Question

Assume that a procedure yields a binomial distribution with a trial repeated n30 times. Use the binomial probability formula to find the probability of x5 successes given the probability p of success on a single trial. Round to three decimal places.

Answer 1

Answer:

The probability is $P(X = 5) = 142506*p^{5}*(1 - p)^{25}$, in which p is the probability of a success on a single trial.

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.

The trial is repeated 30 times:

This means that $n = 30$

Find the probability of 5 successes given the probability p of success on a single trial

This is $P(X = 5)$. So

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

$P(X = 5) = C_{30,5}.p^{5}.(1 - p)^{25}$

$P(X = 5) = 142506*p^{5}*(1 - p)^{25}$

The probability is $P(X = 5) = 142506*p^{5}*(1 - p)^{25}$, in which p is the probability of a success on a single trial.