Maths Q9 – JEE-MAIN 2025

Let a random variable

X

take values

0, 1, 2, 3

with

P (X = 0) = P (X = 1) = p

P (X = 2) = P (X = 3)

and

E (X^{2}) = 2 E (X)

. Then the value of

8 p - 1

is:

Remember that the sum of probabilities for all possible outcomes of a random variable must equal 1.

Step 1: Define probabilities and sum to 1✦ Active

Let $P (X = 0) = P (X = 1) = p$ . Let $P (X = 2) = P (X = 3) = q$ . The sum of all probabilities must be 1.

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

p + p + q + q = 1 ⟹ 2 p + 2 q = 1 (1)

Step 2: Calculate expectations and use the given relation○ Expand

Calculate the expected value $E (X)$ and $E (X^{2})$ using their definitions.

E (X) = 0 \cdot p + 1 \cdot p + 2 \cdot q + 3 \cdot q = p + 5 q

E (X^{2}) = 0^{2} \cdot p + 1^{2} \cdot p + 2^{2} \cdot q + 3^{2} \cdot q = p + 4 q + 9 q = p + 13 q

Use the given relation $E (X^{2}) = 2 E (X)$ to form a second equation.

p + 13 q = 2 (p + 5 q)

p + 13 q = 2 p + 10 q

3 q = p (2)

Step 3: Solve for

p

and calculate the final expression○ Expand

Substitute $p = 3 q$ from equation (2) into equation (1) to find $q$ , then find $p$ .

2 (3 q) + 2 q = 1 ⟹ 6 q + 2 q = 1 ⟹ 8 q = 1 ⟹ q = \frac{1}{8}

p = 3 q = 3 \cdot \frac{1}{8} = \frac{3}{8}

Finally, calculate the value of $8 p - 1$ .

8 p - 1 = 8 (\frac{3}{8}) - 1 = 3 - 1 = 2

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Maths Question 9 – JEE-MAIN 2025