Question

Two dice are thrown. The probability that the sum of the numbers coming up on them is $9$, if it is known that the number $5$ always occurs on the first die, is

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{5}{6}$

Answer 1

The correct option is A $\frac{1}{6}$

Let $S$ be the sample space
$\therefore S=\{1,2,3,4,5,6\}\times \{1,2,3,4,5,6\}\therefore n\left(S\right)=36$
Let $E_1\equiv$ the event that the sum of the numbers coming up is $9$
and $E_2\equiv$ the event of occurrence of $5$ on the first die.
$E_1\equiv \{(3,6),(6,3),(4,5),(5,4)\}\therefore n\left(E_1\right)=4E_2\equiv \{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}\therefore n\left(E_2\right)=6E_1\cap E_2=\left\{(5,4)\right\}\therefore n(E_1\cap E_2)=1$
Now $P(E_1\cap E_2)=\frac{n(E_1\cap E_2)}{n\left(S\right)}=\frac{1}{36}$ and $P\left(E_2\right)=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$
$\therefore$ Required probability $P\left(\frac{E_1}{E_2}\right)=\frac{P(E_1\cap E_2)}{P\left(E_2\right)}=\frac{\frac{1}{36}}{\frac{1}{6}}=\frac{1}{6}$