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  4. A biased dice is tossed and the respective probabilities for various faces to turn up are given below: Face 1 2 3 4 5 6 Probability 0.1 0.24 0.19 0.18 0.15 0.14 If an even face has turned up, then the probability that it is face 2 or face 4 is

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Q. A biased dice is tossed and the respective probabilities for various faces to turn up are given below:
Face 1 2 3 4 5 6
Probability 0.1 0.24 0.19 0.18 0.15 0.14

If an even face has turned up, then the probability that it is face $2$ or face $4$ is

Probability - Part 2

Solution:

Let $A$ be the event that even face turns up and $B$ be the event that it is $2$ or $4$ .
Then $P(A)=P(2)+P(4)+P(6)=0.24+0.18+0.14=0.56$
$P(B)=P(2)+P(4)=0.24+0.18=0.42$
So, $P(B / A)=\frac{P(B \cap A)}{P(A)}=\frac{P(B)}{P(A)}=\frac{0.42}{0.56}=0.75$