A random variable X has the following probability distribution: X 0 1 2 3 4 P X k 2 k 4 k 6 k 8 k The value…

A random variable X has the following probability distribution:

X 0 1 2 3 4
PX k 2k 4k 6k 8k

The value of P1<x<4x2is equal to

  1. 47
  2. 23
  3. 37
  4. 45

Solution

To find P1<x<4x2 or PAB

We know that

 PAB=PABPB

Given

x 0 1 2 3 4
Px k 2k 4k 6k 8k

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.

Asked in: JEE Main 2022 (24 Jun Shift 2)

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