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Let X follows binomial distribution with mean and variance are 4 and . What is probability of (x ≥ 1) ?
Question

Let X follows binomial distribution with mean and variance are 4 and . What is probability of (x ≥ 1) ?

A.

0.00137

B.

0.71256

C.

0.81343

D.

0.99863

Correct option is D

The correct answer is (d).

Introduction 

The binomial distribution measures the probability of successes in a sequence of independent experiments.

For this given problem, the correct option mathematically evaluates the probability of getting at least one success by subtracting the probability of zero successes from the total probability of one.

Using the properties of mean and variance, we derive the necessary parameters to find the exact probability.

Information Booster
  • The binomial distribution is a discrete probability distribution applicable for binary outcomes with success or failure.
  • The mean of a binomial distribution is given by the formula μ=np\mu = np​​
  • The variance of a binomial distribution is given by σ2=npq \sigma^2 = npq​, where q = 1 - p.
  • The probability mass function is P(X=x)=(nx)pxqnxP(X=x) = \binom{n}{x} p^x q^{n-x}​​
  • Assuming the missing variance is 4/3 based on the correct answer, we get p=2/3,q=1/3p = 2/3, q = 1/3​, and n = 6.
  • To find P(X1)P(X \ge 1)​, it is mathematically simpler to calculate 1 - P(X = 0).
  • Substituting the derived values gives 1(1/3)61 - (1/3)^6​, resulting in approximately 0.99863.
Additional Knowledge
  • Option A (0.00137): This represents the probability of getting exactly zero successes, which is P(X=0).
  • Option B (0.71256): Represents a miscalculated cumulative probability value under incorrect variance assumptions.
  • Option C (0.81343): An incorrect probability value that does not correspond to the binomial expansion for the derived values of n and p.

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