Question

find the probability of at least 6 failures in 7 trials of a binomial experiment in which the probability of success in any trial is 9%​

Answer 1

0.8746 = 87.46% probability of at least 6 failures in 7 trials.

Explanation:

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.

7 trials

This means that
`n = 7`

The probability of success in any trial is 9%​?

So the probability of a failure is 100 - 9 = 91%, which means that
`p = 0.91`

Probability of at least 6 failures in 7 trials?

This is:

`P(X \geq 6) = P(X = 6) + P(X = 7)`

In which

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

`P(X = 6) = C_(7,6).(0.91)^(6).(0.09)^(1) = 0.3578`

`P(X = 7) = C_(7,7).(0.91)^(7).(0.09)^(0) = 0.5168`

`P(X \geq 6) = P(X = 6) + P(X = 7) = 0.3578 + 0.5168 = 0.8746`

0.8746 = 87.46% probability of at least 6 failures in 7 trials.