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Binomial Distribution - JEE Chapter Free MCQ Test with solutions


MCQ Practice Test & Solutions: Test: Binomial Distribution (10 Questions)

You can prepare effectively for JEE Chapter-wise Tests for JEE Main & Advanced with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Binomial Distribution ". These 10 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 10

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Test: Binomial Distribution - Question 1
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The variance of the distribution is

​

Detailed Solution: Question 1

Mean= E(X) = 
= 0×0.3+1×0.7
=0.7

= 02×0.3+12×0.7 =0.7
Now, ∴ Var(X)= E(X2)-(E(X))2
= 0.7−(0.7)2
= 0.7 − 0.49
= 0.21

Test: Binomial Distribution - Question 2
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A die is tossed twice. Getting a number greater than 4 is considered a success. Then the variance of the probability distribution of the number of successes is:​

Detailed Solution: Question 2

In a single toss, P(success) = 2/6 = 1/3 and P(non-success) = (1 − 1/3) = 2/3.
P(X=0) = P(non-success in the 1st draw and non-success in the second)
(2/3 × 2/3) = 4/9.
P(X=1) = P(success in the 1st toss and non-success in the 2nd) or (non-success in the 1st toss and success in the 2nd)]
(1/3 x 2/3) + (2/3 × 1/3) = 4/9.

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Test: Binomial Distribution - Question 3
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A man make attempts to hit the target. The probability of hitting the target is 3/5 . Then the probability that A hit the target exactly 2 times in 5 attempts is:​

Detailed Solution: Question 3

Probability of success(p) = ⅗
q = 1 - p
= 1 - (⅗)  ⇒ ⅖
Probability of 2 hit in 5 attempts = 5C2 (⅗)2 (⅖)3 
= 144/625

Test: Binomial Distribution - Question 4
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Trials of a random experiment are called Bernoulli trials, if they satisfy the condition/s:​

Test: Binomial Distribution - Question 5
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In binomial probability distribution, mean is 3 and standard deviation is 3/2 . Then the probability distribution is:

Detailed Solution: Question 5

For a binomial distribution:

Mean = n p
Standard deviation = √(n p q), where q = 1 − p

Given
Mean = 3
Standard deviation = 3/2

Step 1: Use mean
n p = 3 …(1)

Step 2: Use standard deviation
√(n p q) = 3/2
⇒ n p q = 9/4

From (1), n p = 3
So, 3q = 9/4
⇒ q = 3/4
⇒ p = 1/4

Step 3: Find n
n p = 3
⇒ n × 1/4 = 3
⇒ n = 12

So the binomial probability distribution is:

(p + q)n = (1/4 + 3/4)12

Correct option: (c)

Test: Binomial Distribution - Question 6
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A random variable is a real valued function whose domain is the.

Test: Binomial Distribution - Question 7
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In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is​

Detailed Solution: Question 7

Probability of getting first non-defective bulb =90/100
Probablity of getting second non-defective bulb =89/99
Probablity of getting third non-defective bulb =88/98
Probablity of getting fourth non-defective bulb =87/97
Probablity of getting fifth non-defective bulb =86/96
So, probablity of getting all non-defective bulbs in a sample of 5 bulbs =
(90/100)∗(89/99)∗(88/98)∗(87/97)∗(86/96)
= closest option is b

Test: Binomial Distribution - Question 8
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The mean and variance of a binomial distribution are 4 and 3 respectively, then the probability of getting exactly six successes in this distribution is:

Detailed Solution: Question 8

Let X˜B(n,p) be a binomial variate with mean 4 and variance 3. Then,
np=4,and npq=3
⇒q=3/4,p=1/4 and n=16
∴P(X=r)=.16Cr(1/4)r (3/4)(16−r) ,r=0,1,2,....,16
⇒P(X=6)= 16C6(1/4)6 ((3)/(4))10

Test: Binomial Distribution - Question 9
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In a box of 10 electric bulbs, two are defective. Two bulbs are selected at random one after the other from the box. The first bulb is put back in the box before making the second selection. The probability that both the bulbs are not defective is:​

Test: Binomial Distribution - Question 10
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Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. The probability that there is at least one defective egg is:

Detailed Solution: Question 10

Probability of an egg being defective =10/100=110
So, probability of an egg being non-defective=1−0.1=0.9
10 eggs are drawn successively with replacement.
So, the probability of getting no defective egg =(0.9)10
Hence, the probability that there is at least one defective egg = 1−(0.9)10

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About this JEE Practice Test

This test contains 10 MCQs on Binomial Distribution with detailed solutions for each question. It is part of the Chapter-wise Tests for JEE Main & Advanced course on EduRev, designed to align with the current JEE syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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