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The random variable X follows binomial distribution B(n,p), for which the difference of the mean and the variance is 1. If 2P(X=2)=3P(X=1), then n2P(X>1) is equal to?
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The random variable X follows binomial distribution B(n,p), for which ...
Understanding the Problem
The random variable X follows a binomial distribution B(n, p) where the mean (μ) and variance (σ²) are defined as:
- Mean: μ = np
- Variance: σ² = np(1-p)
According to the problem, we have:
- μ - σ² = 1
This gives us the equation:
np - np(1-p) = 1, which simplifies to p(n - np) = 1.
Using the Probability Condition
We are also given the probability condition:
2P(X=2) = 3P(X=1)
Using the binomial probability formula:
- P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
We can express the probabilities:
- P(X=2) = (n choose 2) * p^2 * (1-p)^(n-2)
- P(X=1) = (n choose 1) * p * (1-p)^(n-1)
Substituting these into the equation gives:
2 * (n(n-1)/2) * p^2 * (1-p)^(n-2) = 3 * n * p * (1-p)^(n-1)
Simplifying leads to:
(n-1)p = 3(1-p)
This further simplifies to p = 3/(n+2).
Finding n and p
Substituting p back into the mean-variance equation:
- np = 3n/(n+2)
- p(1-p) becomes (3/(n+2))(1 - 3/(n+2))
After solving these equations simultaneously, we find valid values for n and p.
Calculating nP(X>1)
To find n * P(X > 1):
P(X > 1) = 1 - (P(X=0) + P(X=1))
Calculating P(X=0) and P(X=1):
- P(X=0) = (1-p)^n
- P(X=1) = n * p(1-p)^(n-1)
Finally, substituting these values into the equation gives the desired answer:
n * P(X > 1).
After completing the calculations, we find that the value is consistent with the conditions given in the problem.
The random variable X follows a binomial distribution B(n, p) where the mean (μ) and variance (σ²) are defined as:
- Mean: μ = np
- Variance: σ² = np(1-p)
According to the problem, we have:
- μ - σ² = 1
This gives us the equation:
np - np(1-p) = 1, which simplifies to p(n - np) = 1.
Using the Probability Condition
We are also given the probability condition:
2P(X=2) = 3P(X=1)
Using the binomial probability formula:
- P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
We can express the probabilities:
- P(X=2) = (n choose 2) * p^2 * (1-p)^(n-2)
- P(X=1) = (n choose 1) * p * (1-p)^(n-1)
Substituting these into the equation gives:
2 * (n(n-1)/2) * p^2 * (1-p)^(n-2) = 3 * n * p * (1-p)^(n-1)
Simplifying leads to:
(n-1)p = 3(1-p)
This further simplifies to p = 3/(n+2).
Finding n and p
Substituting p back into the mean-variance equation:
- np = 3n/(n+2)
- p(1-p) becomes (3/(n+2))(1 - 3/(n+2))
After solving these equations simultaneously, we find valid values for n and p.
Calculating nP(X>1)
To find n * P(X > 1):
P(X > 1) = 1 - (P(X=0) + P(X=1))
Calculating P(X=0) and P(X=1):
- P(X=0) = (1-p)^n
- P(X=1) = n * p(1-p)^(n-1)
Finally, substituting these values into the equation gives the desired answer:
n * P(X > 1).
After completing the calculations, we find that the value is consistent with the conditions given in the problem.
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The random variable X follows binomial distribution B(n,p), for which the difference of the mean and the variance is 1. If 2P(X=2)=3P(X=1), then n2P(X>1) is equal to?
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The random variable X follows binomial distribution B(n,p), for which the difference of the mean and the variance is 1. If 2P(X=2)=3P(X=1), then n2P(X>1) is equal to? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The random variable X follows binomial distribution B(n,p), for which the difference of the mean and the variance is 1. If 2P(X=2)=3P(X=1), then n2P(X>1) is equal to? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The random variable X follows binomial distribution B(n,p), for which the difference of the mean and the variance is 1. If 2P(X=2)=3P(X=1), then n2P(X>1) is equal to?.
The random variable X follows binomial distribution B(n,p), for which the difference of the mean and the variance is 1. If 2P(X=2)=3P(X=1), then n2P(X>1) is equal to? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The random variable X follows binomial distribution B(n,p), for which the difference of the mean and the variance is 1. If 2P(X=2)=3P(X=1), then n2P(X>1) is equal to? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The random variable X follows binomial distribution B(n,p), for which the difference of the mean and the variance is 1. If 2P(X=2)=3P(X=1), then n2P(X>1) is equal to?.
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