Question

Assume that a procedure yields a binomial distribution with a trial repeated nequals30 times. Use the binomial probability formula to find the probability of xequals5 successes given the probability pequalsone fifth of success on a single trial. Round to three decimal places.

Answer 1

Answer:

$P(X = 5) = C_{30,5}.(0.2)^{5}.(0.8)^{25} = 0.172$

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 $C_{n,x}[\tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.[tex]C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

$n = 30, p = \frac{1}{5} = 0.2$

Use the binomial probability formula to find the probability of xequals5 successes given the probability pequalsone fifth of success on a single trial.

This is P(X = 5).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{30,5}.(0.2)^{5}.(0.8)^{25} = 0.172$