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)=8/e^4 D. P(X=3)=32/e^4 Choose the correct answer from the options given below: Option 1: A and C only Option 2: B and D only Option 3: B and C only Option 4: A and D only

Answer: A and C only. Solution: For a Poisson variable, 2 e^(-)=(^2e^(-))/2 gives =4, so the mean is 4 (A true, B false). Then P(X=2)=(16e^(-4))/2=8/e^4 (C true), while P(X=3)=(64e^(-4))/6=32/3e^4, so D is false.

Applied Mathematics

Probability Distribution

Medium

applied

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:

A and C only

B and D only

B and C only

A and D only

✅ Correct Option: 1

For a Poisson variable, $2\lambda e^{-\lambda}=\frac{\lambda^2e^{-\lambda}}{2}$ gives $\lambda=4$ , so the mean is $4$ (A true, B false).

Then $P(X=2)=\frac{16e^{-4}}{2}=\frac{8}{e^4}$ (C true), while $P(X=3)=\frac{64e^{-4}}{6}=\frac{32}{3e^4}$ , so D is false.

2026: 18 May Shift 2

For 6 trials of an experiment, let X be a binomial variate which satisfies the relation $9P(X=4)=P(X=2)$ , then the probability of success is

2026: 29 May Shift 1

If X is a Poisson variate such that $3P(X=2)=2P(X=1)$ , then $P(X=0)$ is equal to: (Given $e^{-4/3}=0.264$ )

2023: 11 Aug Shift 1

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

The value of k is

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 $k>0$ )

2026: 12 May Shift 2

If a coin is tossed 8 times and getting a head is considered as success, then which of the following statements are correct ?

(A) The probability of getting 4 heads is $\dfrac{35}{128}$

(B) The mean of the number of successes is 4.

(D) The probability of getting 6 heads is $\dfrac{9}{64}$ .

Choose the correct answer from the options given below:

2023: 7 June Shift 1

Match list I with list II. 4 defective pens are mixed with 10 normal pens. 3 pens are drawn one by one with replacement , then the probability distribution of the number of defective pens is :

List-I	List - II
A. P(X=0)	8/343
B. P(X=1)	60/343
C. P(X=2)	125/343
D. P(X=3)	150/343

Choose the correct option below :

Solution

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