Question 13 Marks
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. Form the sample space, calculate how many different amounts of money he can have after four tosses and the probability of having each of these amounts.
Answer
View full question & answer→Here a coin is tossed four times. So number of elements in the sample space (S) will be $2^4 = 16$. n(S) = 16.
The sample space,
S = {HHHH, HHHT, HHTH, HTHH, HTTH, HTHT, HHTT, HTTT, THHH, THHT, THTH, TTHH, TTTH, TTHT, THTT, TTTT}
Amounts:
(i) When 4 heads turns up = Rs( 1 + 1+ 1 + 1)= Rs. 4. i.e.,Person wins Rs. 4
(ii)When 3 heads and 1 tail turns up = Rs(1+ 1+1 –1.50 =Rs. 1.50. i.e.,Person wins Rs. 1.50
(iii) When2 heads and 2 tails turns up = Rs(1 + 1– 1.50– 1.50) =– Rs. 1. i.e., Person losesRs. 1
(iv) When1 head and 3 tails turns up = Rs(1– 1.50– 1.50– 1.50 )=– Rs 3.50. i.e., Person losesRs. 3.50
(v) When4 tails turns up= Rs(– 1.50– 1.50– 1.50– 1.50) =– Rs 6. i.e., Person loses Rs. 6
Let the events for which the person wins Rs 4, wins Rs 1.50, loses Re1, loses Rs 3.50 and loses Rs 6
be denoted by $E_1, E_2, E_3, E_4$ and $E_5$.
i.e., $E_1 = {HHHH},$
$E_2 = {HHHT, HHTH, HTHH, THHH} $
$E_3={HHTT,HTHT,HTTH,THTH,THHT,TTHH}$
$E_4 = {HTTT, TTTH, THTT, TTHT}, $
$E_5 = {TTTT}$
Here, $n(E_1) = 1, n(E_2) = 4, n(E_3) = 6, n(E_4) = 4$ and $n(E_5) = 1$.
Hence,$\style{font-size:26px}{P(E_1)=\frac{n({E_1)}}{n(S)}=\frac1{16}}$,$\style{font-size:26px}{P(E_2)=\frac{n({E_2)}}{n(S)}=\frac4{16}=\frac14}$
$\style{font-size:26px}{P(E_3)=\frac{n({E_3)}}{n(S)}=\frac6{16}=\frac38}$
$\style{font-size:26px}{P(E_4)=\frac{n({E_4)}}{n(S)}=\frac4{16}=\frac14}$
and $\style{font-size:26px}{P(E_5)=\frac{n({E_5)}}{n(S)}=\frac1{16}=\frac1{16}}$
The sample space,
S = {HHHH, HHHT, HHTH, HTHH, HTTH, HTHT, HHTT, HTTT, THHH, THHT, THTH, TTHH, TTTH, TTHT, THTT, TTTT}
Amounts:
(i) When 4 heads turns up = Rs( 1 + 1+ 1 + 1)= Rs. 4. i.e.,Person wins Rs. 4
(ii)When 3 heads and 1 tail turns up = Rs(1+ 1+1 –1.50 =Rs. 1.50. i.e.,Person wins Rs. 1.50
(iii) When2 heads and 2 tails turns up = Rs(1 + 1– 1.50– 1.50) =– Rs. 1. i.e., Person losesRs. 1
(iv) When1 head and 3 tails turns up = Rs(1– 1.50– 1.50– 1.50 )=– Rs 3.50. i.e., Person losesRs. 3.50
(v) When4 tails turns up= Rs(– 1.50– 1.50– 1.50– 1.50) =– Rs 6. i.e., Person loses Rs. 6
Let the events for which the person wins Rs 4, wins Rs 1.50, loses Re1, loses Rs 3.50 and loses Rs 6
be denoted by $E_1, E_2, E_3, E_4$ and $E_5$.
i.e., $E_1 = {HHHH},$
$E_2 = {HHHT, HHTH, HTHH, THHH} $
$E_3={HHTT,HTHT,HTTH,THTH,THHT,TTHH}$
$E_4 = {HTTT, TTTH, THTT, TTHT}, $
$E_5 = {TTTT}$
Here, $n(E_1) = 1, n(E_2) = 4, n(E_3) = 6, n(E_4) = 4$ and $n(E_5) = 1$.
Hence,$\style{font-size:26px}{P(E_1)=\frac{n({E_1)}}{n(S)}=\frac1{16}}$,$\style{font-size:26px}{P(E_2)=\frac{n({E_2)}}{n(S)}=\frac4{16}=\frac14}$
$\style{font-size:26px}{P(E_3)=\frac{n({E_3)}}{n(S)}=\frac6{16}=\frac38}$
$\style{font-size:26px}{P(E_4)=\frac{n({E_4)}}{n(S)}=\frac4{16}=\frac14}$
and $\style{font-size:26px}{P(E_5)=\frac{n({E_5)}}{n(S)}=\frac1{16}=\frac1{16}}$