Statistics And Probability Question 160
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:
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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
