Binomial distribution Practice Questions - NDA Mathematics Question Bank

Question 1 of 5 NDA MCQ

Consider a random variable X which follows Binomial distribution with parameters n = 10 and p = 1/5. Then Y = 10 − X follows Binomial distribution with parameters n and p respectively given by

Detailed Explanation

Solution

Option (d) is correct.

Explanation:

A binomial random variable \( X \) follows the distribution with parameters \( n = 10 \) and \( p = \frac{1}{5} \).

The complementary variable \( Y = 10 - X \) counts the number of failures in \( 10 \) trials:

\[ Y \sim \text{Binomial}(n, p') \]

Where \( p' = 1 - p \):

\[ p' = 1 - \frac{1}{5} = \frac{4}{5} \]

Thus, \( Y \) follows a Binomial distribution with:

\[ Y \sim \text{Binomial}(10, \frac{4}{5}) \]

Final parameters for \( Y \) are \( n = 10 \) and \( p = \frac{4}{5} \).

📌 Hints / Properties Used:

Binomial Distribution: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
Complementary probability: \( p' = 1 - p \)

📊 Visual Diagram Suggestion:

A diagram showing two binomial distributions for \( X \) and \( Y \) on a graph could illuminate how \( Y \) represents the distribution of failures while \( X \) represents successes.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Mathematics, Chapter 13 (Probability).

Question 2 of 5 NDA MCQ

For a Binomial distribution with mean 6 and standard deviation √2, what is the value of P(X = 0)?

Detailed Explanation

Solution

Option (a) is correct.

Explanation:

Given a binomial distribution with mean \( \mu \) and standard deviation \( \sigma \): \[ \mu = n \cdot p = 6 \] \[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{2} \] Squaring the standard deviation: \[ n \cdot p \cdot (1 - p) = 2 \] We have two equations: \[ 1. n \cdot p = 6 \quad (1) \] \[ 2. n \cdot p \cdot (1 - p) = 2 \quad (2) \] From equation (1), we can express \( n \): \[ n = \frac{6}{p} \] Substituting into equation (2): \[ \frac{6}{p} \cdot p \cdot \left( 1 - p \right) = 2 \] \[ 6(1 - p) = 2 \] \[ 6 - 6p = 2 \] \[ 6p = 4 \\ p = \frac{2}{3} \] Now substituting \( p \) back to find \( n \): \[ n = \frac{6}{\frac{2}{3}} = 9 \] Thus, \( n = 9 \) and \( p = \frac{2}{3} \). Now calculating \( P(X = 0) \): \[ P(X = 0) = (1 - p)^n = \left( \frac{1}{3} \right)^9 \] \[ = \frac{1}{19683} \] In conclusion: \[ P(X = 0) = \left( \frac{1}{3} \right)^9 \quad (Final) \]

📌 Hints / Properties Used:

Binomial Mean: \( \mu = n \cdot p \)
Binomial Variance: \( \sigma^2 = n \cdot p \cdot (1 - p) \)
Probability for Binomial Distribution: \( P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} \)

📊 Visual Diagram Suggestion:

A flowchart depicting how to compute the mean and standard deviation for a binomial distribution, along with probability calculations would aid understanding.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Mathematics, Chapter on Probability.

Question 3 of 5 NDA MCQ

Let X be a random variable following binomial distribution with parameters n = 5 and p = k. Further, P(X = 1)= 0.4096 and P(X = 2) = 0.2048. What is the value of k?

Detailed Explanation

Question 4 of 5 NDA MCQ

Under which of the following conditions may binomial distribution be used?

I. The number of trials is infinite and not fixed.

II. The trials are independent. III. Each trial has two possible outcomes.

Select the correct answer using the code given below.

Detailed Explanation

Explanation: In binomial distribution number

trial is finite and fixed and independent. Each

trial has two possible outcomes (success and

failure).

So, statement II and III are correct.

Question 5 of 5 NDA MCQ

Direction for Questions (97-98): Consider the following for the two (02) items that follow: Let X be a random variable following binomial distribution with parameters n = 6 and p = k. Further, 9P(X = 4) = P(X = 2).

What is the value of P(X = 3)?

Detailed Explanation

Binomial distribution

Solution

Solution

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