Statistics And Probability Question 99
Question: For a biased dice, the probability for the different faces to turn up are
Face 1 2 3 4 5 6 P 0.01 0.32 0.21 0.15 0.05 0.147 The dice is tossed and it is told that either the face 1 or face 2 has shown up, then the probability that it is face, 1, is
Options:
A) $ \frac{16}{21} $
B) $ \frac{1}{10} $
C) $ \frac{5}{16} $
D) $ \frac{5}{21} $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] Let E: ?face 1 comes up? and F: ?face 1 or 2 comes up?
$ \Rightarrow E\cap F=E $ $ (\therefore E\subset F) $
$ \therefore P(E)=0.10 $ and $ P(F)=P(1)+P(2)=0.10+0.32=0.42 $ Hence, required probability $ =P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{P(E)}{P(F)}=\frac{0.10}{0.42}=\frac{5}{21} $
BETA
