50=1-4.50=-3.50$ |
| TTHH | $-1.50-1.50+1+1=2-3.00=-1.00$ |
| TTHT | $-1.50-1.50+1-1.50=1-4.50=-3.50$ |
| TTTH | $-1.50-1.50-1.50+1=-4.50+1=-3.50$ |
| TTTT | $-1.50-1.50-1.50-1.50=-6.00$ |
| Amounts | Number of times occurance |
|---|---|
| 4.00 | 1 |
| 1.50 | 4 |
| -1.00 | 6 |
| -3.50 | 4 |
| -6.00 | 1 |
| Total | 16 |