Respuesta :
Answer:
What is the probability that one or more is scrap?
There is a 40.13% probability that one or more is scrap.
What is the probability that eight or more are not scrap?
There is a 91.38% probability that eight or more are not scrap.
What is the probability that more than two are either downgraded or scrap?
There is a 32.22% probability that more than two are either downgraded or scrap.
What is the probability that exactly nine are either conforming or downgraded?
There is a 31.51% probability that exactly nine are either conforming or downgraded.
Step-by-step explanation:
Each of these problems can be solved using concepts of the binomial probability distribution.
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 combinatios 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.
In this problem
Suppose that 80% of the gears produced are conforming, 15% are downgraded, and 5% are scrap.
Ten gears are selected. This means that [tex]n = 10[/tex].
What is the probability that one or more is scrap?
For each gear, there are only two possible outcomes. Either it is scrap, or it is not. This is why we use the binomial probability distribution.
5% are scrap. So [tex]p = 0.05[/tex].
Either none are scrap, or one or more are. The sum of the decimal probabilities of these events is:
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.05)^{0}.(0.95)^{10} = 0.5987[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.5987 = 0.4013[/tex]
There is a 40.13% probability that one or more is scrap.
What is the probability that eight or more are not scrap?
This is the same as the probability that there is at most one scrap. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.05)^{0}.(0.95)^{10} = 0.5987[/tex]
[tex]P(X = 1) = C_{10,1}.(0.05)^{1}.(0.95)^{9} = 0.3151[/tex]
[tex]P = P(X = 0) + P(X = 1) = 0.5987 + 0.3151 = 0.9138[/tex]
There is a 91.38% probability that eight or more are not scrap.
What is the probability that more than two are either downgraded or scrap?
15% are downgraded and 5% are scrap, so [tex]p = 0.2[/tex].
Either at most two are downgraded or scrap, or more than two are. The sum of the decimal probability of these events is decimal 1. So
[tex]P(X \leq 2) + P(X > 2) = 1[/tex]
[tex]P(X > 2) = 1 - P(X \leq 2)[/tex]
In which
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.20)^{0}.(0.80)^{10} = 0.1074[/tex]
[tex]P(X = 1) = C_{10,1}.(0.20)^{1}.(0.80)^{9} = 0.2684[/tex]
[tex]P(X = 2) = C_{10,2}.(0.20)^{2}.(0.80)^{8} = 0.3020[/tex]
So
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1074 + 0.2684 + 0.3020 = 0.6778[/tex]
[tex]P(X > 2) = 1 - P(X \leq 2) = 1 - 0.6778 = 0.3222[/tex]
There is a 32.22% probability that more than two are either downgraded or scrap.
What is the probability that exactly nine are either conforming or downgraded?
This is the probability that exactly one is scrap.
From the second question, that is 31.51%.
So, there is a 31.51% probability that exactly nine are either conforming or downgraded.
