Question
Download Solution PDFComprehension
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) ^ 0
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
Given,
Let X be a random variable following a binomial distribution with parameters n = 6 and p = \(\frac{1}{4} \)
The probability mass function for a binomial distribution is:
\( P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x} \)
For \( P(X = 3) \), we have:
\( P(X = 3) = \binom{6}{3} \left( \frac{1}{4} \right)^3 \left( \frac{3}{4} \right)^3 \)
Substitute the values:
\( P(X = 3) = 20 \times \frac{1}{64} \times \frac{27}{64} \)
Now simplify the equation:
\( P(X = 3) = 20 \times \frac{27}{4096} = \frac{540}{4096} \)
Simplifying the fraction:
\( P(X = 3) = \frac{135}{1024} \)
Hence, the correct answer is Option 1
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.