List of top Mathematics Questions on Random Variables asked in MHT CET

The c.d.f. of a discrete random variable \( X \) is Then \( P(X = -3) / P(X \leq 0) \) is:

A random variable \( X \) has p.m.f. \( P(X = x) = \frac{{}^{4}C_x}{2^4}, \quad x = 0, 1, 2, 3, 4 \), and \( \mu \) and \( \sigma^2 \) are the mean and variance respectively of the random variable \( X \), then: 

Consider the probability distribution:
Then the value of $P(X > 2)$ is ______.
The cumulative distribution function of a discrete random variable X is given. Then $\frac{P(X \le 0)}{P(X > 0)} = $ ______.

A random variable X has the distribution: $P(X=1,2,3,4) = 0.1, 0.2, 0.3, 0.4$. The mean and standard deviation are:
A random variable X has the distribution: $P(X=1,2,3,4) = 0.1, 0.2, 0.3, 0.4$. The mean and standard deviation are:

The p.d.f. of a continuous random variable \(X\) is \(f(x)\).
\[ f(x)= \begin{cases} \dfrac{x^2}{18}, & -3 \le x \le 3 \\ 0, & \text{otherwise} \end{cases} \] Then find \(P(|X|<2)\).

The integral \( \int e^x \frac{2 + \sin 2x}{1 + \cos 2x} \, dx \) is equal to
If the vector equation of the line \[ \frac{x - 2}{2} = \frac{2y - 5}{-3} = z + 1, \] is given by: \[ \vec{r} = \left(2\hat{i} + \frac{5}{2}\hat{j} - \hat{k}\right) + \lambda\left(2\hat{i} - \frac{3}{2}\hat{j} + p\hat{k}\right), \] then \( p \) is equal to:
A random variable X has the following probability distribution

Then P ( 3 < x $\le$ 6 ) =
Let two cards be drawn at random from a pack of 52 playing cards. Let X be the number of aces obtained. Then the value of E(X) is
The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
The probability distribution of a discrete random variable $X$ is given as

Find the value of $P(2 < X < 6)$.
If the probability distribution function of a random variable $X$ is given as

Then $F(0)$ is equal to
If $X$ is a random variable with p.m.f. as follows.
$$P(X = x) = \begin{cases} \frac{5}{16}, & x = 0, 1 \\ \frac{kx}{48}, & x = 2 \\ \frac{1}{4}, & x = 3 \end{cases}$$
then $E(X) =$
In a meeting 60% of the members favour and 40% oppose a certain proposal. A member is selected at random and we take $X = 0$ if he opposed and $X = 1$ if he is in favour, then $\text{Var}(X) =$
A bakerman sells 5 types of cakes. Profit due to sale of each type of cake is respectively ₹ 2.5, ₹ 3, ₹ 1.5, ₹ 1 and ₹ 2. The demands for these cakes are 20%, 5%, 10%, 50% and 15% respectively, then the expected profit per cake is
Rajesh has just bought a VCR from Maharashtra Electronics and the shop offers after sales service contract for Rs. 1000 for the next five years. Considering the experience of VCR users, the following distribution of maintenance expenses for the next five years is formed.

The expected value of maintenance cost is :
A random variable X has the following probability distribution

then F(4) =
If $X$ is a random variable with p.m.f. as follows.
$$P(X = x) = \begin{cases} \frac{5}{16}, & x = 0, 1 \\ \frac{kx}{48}, & x = 2 \\ \frac{1}{4}, & x = 3 \end{cases}$$
then $E(X) =$