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MathematicsProbabilityRandom Variable and its Probability DistributionMedium2 minPYQ_2022
MathematicsMediumsingle choice

A random variable X has the following probability distribution:

X01234
PXk2k4k6k8k

The value of P1<x<4x2is equal to

Options:

Answer:
A
Solution:

To find P1<x<4x2 or PAB

We know that

 PAB=PABPB

Given

x01234
Pxk2k4k6k8k

PA=2,3

PB=0,1,2

PAB=Px=2

PB=Px=0+Px=1+Px=2

So   PAB=Px=2Px=0+Px=1+Px=2

=4kk+2k+4k=4k7k=47

Hence option A is correct.

Stream:JEESubject:MathematicsTopic:ProbabilitySubtopic:Random Variable and its Probability Distribution
2mℹ️ Source: PYQ_2022

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