Applied Mathematics

Probability Distribution

Medium

If the mean of a binomial distribution is 24 and its standard deviation is 4 , then the probability of getting success is :

2/3

1/5

1/6

1/3

✅ Correct Option: 4

For binomial distribution: mean $np = 24$ and variance $npq = 16$ (since SD = 4).

Dividing: $\frac{npq}{np} = q = \frac{16}{24} = \frac{2}{3}$ .

Probability of success $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$ .

2026: 30 May Shift 2

Match List-I with List-II

List-I	List-II
(A) Mean of a normal variate X is 12 and standard deviation is 4, then the Z-score of data point 20 is	(I) 1
(B) If X is Poisson variate such that P(X=1) = 2P(X=2), then the mean of X is	(II) 8
(C) In a binomial distribution, if mean is 5 and variance is 4, then the number of trials is	(III) 2
(D) In a binomial distribution, if n = 20 and q = 0.6, then its mean is	(IV) 25

Choose the correct answer from the options given below:

2023: 11 Aug Shift 1

A random variable X has a probability distribution P(X) of the following form, where k is some unknown constant:

P(X = 0) = k

P(X = 1) = 2k

P(X = 2) = 3k

P(X = other values) = 0

Then, find the value of 1/k.

2022: 25 May Shift 1

Match List I with List II

LIST I	LIST II
A. The variance of a Poisson distribution with mean $\lambda$ is	I. $\sqrt{\lambda}$
B. The standard deviation of a Poisson distribution with mean $\lambda$ is	II. 4
C. In a Poisson distribution, if mean is 4, then the standard deviation is	III. $\lambda$
D. In a Poisson distribution, if mean is 4, then the variance is	IV. 2

Choose the correct answer from the options given below:

2023: 23 May Shift 3

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

2026: 29 May Shift 1

A random variable X has the following probability distribution:

X	0	1	2	3	4
P(X)	0.1	0.15	0.30	0.30	0.15

Which of the following are correct?

(A) $P(X \le 1)=0.25$

(B) $P(X>3)=0.15$

(D) $P(X \le 2)=0.55$

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 $k>0$ )

Solution