Question

Question 23
Two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9, separately.

Answer 1

Number of total outcomes = n(s) = 36
(i) Let $E_1$ = Event of getting sum 2 = {(1,1), (1,1)}
$n\left(E_1\right)=2$
$\therefore P\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{2}{36}=\frac{1}{18}$
(ii) Let $E_2$ = Event of getting sum = 3 {(1,2), (1,2),(2,1), (2,1)}
$n\left(E_2\right)=4$
$\therefore P\left(E_2\right)=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{4}{36}=\frac{1}{9}$
(iii) Let $E_3$ = Event of getting sum 4 = {(2,2), (2,2), (3,1), (3,1), (1,3), (1,3)}
$\therefore n\left(E_3\right)=6$
$\therefore P\left(E_3\right)=\frac{n\left(E_3\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$.
(iv) Let $E_4$ = Event of getting sum 5 = {(2,2),(2,3)(4,1), (4,1), (3,2), (3,2)}
$\therefore n\left(E_4\right)=6$
$\therefore P\left(E_4\right)=\frac{n\left(E_4\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$
(v) Let $E_5$ = Event of getting sum 6 = {(3,3), (3,3), (4,2), (4,2), (5,1), (5,1)}
$n\left(E_5\right)=6$
$\therefore P\left(E_5\right)=\frac{n\left(E_5\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$
(vi) Let $E_6$ = Event of getting sum 7 = {(4,3), (4,3), (5,2), (5,2), (6,1), (6,1)}
$\therefore n\left(E_6\right)=6$
$\therefore P\left(E_6\right)=\frac{n\left(E_6\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$
(vii) Let $E_7$ = Event of getting sum 8 = {(5,3), (5,3), (6,2), (6,2)}
$\therefore n\left(E_7\right)=4$
$\therefore P\left(E_7\right)=\frac{n\left(E_7\right)}{n\left(S\right)}=\frac{4}{36}=\frac{1}{9}$
(viii) Let $E_8$ = Event of getting sum 9 = {(6,3), (6,3)}
$\therefore n\left(E_8\right)=2$
$\therefore P\left(E_8\right)=\frac{n\left(E_8\right)}{n\left(S\right)}=\frac{2}{36}=\frac{1}{18}$