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CUET Mathematics 2025 PYQs22 May Shift 2. Free, no login required.

Algebra

past year questions, CUET Mathematics 2025 22 May Shift 2

Probability

past year questions, Algebra, CUET Mathematics 2025 22 May Shift 2

Medium

applied

Suppose X has Poisson distribution such that 3P(X=1)=2P(X=2)3 P(X=1) = 2 P(X=2) then P(X>0)P(X>0) is:

Solution

✅ Correct Option: 3

For a Poisson distribution with parameter λ:

P(X=k)=e−λ⋅λkk!P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!}

Given condition: 3P(X=1)=2P(X=2)3P(X=1) = 2P(X=2)

For X=1X = 1:

P(X=1)=e−λ⋅λ11!P(X=1) = \frac{e^{-\lambda} \cdot \lambda^1}{1!}

P(X=1)=λe−λP(X=1) = \lambda e^{-\lambda}

For X=2X = 2:

P(X=2)=e−λ⋅λ22!P(X=2) = \frac{e^{-\lambda} \cdot \lambda^2}{2!}

P(X=2)=λ2e−λ2P(X=2) = \frac{\lambda^2 e^{-\lambda}}{2}

Using the given condition:

3(λe−λ)=2(λ2e−λ2)3(\lambda e^{-\lambda}) = 2\left(\frac{\lambda^2 e^{-\lambda}}{2}\right)

3λe−λ=λ2e−λ3\lambda e^{-\lambda} = \lambda^2 e^{-\lambda}

Dividing both sides by λe−λ\lambda e^{-\lambda}:

3=λ3 = \lambda

Using the complement rule:

P(X>0)=1−P(X=0)P(X>0) = 1 - P(X=0)

For λ=3\lambda = 3:

P(X=0)=e−3⋅300!P(X=0) = \frac{e^{-3} \cdot 3^0}{0!}

P(X=0)=e−3⋅11P(X=0) = \frac{e^{-3} \cdot 1}{1}

P(X=0)=e−3P(X=0) = e^{-3}

Therefore:

P(X>0)=1−e−3P(X>0) = 1 - e^{-3}

Related questions:

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The random variable X has a probability distribution P(X) of the following form, where k is some number.

P(X=x)={k,if x=02k,if x=13k,if x=20,otherwiseP(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 P(x≤2)P(x \leq 2) is :

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Choose the correct answer from the options given below:

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The probability distribution of a discrete random variable X is given as :

x01234
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