Question

Let the probability of success on a Bernoulli trial be 0.26. a. In five Bernoulli trials, what is the probability that there will be 4 failures

Answer 1

0.3898 = 38.98% probability that there will be 4 failures

Explanation:

A sequence of Bernoulli trials forms the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

Let the probability of success on a Bernoulli trial be 0.26.

This means that
`p = 0.26`

a. In five Bernoulli trials, what is the probability that there will be 4 failures?

Five trials means that
`n = 5`

4 failures, so 1 success, and we have to find P(X = 1).

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 1) = C_(5,1).(0.26)^(1).(0.74)^(4) = 0.3898`

0.3898 = 38.98% probability that there will be 4 failures