X is a binomial variable such that 2P(X=2)=P(X=3) and Mean of X is known to be 10/3 . What would be the probability that X assumes at most the value 2 ?

Option b: 17/81. Solution: Given: 2P(X=2) = P(X=3) and Mean = np = 10/3 . Step 1: Use the condition 2P(X=2) = P(X=3) : 2 × (n choose 2) p2 qn-2 = (n choose 3) p3 qn-3 Dividing both sides: 2 × (n/2 choose )3 = /q (n/2 choose )3 = 3!(n-3)!/2!(n-2)! = 3/n-2 So: 6n-2 = Step 2: Also np = 103 . Try n = 5 : /q = 6/5-2 = 6/3 = 2 ⟹ p = 2q Since p + q = 1 : 2q + q = 1 ⟹ q = 13 , p = 23 Check: np = 5 × 23 = 103 ✓ Step 3: Find P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2) : P(X=0) = (1/3)5 = 1/243 P(X=1) = (5 choose 1)(2/3)1(1/3)4 = 5 × 2/3 × 1/81 = 10/243 P(X=2) = (5 choose 2)(2/3)2(1/3)3 = 10 × 4/9 × 1/27 = 40/243 P(X ≤ 2) = 1 + 10 + 40/243 = 51/243 = 17/81 Hence, Option B is the correct answer.

$X$ is a binomial variable such that $2P(X=2)=P(X=3)$ and Mean of $X$ is known to be $\\frac{10}{3}$. What would be the probability that $X$ assumes at most the value $2$?

The correct answer is Option b: $\\frac{17}{81}$. Given: $2P(X=2) = P(X=3)$ and Mean $= np = \\frac{10}{3}$. **Step 1:** Use the condition $2P(X=2) = P(X=3)$: $$2 \\cdot \\binom{n}{2} p^2 q^{n-2} = \\binom{n}{3} p^3 q^{n-3}$$ Dividing both sides: $$2 \\cdot \\frac{\\binom{n}{2}}{\\binom{n}{3}} = \\frac{p}{q}$$ $$\\frac{\\binom{n}{2}}{\\binom{n}{3}} = \\frac{3!(n-3)!}{2!(n-2)!} = \\frac{3}{n-2}$$ So: $\\dfrac{6}{n-2} = \\dfrac{p}{q}$ **Step 2:** Also $np = \\dfrac{10}{3}$. Try $n = 5$: $$\\frac{p}{q} = \\frac{6}{5-2} = \\frac{6}{3} = 2 \\Rightarrow p = 2q$$ Since $p + q = 1$: $2q + q = 1 \\Rightarrow q = \\dfrac{1}{3}$, $p = \\dfrac{2}{3}$ Check: $np = 5 \\times \\dfrac{2}{3} = \\dfrac{10}{3}$ ✓ **Step 3:** Find $P(X \\leq 2) = P(X=0) + P(X=1) + P(X=2)$: $$P(X=0) = \\left(\\frac{1}{3}\\right)^5 = \\frac{1}{243}$$ $$P(X=1) = \\binom{5}{1}\\left(\\frac{2}{3}\\right)^1\\left(\\frac{1}{3}\\right)^4 = 5 \\cdot \\frac{2}{3} \\cdot \\frac{1}{81} = \\frac{10}{243}$$ $$P(X=2) = \\binom{5}{2}\\left(\\frac{2}{3}\\right)^2\\left(\\frac{1}{3}\\right)^3 = 10 \\cdot \\frac{4}{9} \\cdot \\frac{1}{27} = \\frac{40}{243}$$ $$P(X \\leq 2) = \\frac{1 + 10 + 40}{243} = \\frac{51}{243} = \\frac{17}{81}$$ Hence, **Option B** is the correct answer.

Theoretical DistributionsMCQMTP Sep 24 Series IIQuestion 3480 of 230

All Questions

$\displaystyle X$ is a binomial variable such that $\displaystyle 2P(X=2)=P(X=3)$ and Mean of $\displaystyle X$ is known to be $\displaystyle \\frac{10}{3}$ . What would be the probability that $\displaystyle X$ assumes at most the value $\displaystyle 2$ ?

Options

\displaystyle \\frac{16}{81}

\displaystyle \\frac{17}{81}

\displaystyle \\frac{47}{243}

\displaystyle \\frac{46}{243}

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Correct Answer

✅ Option b — $\displaystyle \\frac{17}{81}$

All Options:

A $\displaystyle \\frac{16}{81}$
B $\displaystyle \\frac{17}{81}$
C $\displaystyle \\frac{47}{243}$
D $\displaystyle \\frac{46}{243}$

Detailed Solution & Explanation

Given:

\displaystyle 2P(X=2) = P(X=3)

and Mean

\displaystyle = np = \\frac{10}{3}

. **Step 1:** Use the condition

\displaystyle 2P(X=2) = P(X=3)

2 \\cdot \\binom{n}{2} p^2 q^{n-2} = \\binom{n}{3} p^3 q^{n-3}

Dividing both sides:

2 \\cdot \\frac{\\binom{n}{2}}{\\binom{n}{3}} = \\frac{p}{q}

\\frac{\\binom{n}{2}}{\\binom{n}{3}} = \\frac{3!(n-3)!}{2!(n-2)!} = \\frac{3}{n-2}

So:

\displaystyle \\dfrac{6}{n-2} = \\dfrac{p}{q}

**Step 2:** Also

\displaystyle np = \\dfrac{10}{3}

. Try

\displaystyle n = 5

\\frac{p}{q} = \\frac{6}{5-2} = \\frac{6}{3} = 2 \\Rightarrow p = 2q

Since

\displaystyle p + q = 1

\displaystyle 2q + q = 1 \\Rightarrow q = \\dfrac{1}{3}

\displaystyle p = \\dfrac{2}{3}

Check:

\displaystyle np = 5 \\times \\dfrac{2}{3} = \\dfrac{10}{3}

✓ **Step 3:** Find

\displaystyle P(X \\leq 2) = P(X=0) + P(X=1) + P(X=2)

P(X=0) = \\left(\\frac{1}{3}\\right)^5 = \\frac{1}{243}

P(X=1) = \\binom{5}{1}\\left(\\frac{2}{3}\\right)^1\\left(\\frac{1}{3}\\right)^4 = 5 \\cdot \\frac{2}{3} \\cdot \\frac{1}{81} = \\frac{10}{243}

P(X=2) = \\binom{5}{2}\\left(\\frac{2}{3}\\right)^2\\left(\\frac{1}{3}\\right)^3 = 10 \\cdot \\frac{4}{9} \\cdot \\frac{1}{27} = \\frac{40}{243}

P(X \\leq 2) = \\frac{1 + 10 + 40}{243} = \\frac{51}{243} = \\frac{17}{81}

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.

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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X\displaystyle XX is a binomial variable such that 2P(X=2)=P(X=3)\displaystyle 2P(X=2)=P(X=3)2P(X=2)=P(X=3) and Mean of X\displaystyle XX is known to be frac103\displaystyle \\frac{10}{3}frac103. What would be the probability that X\displaystyle XX assumes at most the value 2\displaystyle 22?