Question

For a biased die the probabilities for different faces to turn up are given below: The die is tossed and you are told that either face 1 or 2 has turned up. Then the probability that it is face 1 is .....

Face	1	2	3	4	5	6
Probability	0.1	0.32	0.21	0.15	0.05	0.17.

• A. $\frac{5}{42}$
• B. $\frac{16}{21}$
• C. $\frac{5}{21}$
• D. $\frac{16}{42}$

Answer 1

The correct option is C $\frac{5}{21}$

Let $A$ be the event that face 1 turns up
$B$ be the event that face 2 turns up.
Then $P\left(A\right)=0.10$ and $P\left(B\right)=0.32,$
Since A, B are mutually exclusive, we have
$P(A\cup B)=P\left(A\right)+P\left(B\right)=0.10+0.32=\mathrm{0.42.}$
We are to find $P(A|A\cup B)$
But $P(A|A\cup B)=\frac{P[A\cap (A\cup B)]}{P(A\cup B)}$
$=\frac{P\left(A\right)}{P(A\cup B)}$
Hence $P(A|A\cup B)=\frac{0.10}{0.42}=\frac{5}{21}$