Respuesta :
Answer:
a) 96.04% probability of getting no defectives.
b) 3.92% probability of getting 1 defective.
c) 0.04% probability of getting 2 defectives.
You pick 2 plugs, so either you get no defective, 1 defective, or both defective. The sum of these probabilities must be 100%, which is what we get.
Step-by-step explanation:
For each plug, there are only two possible outcomes. Either it is defective, or it is not. The probability of a plug being defective is independent of other plugs. So we use the binomial probability distribution to solve this question.
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.
2 % over a long time and that this process is controlled every half hour by drawing and inspecting two just produced.
This means that [tex]p = 0.02, n = 2[/tex]
(a) no defectives,
P(X = 0)
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{2,0}.(0.02)^{0}.(0.98)^{2} = 0.9604[/tex]
96.04% probability of getting no defectives.
(b)1 defective,
P(X = 1)
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{2,1}.(0.02)^{1}.(0.98)^{1} = 0.0392[/tex]
3.92% probability of getting 1 defective.
(c) 2 defectives.
P(X = 2)
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{2,2}.(0.02)^{2}.(0.98)^{0} = 0.0004[/tex]
0.04% probability of getting 2 defectives.
What is the sum of these probabilities?
You pick 2 plugs, so either you get no defective, 1 defective, or both defective. The sum of these probabilities must be 100%, which is what we get.
