Question

Two dices 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. $\frac{2}{3}$
• D. $\frac{1}{2}$

Answer 1

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

$S=\{1,2,3,4,5,6\}\times \{1,2,3,4,5,6\}$
$\therefore n\left(S\right)=36$
& Let $E_1$ $\equiv$ the event that the sum of the numbers coming up is 9.
$E_1\equiv \left\{\right(3,6),(6,3),(4,5),(5,4\left)\right\}$
$\therefore n\left(E_1\right)=4$ and
$E_2=\left\{\right(5,1),(5,2),(5,3),(5,4),(5,5),(5,6\left)\right\}$
$\therefore n\left(E_2\right)=6$
$E_1\cap E_2=\left\{\right(5,4\left)\right\}\therefore (E_1\cap E_2)=1$
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
$=P\left(\frac{E_1}{E_2}\right)=\frac{P(E_1\cap E_2)}{P\left(E_2\right)}=\frac{\frac{1}{36}}{\frac{1}{6}}=\frac{1}{6}$