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 . Then the value 9p – 1 is : [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
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 8p – 1 is : [2025]

1 0

2 3

3 1

4 2

Ans.
(4)

We have, P(X = 0) = P(X = 1) = p and let P(X = 2) = P(X = 3)= q

$E (X) = \sum_{i = 0}^{3} x_{i}^{} P (X = x_{i}) = 0 + 1 p + 2 q + 3 q = p + 5 q$

$E (X^{2}) = \sum_{i = 0}^{3} x_{i}^{2} P (X = x_{i})$

$= 0 \cdot p + 1 \cdot p + 4 \cdot q + 9 \cdot q = p + 13 q$

Now, $E (X^{2}) = 2 E (X)$ [Given]

$\Rightarrow p + 13 q = 2 (p + 5 q)$

$\Rightarrow p = 3 q$ ... (i)

Also, $\sum_{i = 0}^{3} P (X = x_{i}) = 1$

$\Rightarrow 2 p + 2 q = 1$

$\Rightarrow 2 (p + q) = 1$ ... (ii)

On solving (i) and (ii), we get

$q = \frac{1}{8} and p = \frac{3}{8}$

$∴ 8 p - 1 = 3 - 1 = 2$ .

Want Us to Email you About Special Offers & Updates?