Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Probability

Question:

A random variable X has the following probability distributions:

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 a prime number}, F = { X < 4}, the probability P(E ∪ F), is

Options:

0.50

0.77

0.35

0.87

Correct Answer:

0.77

Explanation:

We have,

$P(E)=P(X=2 \, or \, X=3 \, or \, X=5 \, or \, X = 7)$

$= P(X=2) +P(X=3) +P(X=5)+P(X=7)$

$= 0.23+0.12+0.20+0.07=0.62$

$P(F) =P(X < 4)$

$= P(X=1)+P(X=2) +P(X=3)$

$= 0.15 + 0.23 + 0.12 = 0.50$

$ P(E ∩ F)=$ P( X is a prime number less than 4)

$=P(X=2)+P(X=3) =0.23 +0.12=0.35$

$∴ P(E ∪ F)= P(E) +P(F) -P(E ∩ F)$

$= 0.62 + 0.50 - 0.35 = 0.77$