Algebra

Probability

Medium

common

A random variable X has the following probability distribution:

X 0 1 2 3 4 5 6 7
P(X) 0 k 2k 2k 3k k² 2k² 7k² + k

The value of $P(4 < x < 7)$ is equal to

X	0	1	2	3	4	5	6	7
P(X)	0	k	2k	2k	3k	k²	2k²	7k² + k

$\frac{3}{100}$

$\frac{1}{2}$

$\frac{5}{100}$

$\frac{7}{100}$

✅ Correct Option: 1

The probability distribution has an unknown value $k$ . The sum of all probabilities must equal 1.

$P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = 1$

$0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1$

Combining like terms:

$10k^2 + 9k = 1$

$10k^2 + 9k - 1 = 0$

Using the quadratic formula where $a = 10$ , $b = 9$ , $c = -1$ :

$k = \dfrac{-9 \pm \sqrt{81 - 4(10)(-1)}}{2(10)}$

$k = \dfrac{-9 \pm \sqrt{81 + 40}}{20}$

$k = \dfrac{-9 \pm \sqrt{121}}{20}$

$k = \dfrac{-9 \pm 11}{20}$

This gives $k = \dfrac{2}{20} = \dfrac{1}{10}$ or $k = \dfrac{-20}{20} = -1$

Since probability cannot be negative, $k = \dfrac{1}{10}$ .

The notation $4 < X < 7$ means $X$ is strictly between 4 and 7, so $X$ can only be 5 or 6.

$P(4 < X < 7) = P(X=5) + P(X=6)$

$P(4 < X < 7) = k^2 + 2k^2$

$P(4 < X < 7) = 3k^2$

$P(4 < X < 7) = 3 \times \left(\dfrac{1}{10}\right)^2$

$P(4 < X < 7) = 3 \times \dfrac{1}{100}$

$P(4 < X < 7) = \dfrac{3}{100}$

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: 19 May Shift 1

If A and B are independent events, then which of the following statements are TRUE?

(A) $P(A \cap B) = P(A).P(B)$

(B) $P(A \cap B) = P(A) - P(B)$

(D) $P(A \cap B) = P(A). P(B|A)$

Choose the correct answer from the options given below:

2026: 22 May Shift 2

Two dice are thrown simultaneously. If X denotes the number of sixes appearing on two dice, then which of the following are correct?

(A) $P(X=0)=\frac{1}{36}$

(B) $P(X=1)=\frac{10}{36}$

(D) $P(X=1)+P(X=2)=\frac{11}{36}$

Choose the correct answer from the options given below:

2026: 18 May Shift 1

A bag contains 3 white and 2 black balls, another bag contains some white and 4 black balls. One bag is chosen at random. From the selected bag, one ball is drawn. If the probability of getting a black ball from the selected bag is 8/15, then the number of white balls in the second bag is

2025: 19 May Shift 1

Match List-I with List-II

Let A and B be any two events

List-I	List-II
(A) $P(A')$	(I) $\frac{P(A \cap B)}{P(A)}; P(A) \neq 0$
(B) $P(\phi)$	(II) $\frac{P(A \cap B)}{P(B)}; P(B) \neq 0$
(C) $P(A\vert B)$	(III) $1 - P(A)$
(D) $P(B\vert A)$	(IV) 0

Choose the correct answer from the options given below:

2023: 22 May Shift 3

For the following probability distribution :

X	1	2	3	4
P(X)	1/10	1/5	3/10	2/5

$E(X^2)$ is equal to :

A random variable X has the following probability distribution: X01234567P(X)0k2k2k3kk²2k²7k² + k The value of P(4<x<7)P(4 < x < 7)P(4<x<7) is equal to

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

A random variable X has the following probability distribution:

X 0 1 2 3 4 5 6 7
P(X) 0 k 2k 2k 3k k² 2k² 7k² + k

The value of $P(4 < x < 7)$ is equal to