Question

An unbalanced dice (with six faces numbered 1 to 6) is thrown. The probability that face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is 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. 0.468

Answer 1

The correct option is A 0.468
Given P(2) = P(4) = P(6) = x (let)

$AgainP\left(oddface\right)=\frac{90}{100}P\left(evenface\right)$

$P\left(\frac{evenface}{face>3}\right)=0.75,P(face>3)=?$

We know that P(odd f) + P(even f) = 1

$\Rightarrow \frac{90}{100}P\left(evenf\right)+P\left(evenf\right)=1$

$P\left(evenf\right)=\frac{10}{19}$

$P\left(2\right)+P\left(4\right)+P\left(6\right)=\frac{10}{19}\Rightarrow x=\frac{10}{3\times 19}$

$NowP\left[\frac{evenf}{f>3}\right]=\frac{P(evenf\cap f>3)}{P(f>3)}$

$0.75=\frac{P\left(4\right)+P\left(6\right)}{P(f>3)}$

$\Rightarrow P(f>3)=\frac{x+x}{0.75}=\frac{2x}{0.75}=0.468$