Respuesta :
Answer:
0.4718 = 47.18% probability that there will be more than the expected number of failures.
Step-by-step explanation:
The binomial probability distribution is a sequence of Bernoulli trials.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]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)!}[/tex]
And p is the probability of X happening.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
In this problem we have that:
Let the probability of success on a Bernoulli trial be 0.30. Five Bernoulli trials. This means that [tex]n = 5, p = 0.3[/tex]
In five Bernoulli trials, what is the probability that there will be more than the expected number of failures?
The expected number of failures is:
[tex]E(X) = np = 5*0.3 = 1.5[/tex]
So, either there are less than 1.5 failures, or there are more than 1.5 failures. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 1.5) + P(X \geq 1.5) = 1[/tex]
The number of failures is a discrete number, so
[tex]P(X < 1.5) = P(X \leq 1)[/tex]
and
[tex]P(X \geq 1.5) = P(X \geq 2)[/tex]
So
[tex]P(X \leq 1) + P(X \geq 2) = 1[/tex]
We want to find [tex]P(X \geq 2)[/tex]
So
[tex]P(X \geq 2) = 1 - P(X \leq 1)[/tex]
In which
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5} = 0.16807[/tex]
[tex]P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{4} = 0.36015[/tex]
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.16807 + 0.36015 = 0.52822[/tex]
So
[tex]P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.52822 = 0.47178[/tex]
Rounded to four decimal places
0.4718 = 47.18% probability that there will be more than the expected number of failures.
