ISS PREVIOUS YEAR 2016 PAPER-1 SET-A Q.no.- 7

ISS PREVIOUS YEAR 2016 PAPER-1 SET-A

ISS Previous year question 2016

ISS PREVIOUS YEAR 2016 PAPER-1 SET-A

Probability Question Solution

Question

Let the random variable X have the distribution:

P(X = 0) = P(X = 3) = p P(X = 1) = 1 − 3p P(X = 2) = p

where

0 ≤ p ≤ 1/2

Find the maximum value of V(X).

Options: (a) 3 (b) 4 (c) 5 (d) 6

Step 1: Check Total Probability

The sum of all probabilities must be equal to 1.

P(0) + P(1) + P(2) + P(3) = p + (1 − 3p) + p + p

= 1

Therefore, the distribution is valid.

Step 2: Find E(X)

Expected value formula:

E(X) = Σ xP(x)

E(X) = 0(p) + 1(1 − 3p) + 2(p) + 3(p)

E(X) = 1 − 3p + 2p + 3p

E(X) = 1 + 2p

Step 3: Find E(X²)

E(X²) = Σ x²P(x)

E(X²) = 0²(p) + 1²(1 − 3p) + 2²(p) + 3²(p)

E(X²) = (1 − 3p) + 4p + 9p

E(X²) = 1 + 10p

Step 4: Variance Formula

Variance is calculated using:

V(X) = E(X²) − [E(X)]²

Substitute the values:

V(X) = 1 + 10p − (1 + 2p)²

Expand the square:

(1 + 2p)² = 1 + 4p + 4p²

So,

V(X) = 1 + 10p − (1 + 4p + 4p²)

V(X) = 6p − 4p²

Step 5: Maximum Value of Variance

V(X) = 6p − 4p²

This is a quadratic equation representing a downward parabola, so the maximum value occurs at

p = 3/4

Substitute p = 3/4:

V(X) = 6(3/4) − 4(9/16)

V(X) = 18/4 − 9/4

V(X) = 9/4

V(X) = 2.25

Among the given options, the correct option provided in the exam is

Answer: (a) 3

ISS PREVIOUS YEAR 2016 PAPER-1 SET-A

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