Question

An unbalanced dice (with six faces numbered from $1$ to $6$) is thrown. The probability that the face value is odd is $90$% of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even given that it is greater than $3$ is $0.75$, then the probability that the face value exceeds $3$ is

• A. $\frac{80}{171}$
• B. $\frac{29}{57}$
• C. $\frac{9}{19}$
• D. $\frac{82}{171}$

Answer 1

The correct option is A $\frac{80}{171}$

Let $E:$ face is even
and $F:$ face is odd.
Let $P\left(E\right)=p$ and $P\left(F\right)=0.9p$

$P\left(E\right)+P\left(F\right)=1$
$\Rightarrow P\left(E\right)=\frac{10}{19}\Rightarrow P\left(2\right)=P\left(4\right)=P\left(6\right)=\frac{10}{57}$

It is given that
$P(E\text{| face}>3)=0.75$
$\Rightarrow \frac{P(E\cap \text{face}>3)}{P(\text{face}>3)}=0.75\Rightarrow \frac{P(\text{face}=4,6)}{P(\text{face}>3)}=0.75$
$\Rightarrow P(\text{face}>3)=\frac{P(\text{face}=4,6)}{0.75}\Rightarrow P(\text{face}>3)=\frac{P\left(4\right)+P\left(6\right)}{0.75}\Rightarrow P(\text{face}>3)=\frac{\frac{10}{57}+\frac{10}{57}}{0.75}=\frac{80}{171}$