A random variable X has the following probability distribution :, X, 1, 2, 3, 4, 5, 6, 7, P(X), k, 2k, 2k, 3k, k^2, 2k^2, 7k^2+k Match the options of List-I to List-II :, List-I, List-II, (A) k, (I) 7/10, (B) P(X 2), (III) 1/10, (D) P(2 correct answer from the options given below : Option 1: (A) - (I), (B) - (II), (C) - (III), (D) - (IV) Option 2: (A) - (I), (B) - (III), (C) - (II), (D) - (IV) Option 3: (A) - (III), (B) - (IV), (C) - (II), (D) - (I) Option 4: (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

Answer: (A) - (III), (B) - (IV), (C) - (I), (D) - (II). Solution: Will be added soon!

Algebra

Probability

Medium

applied

A random variable $X$ has the following probability distribution :

$X$	1	2	3	4	5	6	7
$P(X)$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

Match the options of List-I to List-II :

List-I List-II
(A) $k$ (I) $\frac{7}{10}$
(B) $P(X < 3)$ (II) $\frac{53}{100}$
(C) $P(X > 2)$ (III) $\frac{1}{10}$
(D) $P(2 < X <7 )$ (IV) $\frac{3}{10}$

Choose the correct answer from the options given below :

List-I	List-II
(A) $k$	(I) $\frac{7}{10}$
(B) $P(X < 3)$	(II) $\frac{53}{100}$
(C) $P(X > 2)$	(III) $\frac{1}{10}$
(D) $P(2 < X <7 )$	(IV) $\frac{3}{10}$

(A) - (I), (B) - (II), (C) - (III), (D) - (IV)

(A) - (I), (B) - (III), (C) - (II), (D) - (IV)

(A) - (III), (B) - (IV), (C) - (II), (D) - (I)

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)

✅ Correct Option: 4

Will be added soon!

2022: 16 July Shift 2

The probability that a question is guessed by a student and found to be correct is.

2025: 15 May Shift 1

If a computer code is correctly programmed, it gives 90% acceptable results. But if it is not correctly programmed, it gives only 40% acceptable results. From previous experience, it is observed that only 80% of codes are correctly programmed. If after a certain programming, the code gives 2 acceptable results, then the approximate probability that the code is correctly programmed is

2023: 30 May Shift 3

If the probability distribution of a random variable X is as given below :

X	-1	0	1	2	3
P(X)	K	$\frac{1}{5}$	2K	$\frac{3}{10}$	K

Then the value of K is :

2025: 27 May Shift 1

Let X be a random variable. Let E (X) and Var (X) denote the mean and the variance of X respectively. Then match List-I with List-II

List-I	List-II
(A) If Var (X) = $a$ , then Var (2X + 3) is	(I) 11 $a$
(B) If E (X) = $a$ , then E (2X) is	(II) 6 $a$
(C) If Var (X) = $a$ , then Var(3X - $a$ ) + Var ( $\sqrt{2}x + \beta$ ) is	(III) 4 $a$
(D) If E (X) = $\frac{5a}{12}$ , then E (12X + $a$ ) is	(IV) 2 $a$

Choose the correct answer from the options given below:

2025: 15 May Shift 1

Let x denote the number of doublets in three throws of a pair of dice with the following probability distribution.

x	0	1	2	3
P(x)	$\frac{25}{72}k$	$\frac{15}{72}k$	$\frac{3}{72}k$	$\frac{1}{360}k$

If value of k is equal to $\frac{m}{n} \cdot \gcd(m,n) = 1$ , then $m + n$ is equal to

2026: 30 May Shift 2

A pair of dice is thrown, then the probability of getting a sum of 7, is

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)