Statistics And Probability Question 580
Question: A random variable X has the probability distribution
X 1 2 3 4 5 6 7 8 P(X) 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05 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
