The random variable X has a probability distribution P(X) of the following form where k is a scalar and P(X = x) = cases k, & if x = 0 \\ 2k, & if x = 1 \\ 3k, & if x = 2 \\ 0, & otherwise cases then value of P(X < 2) = ___. Option 1: 5/6 Option 2: 3/4 Option 3: 1 Option 4: 1/2

Answer: 1/2. Solution: Total probability is 1: k + 2k + 3k = 6k = 1, so k = 1/6. Then P(X<2) = P(X=0) + P(X=1) = k + 2k = 3k = 3/6 = 1/2.

Applied Mathematics

Probability Distribution

Easy

applied

The random variable X has a probability distribution P(X) of the following form where k is a scalar and

$P(X = x) = \begin{cases} k, & \text{if } x = 0 \\ 2k, & \text{if } x = 1 \\ 3k, & \text{if } x = 2 \\ 0, & \text{otherwise} \end{cases}$

then value of P(X < 2) = _______.

$\frac{5}{6}$

$\frac{3}{4}$

$\frac{1}{2}$

✅ Correct Option: 4

Total probability is 1: $k + 2k + 3k = 6k = 1$ , so $k = 1/6$ . Then $P(X<2) = P(X=0) + P(X=1) = k + 2k = 3k = 3/6 = 1/2$ .

2026: 14 May Shift 2

If X has a Poisson distribution such that $2P(X=1)=P(X=2)$ , then which of the following are correct?

A. The mean of X is 4

B. The mean of X is 2

C. $P(X=2)=\frac{8}{e^4}$

D. $P(X=3)=\frac{32}{e^4}$

Choose the correct answer from the options given below:

2026: 18 May Shift 2

A pair of dice is thrown 5 times. If getting a total of 7 is considered a success, then

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. Probability of no success	I. $26\left(\dfrac{1}{6}\right)^5$
B. Probability of 5 successes	II. $2\left(\dfrac{5}{6}\right)^5$
C. Probability of at least 4 successes	III. $\left(\dfrac{1}{6}\right)^5$
D. Probability of at most one success	IV. $\left(\dfrac{5}{6}\right)^5$

Choose the correct answer from the options given below:

2026: 25 May Shift 1

If $X$ is a random variable such that $X$ can take values 0,1, 2 or 3. Then the expectation of $X$ for the following data is :

$X$	0	1	2	3
$P(X)$	$k$	$k^2$	$1-5k^2$	$k^2$

(where $k>0$ )

2023: 23 May Shift 3

The variance of the Binomial Distribution $B\left(5, \frac{1}{4}\right)$ is:

2026: 12 May Shift 2

If $X$ is a Poisson variate such that $4P(X=1)=3P(X=2)$ , then $P(X=0)=$

2026: 25 May Shift 1

If a random variable X follows poisson's distribution such that $P(X=2) = 9P(X=4) + 90P(X=6)$ , then the mean of X is