Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

A random variable X takes the values 0, 1, 2, 3 and its mean is 1.3. If P(X=3)=2P(X=1) and P(X=2)=0.3 then P(X=0) is :

Options:

0.4

0.3

0.2

0.1

Correct Answer:

0.4

Explanation:

The correct answer is Option (1) → 0.4

$P(X=1)+P(X=2)+P(X+3)+P(X=0)=1$

$⇒P(X=0)+P(X=1)+2P(X=1)+0.3=1$

$P(X=0)+3P(X=1)=0.7$  ...(1)

also $∑x_iP(x_i)=1.3$

$⇒0×P(X=0)+1×P(X=1)+2×P(X=2)+3×P(X+3)=1.3$

$P(X=1)=0.7$

$P(X=1)=0.1$

from (1) $P(X=0)+0.3=0.7$

$⇒P(X=0)=0.4$