Respuesta :
Answer:
a) P=0.0225
b) P=0.057375
Step-by-step explanation:
From exercise we have that 15% of items produced are defective, we conclude that probabiity:
P=15/100
P=0.15
a) We calculate the probabiity that two items are defective:
P= 0.15 · 0.15
P=0.0225
b) We calculate the probabiity that two of three items are defective:
P={3}_C_{2} · 0.85 · 0.15 · 0.15
P=\frac{3!}{2!(3-2)!} · 0.019125
P=3 · 0.019125
P=0.057375
Using the binomial distribution, it is found that there is a:
a) 0.0225 = 2.25% probability that both are defective.
b) 0.0574 = 5.74% probability that two are defective.
For each item, there are only two possible outcomes, either it is defective, or it is not. The probability of an item being defective is independent of any other item, hence, the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem, 15% of items produced are defective, hence [tex]p = 0.15[/tex].
Item a:
Two items, hence [tex]n = 2[/tex].
The probability is P(X = 2), then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{2,2}.(0.15)^{2}.(0.85)^{0} = 0.0225[/tex]
0.0225 = 2.25% probability that both are defective.
Item b:
Three items, hence [tex]n = 3[/tex].
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{3,2}.(0.15)^{2}.(0.85)^{1} = 0.0574[/tex]
0.0574 = 5.74% probability that two are defective.
To learn more about the binomial distribution, you can take a look at https://brainly.com/question/24863377
