Question

for a binomial distribution mean is 4 and variance is 2. find probability of getting atleast 2 successes.

Answer 1

Mean = n p = 4

Variance = n p q = 2

Dividing them, q = 1/2

So, p = 1/2

So, n = 8

Probability of getting atleast two successes = 1 - {P(x = 0) + P(x = 1) }

P(x = 0) = 8 C0 / 2^(8) = 1/2^(8)

P(x = 1) = 8C1/ (2^8) = 8 /2^(8)

P(x = 0) + P(x = 1) = 9/2^(8) = 9/256

Probability of getting atleast two successes = 1 - (9/256) = 247 / 256