Question

Consider a binomially distributed random variable constructed from a series of 8 trials with a 60% chance of success on any given trial. Calculate the probability that there are more than 3 successes.

Answer 1

Solution:

Given,

p = 60% = 0.60

1 - p = 1 - 0.60 = 0.40

n = 8

X follows the Binomial(n = 8 , p = 0.60)

The PMF of the Binomial distribution is

P(X = x) = (_n C _x) * p^x * (1 - p)^{n - x} ; x = 0 ,1 , 2 , ....., n

Now

P[ more than 3 successes]

= P[X > 3]

= 1 - { P[X 3] }

= 1 - {  P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) }

= 1 - { (₈ C ₀) * 0.60⁰ * (0.40)^{8 - 0} + (₈ C ₁) * 0.60¹ * (0.40)^{8 - 1} + (₈ C ₂) * 0.60² * (0.40)^{8 - 2} + (₈ C ₃) * 0.60³ * (0.40)^{8 - 3} }

= 1 - {0.00065536+0.00786432+0.04128768+0.12386304}

= 1 - { 0.1736704 }

= 0.8263296

P[ more than 3 successes] = 0.8263296