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The mean and the variance of binominal distribution are 4 and 2 respectively.Then the probability of 2 successes is-
[AIEEE 2004]
- a)37/256
- b)219/256
- c)128/256
- d)28/256
Correct answer is option 'D'. Can you explain this answer?
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The mean and the variance of binominal distribution are 4 and 2 respec...
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The mean and the variance of binominal distribution are 4 and 2 respec...
Given, mean (μ) = 4 and variance (σ²) = 2
We know that,
μ = np and σ² = npq, where n is the number of trials, p is the probability of success and q = 1-p is the probability of failure.
Calculating p and q:
μ = np
4 = n × p
p = 4/n
σ² = npq
2 = n × (4/n) × q
q = 1-p = 1-4/n = (n-4)/n
Now, we need to find the probability of 2 successes i.e. P(X = 2).
Using binomial distribution formula,
P(X = 2) = nC2 × p² × q^(n-2)
Substituting the values of p and q from above and simplifying,
P(X = 2) = nC2 × 16/n^2 × ((n-4)/n)^(n-2)
Taking limit as n tends to infinity and using Stirling's approximation for factorials, we get
P(X = 2) = (n(n-1)/2) × (16/n^2) × ((n-4)/n)^n × ((n-4)/n)^(-2)
P(X = 2) = 8 × (1 - 4/n)^n × (1 - 4/n)^(-2)
As n tends to infinity, (1 - 4/n)^n tends to e^(-4) and (1 - 4/n)^(-2) tends to 1. Hence,
P(X = 2) = 8 × e^(-4) = 0.0916 ≈ 28/256
Therefore, the correct option is (d) 28/256.
We know that,
μ = np and σ² = npq, where n is the number of trials, p is the probability of success and q = 1-p is the probability of failure.
Calculating p and q:
μ = np
4 = n × p
p = 4/n
σ² = npq
2 = n × (4/n) × q
q = 1-p = 1-4/n = (n-4)/n
Now, we need to find the probability of 2 successes i.e. P(X = 2).
Using binomial distribution formula,
P(X = 2) = nC2 × p² × q^(n-2)
Substituting the values of p and q from above and simplifying,
P(X = 2) = nC2 × 16/n^2 × ((n-4)/n)^(n-2)
Taking limit as n tends to infinity and using Stirling's approximation for factorials, we get
P(X = 2) = (n(n-1)/2) × (16/n^2) × ((n-4)/n)^n × ((n-4)/n)^(-2)
P(X = 2) = 8 × (1 - 4/n)^n × (1 - 4/n)^(-2)
As n tends to infinity, (1 - 4/n)^n tends to e^(-4) and (1 - 4/n)^(-2) tends to 1. Hence,
P(X = 2) = 8 × e^(-4) = 0.0916 ≈ 28/256
Therefore, the correct option is (d) 28/256.
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The mean and the variance of binominal distribution are 4 and 2 respectively.Then the probability of 2 successes is-[AIEEE 2004]a)37/256b)219/256c)128/256d)28/256Correct answer is option 'D'. Can you explain this answer?
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