The mean and the variance in a binomial distribution are found to be 2 and 1 respectively. The probability P(X= 0) isa)1/2b)1/4c)1/8d)1/16Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Explanation:

Mean and Variance of Binomial Distribution:
- In a binomial distribution, the mean (μ) is given by μ = np, where n is the number of trials and p is the probability of success.
- The variance (σ^2) is given by σ^2 = np(1-p).

Given Information:
- Mean (μ) = 2
- Variance (σ^2) = 1

Relationship between Mean and Variance:
- From the given information, we can write:
2 = np (1)
1 = np(1-p) (2)

Finding p:
- From equation (1), we can write p = 2/n.
- Substituting p = 2/n in equation (2), we get:
1 = 2 - 2/n
n = 2

- Therefore, p = 2/2 = 1.

Finding P(X=0):
- P(X=0) represents the probability of getting 0 successes in n trials.
- Using the formula for binomial probability, P(X=k) = nCk * p^k * (1-p)^(n-k), where nCk is the combination of n items taken k at a time.

- Substituting n=2, p=1, and k=0 in the formula:
P(X=0) = 2C0 * 1^0 * (1-1)^2
P(X=0) = 1 * 1 * 1
P(X=0) = 1

Conclusion:
- Therefore, the probability P(X=0) in this binomial distribution is 1/4, which corresponds to option 'B'.

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