Probability Question 257
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:
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
