If X is a Poisson variate such that P(X=3)=2P(X=4), then P(X=2) equals: [given e^(-2)=0.14] Option 1: 0.14 Option 2: 2 Option 3: -2 Option 4: 0.28

Answer: 0.28. Solution: For a Poisson variate, (e^(-)^3)/(3!) = 2(e^(-)^4)/(4!). This reduces to 1/6 = ()/12, so = 2. Then P(X=2) = (e^(-2) 4)/2 = 2e^(-2) = 2(0.14) = 0.28.

Applied Mathematics

Probability Distribution

Medium

applied

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

[given $e^{-2}=0.14$ ]

0.14

-2

0.28

✅ Correct Option: 4

For a Poisson variate, $\dfrac{e^{-\lambda}\lambda^3}{3!} = 2\cdot\dfrac{e^{-\lambda}\lambda^4}{4!}$ .

This reduces to $\dfrac{1}{6} = \dfrac{\lambda}{12}$ , so $\lambda = 2$ .

Then $P(X=2) = \dfrac{e^{-2}\cdot 4}{2} = 2e^{-2} = 2(0.14) = 0.28$ .

2026: 14 May Shift 2

In a random experiment, a collection of trials is called Bernoulli trials if

A. The number of trials is infinite.

B. The trials are independent of each other.

C. Each trial has exactly two outcomes, defined as success and failure.

D. The probability of success will vary in each trial.

Choose the correct answer from the options given below:

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In a binomial distribution the probability of getting success is $\frac{1}{4}$ And standard deviation is 3 then its mean is ?

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

$x_i$	0	1	2	3
$p_i$	$2k^2$	$k^2$	$3k^2$	$k$

Then the mean of X is

2026: 21 May Shift 2

A random variable X follows Poisson distribution. The variance of X is 2. The value of $P(X=3)$ is

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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.

2023: 15 June Shift 2

A discrete random variable X takes the values 0, 1, 2, 3, 4 and its mean is 1.6. If $P(X=1) = 0.4$ , $P(X=4) = P(X=2)$ and $P(X=3) = 2P(X=2)$ , then $P(X=0)$ is :

If XXX is a Poisson variate such that P(X=3)=2P(X=4)P(X=3)=2P(X=4)P(X=3)=2P(X=4), then P(X=2)P(X=2)P(X=2) equals: [given e−2=0.14e^{-2}=0.14e−2=0.14]

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

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

[given $e^{-2}=0.14$ ]