Difference between revisions of "GATE2009 q21"

Revision as of 12:08, 14 July 2014

An unbalanced dice (with $6$ 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 $0.75$, which one of the following options is closest to the probability that the face value exceeds $3$?

(A) 0.453

(B) 0.468

(C) 0.485

(D) 0.492

Solution by Happy Mittal

Let $P(even) = x$, so $P(odd) = 90%$ of $x$

= $9x/10$,

but $P(even) + P(odd) = 1$,

so $x + 9x/10 = 1$, $x = 10/19$.

So P(even) = 10/19. Since prob of any even number is same,

$P(2) = P(4) = P(6) = 10/(19*3) = 10/57$

Now P(even and exceeds 3) = P(exceeds 3) * P(even|exceeds). We have to evaluate P(exceeds 3), so

P(exceeds 3) = P(even and exceeds 3)/P(even|exceeds) = (P(4) + P(6))/0.75 = (20/57)/0.75 = 0.468

So option (B) is correct.