Theoretical DistributionsMCQMTP May 20Question 3443 of 230
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Find the probability of a success for the binomial distribution satisfying the following relation $4 P(x=4) = P(x=2)$ and having the parameter $n$ as six.

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option a — 1/3\displaystyle 1/3

All Options:

  • A1/3\displaystyle 1/3
  • B1/2\displaystyle 1/2
  • C1/5\displaystyle 1/5
  • D1/8\displaystyle 1/8

Detailed Solution & Explanation

**Finding Probability of Success: $4P(X=4) = P(X=2),\displaystyle ,n=6$** For a Binomial distribution with n=6\displaystyle n=6: P(X=r)=binom6rprq6−rP(X=r) = \(6r)\displaystyle \binom{6}{r} p^r q^{6-r} **Setting up the equation:** 4P(X=4)=P(X=2)4P(X=4) = P(X=2) 4cdotbinom64p4q2=binom62p2q44 \⋅\displaystyle \cdot \(64)\displaystyle \binom{6}{4} p^4 q^2 = \(62)\displaystyle \binom{6}{2} p^2 q^4 4times15cdotp4q2=15cdotp2q44 \×\displaystyle \times 15 \⋅\displaystyle \cdot p^4 q^2 = 15 \⋅\displaystyle \cdot p^2 q^4 60p4q2=15p2q460 p^4 q^2 = 15 p^2 q^4 **Dividing both sides by $15p^2q^2$:** 4p2=q24p^2 = q^2 2p=qquad(texttakingpositivesquareroot)2p = q \\quad (\taking positive square root\displaystyle \text{taking positive square root}) **Using } p+q=1\displaystyle p + q = 1:** p+2p=1p + 2p = 1 3p=1Rightarrowp=frac133p = 1 \\Rightarrow p = \13\displaystyle \frac{1}{3} **Verification:** - p=frac13, q=frac23\displaystyle p = \13\displaystyle \frac{1}{3},\ q = \23\displaystyle \frac{2}{3} - P(X=2)=binom62left(frac13right)2left(frac23right)4=15timesfrac19timesfrac1681=frac240729\displaystyle P(X=2) = \(62)\displaystyle \binom{6}{2}\\left(\13\displaystyle \frac{1}{3}\\right)^2\\left(\23\displaystyle \frac{2}{3}\\right)^4 = 15 \×\displaystyle \times \19\displaystyle \frac{1}{9} \×\displaystyle \times \1681\displaystyle \frac{16}{81} = \240729\displaystyle \frac{240}{729} - P(X=4)=binom64left(frac13right)4left(frac23right)2=15timesfrac181timesfrac49=frac60729\displaystyle P(X=4) = \(64)\displaystyle \binom{6}{4}\\left(\13\displaystyle \frac{1}{3}\\right)^4\\left(\23\displaystyle \frac{2}{3}\\right)^2 = 15 \×\displaystyle \times \181\displaystyle \frac{1}{81} \×\displaystyle \times \49\displaystyle \frac{4}{9} = \60729\displaystyle \frac{60}{729} - 4timesP(X=4)=frac240729=P(X=2)\displaystyle 4 \×\displaystyle \times P(X=4) = \240729=P\displaystyle \frac{240}{729} = P(X=2) ✓ Hence, **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.

Key Concepts to Understand

Probability

A numerical measure of the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Formula: P(A) = Number of favorable outcomes / Total number of equally likely outcomes.

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