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Binomial with given mean and variance (or their sum and product): find n, p and a probability

This type of question, asked 6 times: 25 Jul 2022, 26 Jul 2022, 27 Jul 2022, 29 Jul 2022, 1 Feb 2023, 13 Apr 2023. All 7 questions below, oldest first.

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  1. Question 1

    25 Jul 2022

    76
    If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is:

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  2. Question 2

    26 Jul 2022

    76
    The mean and variance of a binomial distribution are α\alpha and α3\frac{\alpha}{3} respectively. If P(X=1)=4243P(X = 1) =\frac{4}{243}, then P(X=4P(X = 4 or 5)) is equal to:

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  3. Question 3

    26 Jul 2022

    168
    Let XX be a binomially distributed random variable with mean 4 and variance 43\frac{4}{3}. Then 54P(X≤2)54P(X \leq 2) is equal to

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  4. Question 4

    27 Jul 2022

    166
    Let XX have a binomial distribution B(n,p)B(n,p) such that the sum and the product of the mean and variance of XX are 24 and 128 respectively. If P(X>n−3)=k2nP(X > n - 3) =\frac{k}{2^{n}}, then kk is equal to

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  5. Question 5

    29 Jul 2022

    171
    The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. Then the number of trials in the binomial distribution is:
  6. Question 6

    1 Feb 2023

    66
    In a binomial distribution B(n,p)B(n,p), the sum and the product of the mean and the variance are 5 and 6 respectively, then 6(n+p−q)6(n + p - q) is equal to

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  7. Question 7

    13 Apr 2023

    164
    The random variable XX follows binomial distribution BB (n,p)(n,p) for which the difference of the mean and the variance is 1 . If 2P(X=2)=3P(X=1)2P(X = 2) = 3P(X = 1), then n2P(X>1)n^{2}P(X > 1) is equal to

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