Question

Assume that a procedure yields a binomial distribution with nequals5 trials and a probability of success of pequals0.05. Use a binomial probability table to find the probability that the number of successes x is exactly 3.

Answer 1

Let X be a binomial random variable with n = 5 and p = 0.05

X ~ Binomial ( n = 5, p = 0.05)

Probability mass function of X is,

P(X = x) = ⁿC_x p^x (1 - p)^n-x

We want to find, P(X = 3)

P(X = 3)

= ⁵C₃ * (0.05)³ * (1 - 0.05)^5-3

= 10 * (0.05)³ * (0.95)²

= 0.0011

=> P(X = 3) = 0.001 ( rounded to 3 decimal places )

Now, using Binomial table we get, P(X = 3) as follows,

P(X = 3) = P(X <= 3) - P(X <= 2) = 1 - 0.999 = 0.001

=> P(X = 3) = 0.001

Therefore, the probability that the number of successes x is exactly 3 is 0.001