Question

A black and a red dice are rolled. The conditional probability of obtaining a sum greater than $9$, given that the black die resulted in a $5$ is:

• A. $\frac{1}{2}$
• B. $\frac{1}{18}$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D $\frac{1}{3}$

Let $B$ denote black coloured die and $R$ denote red coloured die. Then, the sample space for the given experiement will be:
$\Rightarrow S=\left\{\begin{array}{c}(B1,R1),(B1,R2),(B1,R3),(B1,R4),(B1,R5),(B1,R6)\\ (B2,R1),(B2,R2),(B2,R3),(B2,R4),(B2,R5),(B2,R6)\\ (B3,R1),(B3,R2),(B3,R3),(B3,R4),(B3,R5),(B3,R6)\\ (B4,R1),(B4,R2),(B4,R3),(B4,R4),(B4,R5),(B4,R6)\\ (B5,R1),(B5,R2),(B5,R3),(B5,R4),(B5,R5),(B5,R6)\\ (B6,R1),(B6,R2),(B6,R3),(B6,R4),(B6,R5),(B6,R6)\end{array}\right\}$
Let $A$ be the event of 'obtaining a sum greater than $9$' and $B$ be the event of 'getting a $5$ on black die'
Then $A=\left\{\right(B4,R6),(B5,R5),(B5,R6),(B6,R4),(B6,R5),(B6,R6\left)\right\}$
and $B=\left\{\right(B5,R1),(B5,R2),(B5,R3),(B5,R4),(B5,R5),(B5,R6\left)\right\}$
$\Rightarrow A\cap B=\left\{\right(B5,R5),(B5,R6\left)\right\}$
So, $P\left(A\right)=\frac{6}{36}=\frac{1}{6},P\left(B\right)=\frac{6}{36}=\frac{1}{6},P(A\cap B)=\frac{2}{36}=\frac{1}{18}$
Now, we know that by definition of conditional probability,
$P\left(A/B\right)=\frac{P(A\cap B)}{P\left(B\right)}$
Now by substituting the values we get
$\Rightarrow P\left(A/B\right)=\frac{1/18}{1/6}=\frac{6}{18}=\frac{1}{3}$