Question

A biased die 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 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)

$0.75$

Required probability = $\frac{P(A\cap B)}{P\left(B\right)}=P$

B = even face

A = (face = 2 or 4)

$A\cap B$ = (face = 2 or 4)

$\therefore P=\frac{P\left(2or4\right)}{P\left(2or4or6\right)}=\frac{0.24+0.18}{0.24+0.18+0.14}=0.75$