Two dice are numbered 1, 2, 3, 4, 5, 6

Question:

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.

Solution:

Number of total outcomes = 36

(i) Let $E_{1}=$ Event of getting sum $2=\{(1,1),(1,1)\}$

$n\left(E_{1}\right)=2$

$\therefore$  $P\left(E_{1}\right)=\frac{n\left(E_{1}\right)}{n(S)}=\frac{2}{36}=\frac{1}{18}$

(ii) Let $E_{2}=$ Event of getting sum $3=\{(1,2),(1,2),(2,1),(2,1)\}$

$n\left(E_{2}\right)=4$

$\therefore$ $P\left(E_{2}\right)=\frac{n\left(E_{2}\right)}{n(S)}=\frac{4}{36}=\frac{1}{9}$

(iii) Let $E_{3}=$ Event of getting sum $4=\{(2,2),(2,2),(3,1),(3,1),(1,3),(1,3)\}$

$\therefore \quad n\left(E_{3}\right)=6$

$\therefore \quad P\left(E_{3}\right)=\frac{n\left(E_{3}\right)}{n(S)}=\frac{6}{36}=\frac{1}{6}$

(iv) Let $E_{4}=$ Event of getting sum $5=\{(2,3),(2,3),(4,1),(4,1),(3,2),(3,2)\}$

$\therefore \quad n\left(E_{4}\right)=6$

$\therefore \quad P\left(E_{4}\right)=\frac{n\left(E_{4}\right)}{n(S)}=\frac{6}{36}=\frac{1}{6}$

(v) Let $E_{5}=$ Event of getting sum $6=\{(3,3),(3,3),(4,2),(4,2),(5,1),(5,1)\}$

$n\left(E_{5}\right)=6$

$\therefore$ $P\left(E_{5}\right)=\frac{n\left(E_{5}\right)}{n(S)}=\frac{6}{36}=\frac{1}{6}$

(vi) Let $E_{6}=$ Event of getting sum $7=\{(4,3),(4,3),(5,2),(5,2),(6,1),(6,1)\}$

$\therefore \quad n\left(E_{6}\right)=6$

$\therefore \quad P\left(E_{6}\right)=\frac{n\left(E_{6}\right)}{n(S)}=\frac{6}{36}=\frac{1}{6}$

(vii) Let $E_{7}=$ Event of getting sum $8=\{(5,3),(5,3),(6,2),(6,2)\}$

$\therefore \quad n\left(E_{7}\right)=4$

$\therefore$ $P\left(E_{7}\right)=\frac{n\left(E_{7}\right)}{n(S)}=\frac{4}{36}=\frac{1}{9}$

(viii) Let $E_{8}=$ Event of getting sum $9=\{(6,3),(6,3)\}$

$\therefore$ $n\left(E_{8}\right)=2$

$\therefore$ $P\left(E_{8}\right)=\frac{n\left(E_{8}\right)}{n(S)}=\frac{2}{36}=\frac{1}{18}$

 

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