Question

Let X be a binomial random variable with n = 9 and p = 0.2 (nine trials, probability of success = 0.2). Find the probability of 4 successes, i.e., P(X = 4). Give your answer as a decimal to three decimal places, e.g., 0.123. Round as needed.

Answer 1

The probability of 4 successes,

P(X=4) =0.0660

Explanation:

Step 1:-

Let X be a binomial random variable with n = 9 and p = 0.2

probability of successes p = 0.2

probability of failure q = 1-p = 1-0.2 = 0.8

By using binomial distribution

`P(X=r) = n_(Cr) p^(r) q^(n-r)`

Step 2:-

The probability of 4 successes, i.e.,

`P(X=4) = 9_(C4) (0.2)^(4) (0.8)^(9-4)`

by using
`n_(Cr) = (n!)/((n-r)!r!)`

`9_(C4) = (9!)/((9-4)!4!) = (9X8X7X6X5!)/(5!4!)`
`= (9X8X7X6)/(4!) = (9X8X7X6)/(4X3X2X1) =126`

`P(X=4) =126 X (0.0016)(0.32768)`

P(X=4) =0.0660

conclusion:-

The probability of 4 successes, that is P(X=4) =0.0660