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

Answer:

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

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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.