Probability Question 169
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
Options:
A) $ \frac{1}{6} $
B) $ \frac{1}{4} $
C) $ \frac{5}{6} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ A $ be the event that face 4 turns up and $ B $ be the event that face 5 turns up then $ P(A)=0.25, $
$ P(B)=0.05 $ .
Since $ A $ and $ B $ are mutually exclusive, so $ P(A\cup B)=P(A)+P(B)=0.25+0.05=0.30 $ .
We have to find $ P( \frac{A}{A\cup B} ), $ which is equal to $ P\frac{[A\cap (A\cup B)]}{P(A\cup B)}=\frac{P(A)}{P(A\cup B)}=\frac{0.25}{0.30}=\frac{5}{6} $ .
BETA
