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The mean and variance of binomial distribution b(x:n,p) are 4 and 4/3 respectively. What is the probability of getting 2 successes?
- a)20
- b)20/243
- c)1/243
- d)40/243
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The mean and variance of binomial distribution b(x:n,p) are 4 and4/3 r...
Calculation of Probability of getting 2 successes
Given:
Mean (μ) = 4
Variance (σ^2) = 4/3
Formulae:
Mean (μ) = np
Variance (σ^2) = np(1-p)
Step 1:
From the given mean and variance, we get:
4 = n * p
4/3 = n * p * (1-p)
Step 2:
Solving the above equations simultaneously, we get:
n = 4/p
4/3 = 4/p * (1-p)
Step 3:
Solving the second equation, we get:
4/3 = 4 - 4p
4p = 4 - 4/3
4p = 8/3
p = 2/3
Step 4:
Substitute p = 2/3 in n = 4/p, we get:
n = 4 / (2/3)
n = 6
Step 5:
Using the probability mass function of the binomial distribution:
P(X = 2) = nC2 * p^2 * (1-p)^(n-2)
P(X = 2) = 6C2 * (2/3)^2 * (1/3)^(6-2)
P(X = 2) = 15 * (4/9) * (1/81)
P(X = 2) = 20/243
Therefore, the probability of getting 2 successes is 20/243. Hence, the correct answer is option B.
Given:
Mean (μ) = 4
Variance (σ^2) = 4/3
Formulae:
Mean (μ) = np
Variance (σ^2) = np(1-p)
Step 1:
From the given mean and variance, we get:
4 = n * p
4/3 = n * p * (1-p)
Step 2:
Solving the above equations simultaneously, we get:
n = 4/p
4/3 = 4/p * (1-p)
Step 3:
Solving the second equation, we get:
4/3 = 4 - 4p
4p = 4 - 4/3
4p = 8/3
p = 2/3
Step 4:
Substitute p = 2/3 in n = 4/p, we get:
n = 4 / (2/3)
n = 6
Step 5:
Using the probability mass function of the binomial distribution:
P(X = 2) = nC2 * p^2 * (1-p)^(n-2)
P(X = 2) = 6C2 * (2/3)^2 * (1/3)^(6-2)
P(X = 2) = 15 * (4/9) * (1/81)
P(X = 2) = 20/243
Therefore, the probability of getting 2 successes is 20/243. Hence, the correct answer is option B.
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Community Answer
The mean and variance of binomial distribution b(x:n,p) are 4 and4/3 r...
Given,
Binomial distribution b(x:n,p)
So, Mean = np - 4
= 6
= 20/243
So, the probability of getting 2 successes is 20/243.
Hence, the correct option is (B).
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The mean and variance of binomial distribution b(x:n,p) are 4 and4/3 respectively. What is the probability of getting 2 successes?a)20b)20/243c)1/243d)40/243Correct answer is option 'B'. Can you explain this answer?
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