If mean and variance of a Binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is

This question was previously asked in
NDA (Held On: 16 Dec 2015) Maths Previous Year paper
View all NDA Papers >

Answer (Detailed Solution Below)

Option 4 :
Free
UPSC NDA 01/2025 General Ability Full (GAT) Full Mock Test
150 Qs. 600 Marks 150 Mins

Detailed Solution

Download Solution PDF

Concept:

Binomial distribution The binomial distribution formula is:P (X = r) =

P(X) = probability of a success on an individual trial

p = probability of “success”

r = number of success

q = probability of “failure” OR q = 1- p

n = number of trials

  • Mean = np
  • Variance = npq

 

Calculation:

Given: mean = 2, variance = 1

We know p + q = 1

So, p = 1 - ½ = ½

mean = 2

⇒ np = 2

⇒ n = 2/p

= 4

We know P(required) = 1 – P (not required)

P (X > 1) = 1 – P (X = 0) – P (X = 1)

    (P (X = r) = )

         (∵  )

Hence, option (4) is correct

Latest NDA Updates

Last updated on Jul 8, 2025

->UPSC NDA Application Correction Window is open from 7th July to 9th July 2025.

->UPSC had extended the UPSC NDA 2 Registration Date till 20th June 2025.

-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.

->The NDA exam date 2025 has been announced. The written examination will be held on 14th September 2025.

-> The selection process for the NDA exam includes a Written Exam and SSB Interview.

-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100. 

-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential. 

More Binomial Distribution Questions

More Probability Questions

Hot Links: real teen patti teen patti apk download teen patti master update teen patti winner