Question

For a biased die, the probabilities for different faces to turn up are

Face :	1	2	3	4	5	6
Probability:	0.2	0.22	0.11	0.25	0.05	0.17

The die is tossed and you are told that either face 4 or face 5 has turned up. The probability that it is face 4 is

• A. 1/6
• B. ¼
• C. 5/6
• D. None of these

Answer 1

Answer: (C)

5/6

Answer 2

Let A be the event that face 4 turns up and B be the event that face 5 turns up then P(1) = 0.25, P(2) = 0.05. Since A and B are mutually exclusive, so P(AUB) = P(1) + P(2) =

0.25+0.05 = 0.30.

We have no find P, which is equal to

$\mathrm { P } \frac { [ \mathrm { A } \cap ( \mathrm { AUB } ) ] } { \mathrm { P } ( \mathrm { AUB } ) } = \frac { \mathrm { P } ( \mathrm { A } ) } { \mathrm { P } ( \mathrm { AUB } ) } = \frac { 0.25 } { 0.30 } = \frac { 5 } { 6 }$