Question

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

• A. $0.25$
• B. $0.42$
• C. $0.75$
• D. $0.9$

Answer 1

Answer: (C)

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, $P(B/A)=\frac{P(B\cap A)}{P(A)}=\frac{P(B)}{P(A)}=\frac{0.42}{0.56}=0.75$