In a binomial distribution B(n, p) , n = 4 , P(X=2) = 3P(X=3) find p

Option a: 1/3. Solution: Finding p in B(4,p) given P(X=2) = 3P(X=3) Given: n = 4 , P(X=2) = 3P(X=3) Step 1: Write out the binomial probabilities P(X=2) = (4 choose 2) p2 q2 = 6p2q2 P(X=3) = (4 choose 3) p3 q1 = 4p3q Step 2: Apply the given condition P(X=2) = 3 × P(X=3) 6p2q2 = 3 × 4p3q 6p2q2 = 12p3q Step 3: Simplify (dividing both sides by 6p2q , assuming p ≠ 0, q ≠ 0 ) q = 2p Step 4: Use p + q = 1 p + 2p = 1 3p = 1 p = 1/3 Therefore q = 2/3 . Verification: - P(X=2) = 6 × (1/3)2 × (2/3)2 = 6 × 1/9 × 4/9 = 24/81 - P(X=3) = 4 × (1/3)3 × 2/3 = 4 × 1/27 × 2/3 = 8/81 - 3 × P(X=3) = 24/81 = P(X=2) ✓ So p = 1/3 . The textbook lists this as option B ( 2/3 ) in its correctoption field, but the mathematical derivation gives p = 1/3 . Hence, Option A is the correct answer.

In a binomial distribution $B(n, p)$, $n = 4$, $P(X=2) = 3P(X=3)$ find $p$

The correct answer is Option a: $1/3$. **Finding $p$ in $B(4,p)$ given $P(X=2) = 3P(X=3)$** **Given:** $n = 4$, $P(X=2) = 3P(X=3)$ **Step 1: Write out the binomial probabilities** $$P(X=2) = \binom{4}{2} p^2 q^2 = 6p^2q^2$$ $$P(X=3) = \binom{4}{3} p^3 q^1 = 4p^3q$$ **Step 2: Apply the given condition** $$P(X=2) = 3 \cdot P(X=3)$$ $$6p^2q^2 = 3 \times 4p^3q$$ $$6p^2q^2 = 12p^3q$$ **Step 3: Simplify** (dividing both sides by $6p^2q$, assuming $p \neq 0, q \neq 0$) $$q = 2p$$ **Step 4: Use $p + q = 1$** $$p + 2p = 1$$ $$3p = 1$$ $$p = \\frac{1}{3}$$ Therefore $q = \\frac{2}{3}$. **Verification:** - $P(X=2) = 6 \times \left(\\frac{1}{3}\right)^2 \times \left(\\frac{2}{3}\right)^2 = 6 \times \\frac{1}{9} \times \\frac{4}{9} = \\frac{24}{81}$ - $P(X=3) = 4 \times \left(\\frac{1}{3}\right)^3 \times \\frac{2}{3} = 4 \times \\frac{1}{27} \times \\frac{2}{3} = \\frac{8}{81}$ - $3 \times P(X=3) = \\frac{24}{81} = P(X=2)$ ✓ So $p = \\frac{1}{3}$. The textbook lists this as option B ($\\frac{2}{3}$) in its correct_option field, but the mathematical derivation gives $p = \\frac{1}{3}$. Hence, **Option A** is the correct answer.

Theoretical DistributionsMCQPYQ Nov 20Question 3420 of 230

All Questions

In a binomial distribution $\displaystyle B(n, p)$ , $\displaystyle n = 4$ , $\displaystyle P(X=2) = 3P(X=3)$ find $\displaystyle p$

Options

\displaystyle 1/3

\displaystyle 2/3

\displaystyle 6/4

\displaystyle 4/3

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Correct Answer

✅ Option a — $\displaystyle 1/3$

All Options:

A $\displaystyle 1/3$
B $\displaystyle 2/3$
C $\displaystyle 6/4$
D $\displaystyle 4/3$

Detailed Solution & Explanation

**Finding

\displaystyle p

\displaystyle B(4,p)

given

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

** **Given:**

\displaystyle n = 4

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

**Step 1: Write out the binomial probabilities**

P(X=2) = \binom{4}{2} p^2 q^2 = 6p^2q^2

P(X=3) = \binom{4}{3} p^3 q^1 = 4p^3q

**Step 2: Apply the given condition**

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

6p^2q^2 = 3 \times 4p^3q

6p^2q^2 = 12p^3q

**Step 3: Simplify** (dividing both sides by

\displaystyle 6p^2q

, assuming

\displaystyle p \neq 0, q \neq 0

)

q = 2p

**Step 4: Use

\displaystyle p + q = 1

p + 2p = 1

3p = 1

p = \\frac{1}{3}

Therefore

\displaystyle q = \\frac{2}{3}

. **Verification:** -

\displaystyle P(X=2) = 6 \times \left(\\frac{1}{3}\right)^2 \times \left(\\frac{2}{3}\right)^2 = 6 \times \\frac{1}{9} \times \\frac{4}{9} = \\frac{24}{81}

\displaystyle P(X=3) = 4 \times \left(\\frac{1}{3}\right)^3 \times \\frac{2}{3} = 4 \times \\frac{1}{27} \times \\frac{2}{3} = \\frac{8}{81}

\displaystyle 3 \times P(X=3) = \\frac{24}{81} = P(X=2)

✓ So

\displaystyle p = \\frac{1}{3}

. The textbook lists this as option B (

\displaystyle \\frac{2}{3}

) in its correct_option field, but the mathematical derivation gives

\displaystyle p = \\frac{1}{3}

. 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.

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In a binomial distribution B(n,p)\displaystyle B(n, p)B(n,p), n=4\displaystyle n = 4n=4, P(X=2)=3P(X=3)\displaystyle P(X=2) = 3P(X=3)P(X=2)=3P(X=3) find p\displaystyle pp