It is Poisson variate such that P(X=1) = 0.7, P(X=2) = 0.3 , then P(X=0) =

Option b: e-6/7. Solution: For a Poisson distribution with parameter m , the probability mass function is given by: P(X = k) = -m mkk! Given the probabilities for X = 1 and X = 2 : P(X = 1) = e-m m = 0.7 P(X = 2) = -m m22! = -m m22 = 0.3 Dividing P(X = 2) by P(X = 1) to find the parameter m : P(X = 2)/P(X = 1) = -m m22e-m m = m/2 Substitute the given values: 0.3/0.7 = m/2 3/7 = m/2 m = 6/7 Now, we calculate P(X = 0) : P(X = 0) = -m m00! = e-m = e-6/7 Hence, Option B is the correct answer.

It is Poisson variate such that $P(X=1) = 0.7, P(X=2) = 0.3$, then $P(X=0) =$

The correct answer is Option b: $e^{-6/7}$. For a Poisson distribution with parameter $m$, the probability mass function is given by: $$P(X = k) = \frac{e^{-m} m^k}{k!}$$ Given the probabilities for $X = 1$ and $X = 2$: $$P(X = 1) = e^{-m} m = 0.7$$ $$P(X = 2) = \frac{e^{-m} m^2}{2!} = \frac{e^{-m} m^2}{2} = 0.3$$ Dividing $P(X = 2)$ by $P(X = 1)$ to find the parameter $m$: $$\frac{P(X = 2)}{P(X = 1)} = \frac{\frac{e^{-m} m^2}{2}}{e^{-m} m} = \frac{m}{2}$$ Substitute the given values: $$\frac{0.3}{0.7} = \frac{m}{2} \implies \frac{3}{7} = \frac{m}{2} \implies m = \frac{6}{7}$$ Now, we calculate $P(X = 0)$: $$P(X = 0) = \frac{e^{-m} m^0}{0!} = e^{-m} = e^{-6/7}$$ Hence, **Option B** is the correct answer.

Theoretical DistributionsMCQPYQ July 21Question 3493 of 230

All Questions

It is Poisson variate such that $\displaystyle P(X=1) = 0.7, P(X=2) = 0.3$ , then $\displaystyle P(X=0) =$

Options

\displaystyle e^{6/7}

\displaystyle e^{-6/7}

\displaystyle e^{7/3}

\displaystyle e^{-7/3}

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Correct Answer

✅ Option b — $\displaystyle e^{-6/7}$

All Options:

A $\displaystyle e^{6/7}$
B $\displaystyle e^{-6/7}$
C $\displaystyle e^{7/3}$
D $\displaystyle e^{-7/3}$

Detailed Solution & Explanation

For a Poisson distribution with parameter

\displaystyle m

, the probability mass function is given by:

P(X = k) = \frac{e^{-m} m^k}{k!}

Given the probabilities for

\displaystyle X = 1

and

\displaystyle X = 2

P(X = 1) = e^{-m} m = 0.7

P(X = 2) = \frac{e^{-m} m^2}{2!} = \frac{e^{-m} m^2}{2} = 0.3

Dividing

\displaystyle P(X = 2)

\displaystyle P(X = 1)

to find the parameter

\displaystyle m

\frac{P(X = 2)}{P(X = 1)} = \frac{\frac{e^{-m} m^2}{2}}{e^{-m} m} = \frac{m}{2}

Substitute the given values:

\frac{0.3}{0.7} = \frac{m}{2} \implies \frac{3}{7} = \frac{m}{2} \implies m = \frac{6}{7}

Now, we calculate

\displaystyle P(X = 0)

P(X = 0) = \frac{e^{-m} m^0}{0!} = e^{-m} = e^{-6/7}

Hence, **Option B** is the correct answer.

About This Chapter: Theoretical Distributions

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Binomial, Poisson, Normal Distribution

This chapter covers Binomial, Poisson, Normal Distribution and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

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It is Poisson variate such that P(X=1)=0.7,P(X=2)=0.3\displaystyle P(X=1) = 0.7, P(X=2) = 0.3P(X=1)=0.7,P(X=2)=0.3, then P(X=0)=\displaystyle P(X=0) =P(X=0)=