Question

A dice is thrown twice and the sum of numbers appearing is observed to be $6$. The conditional probability that the number $4$ has appeared atleast once is:

• A. $\frac{2}{15}$
• B. $\frac{3}{5}$
• C. $\frac{2}{7}$
• D. $\frac{2}{5}$

Answer 1

The correct option is D $\frac{2}{5}$

Let $E$ be the event that 'number $4$ appears atleast once' and $F$ be the event that 'the sum of numbers appearing is $6$'
Then, $E=\left\{\right(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(1,4),(2,4),(3,4),(5,4),(6,4\left)\right\}$
and $F=\left\{\right(1,5),(2,4),(3,3),(4,2),(5,1\left)\right\}$
We have $P\left(E\right)=\frac{11}{36}$ and $P\left(F\right)=\frac{5}{36}$
Also $E\cap F=\left\{\right(2,4),(4,2\left)\right\}$
Therefore $P(E\cap F)=\frac{2}{36}$
Hence, the required probability
$P\left(E/F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{2}{36}}{\frac{5}{36}}=\frac{2}{5}$