Question

Find the probability of at least 6 failures in 7 trials of a binomial experiment in which the probability of success in any one trial is 9%.
Round to the nearest tenth of a percent

Answer 1

Answer:

$P(x \geq 6)=P(X=6)+P(X=7)$

And we can find the individual probabilities:

$P(X=6)=(7C6)(0.91)^6 (1-0.91)^{7-6}=0.358$

$P(X=7)=(7C7)(0.91)^7 (1-0.91)^{7-7}=0.517$

And replacing we got:

$P(x \geq 6)=P(X=6)+P(X=7)= 0.358+0.517=0.875$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=7, p=1-0.09=0.91)$

The probability associated to a failure would be p =1-0.09 = 0.91

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

And we want to find this probability:

$P(x \geq 6)=P(X=6)+P(X=7)$

And we can find the individual probabilities:

$P(X=6)=(7C6)(0.91)^6 (1-0.91)^{7-6}=0.358$

$P(X=7)=(7C7)(0.91)^7 (1-0.91)^{7-7}=0.517$

And replacing we got:

$P(x \geq 6)=P(X=6)+P(X=7)= 0.358+0.517=0.875$