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MATHS-CBSE-12-2018-->View question

## Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X (MATHS-CBSE-12-2018)

Two numbers are selected at random (without replacement) from the first
five positive integers. Let X denote the larger of the two numbers
obtained. Find the mean and variance of X (MATHS-CBSE-12-2018)

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We have, the first five positive integers arc 1, 2, 3, 4 and 5.
We can selected two numbers from 5 numbers is
5P2 = 20 different ways
Let X be the random variable denoting the largest of two numbers
P(X=2) = (1/5*1/4)+(1/5*1/4) = 2/20
P(X=3) =  (2/5*1/4)+(1/5*2/4) = 4/20
P(X=4) =  (3/5*1/4)+(1/5*3/4) = 6/20
P(X=5) =  (4/5*1/4)+(1/5*4/5) = 8/20
Mean E(X) = 2*2/20 + 3*3/30 + 4*6/20 + 5*8/20
= 4
Therefore mean of probability distribution is 4.
Variance V(X) = E(X^2) - (E(X))^2  =>   17 - 16 = 1
Therefore variance of probability distribution is 1.

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