Theoretical DistributionsMCQMTP Nov 19Question 3441 of 230
All Questions

If X\displaystyle X & Y\displaystyle Y are two independent variables such that XsimB(n1,p)\displaystyle X \\sim B(n_1, p) and YsimB(n2,p)\displaystyle Y \\sim B(n_2, p) then the parameter of Z=X+Y\displaystyle Z = X+Y is

Options

A(n1+n2),p\displaystyle (n_1 + n_2), p
B(n1−n2),p\displaystyle (n_1 - n_2), p
C(n1+n2),2p\displaystyle (n_1 + n_2), 2p
DNone of these
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — (n1+n2),p\displaystyle (n_1 + n_2), p

All Options:

  • A(n1+n2),p\displaystyle (n_1 + n_2), p
  • B(n1−n2),p\displaystyle (n_1 - n_2), p
  • C(n1+n2),2p\displaystyle (n_1 + n_2), 2p
  • DNone of these

Ad

Detailed Solution & Explanation

**Sum of Two Independent Binomial Variables**\n\nGiven: XsimB(n1,p)\displaystyle X \\sim B(n_1, p) and YsimB(n2,p)\displaystyle Y \\sim B(n_2, p), both with the **same probability** p\displaystyle p.\n\n**Reproductive Property of Binomial Distribution:**\nIf X\displaystyle X and Y\displaystyle Y are independent binomial variables with the **same** p\displaystyle p, then their sum Z=X+Y\displaystyle Z = X + Y is also binomially distributed:\nZ=X+YsimB(n1+n2,;p)Z = X + Y \\sim B(n_1 + n_2, \\; p)\n\n**Verification using moments:**\n- E(Z)=E(X)+E(Y)=n1p+n2p=(n1+n2)p\displaystyle E(Z) = E(X) + E(Y) = n_1 p + n_2 p = (n_1 + n_2)p\n- textVar(Z)=textVar(X)+textVar(Y)=n1pq+n2pq=(n1+n2)pq\displaystyle \\text{Var}(Z) = \\text{Var}(X) + \\text{Var}(Y) = n_1 pq + n_2 pq = (n_1 + n_2)pq\n\nThese match a B(n1+n2,p)\displaystyle B(n_1 + n_2, p) distribution. The probability parameter remains p\displaystyle p, **not** 2p\displaystyle 2p.\n\n> **Note:** Option C lists (n1+n2),2p\displaystyle (n_1 + n_2), 2p which is **incorrect** — the probability stays p\displaystyle p. The correct parameters are (n1+n2,p)\displaystyle (n_1 + n_2, p) = Option A.\n\nHence, **Option A** 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.

More Questions from Theoretical Distributions

Find the variance of binomial distribution with n=10\displaystyle n=10, p=0.3\displaystyle p=0.3

When p=0.5\displaystyle p=0.5, the binomial distribution is

If mean and standard deviation of a binomial distribution is 10\displaystyle 10 and 2\displaystyle 2 respectively, q\displaystyle q will be

A random variable X\displaystyle X follows Binomial Distribution With E(x)=2\displaystyle E(x)=2 and V(x)=1.2\displaystyle V(x)=1.2, then the value of n\displaystyle n is

When 'p\displaystyle p' is large than 0.5\displaystyle 0.5, the Binomial Distribution is:

For a Poisson distribution,

Ready to Master Theoretical Distributions?

Practice all 230 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free