Probability Question 162
Question: A random variable X has the probability distribution
| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 |
For the events $ E= X $ is prime number and $ F={X<4} $ , the probability of $ P(E\cup F) $ is
[AIEEE 2004]
Options:
A) 0.50
B) 0.77
C) 0.35
D) 0.87
Show Answer
Answer:
Correct Answer: B
Solution:
$ E=x $ is a prime number
$ P(E)=P(2)+P(3)+P(5)+P(7)=0.62, $
$ F={x<4} $ , $ P(F)=P(1)+P(2)+P(3)=0.50 $
and $ P(E\cap F)=P(2)+P(3)=0.35 $
$ \therefore $ $ P(E\cup F)=P(E)+P(F)-P(E\cap F) $ = 0.62+0.50 - 0.35 = 0.77.
BETA
