Respuesta :
Answer:
0.1616 = 16.16% probability that exactly 3 devices broke down if it is known that at least 1 device broke down
Step-by-step explanation:
To solve this question, we use both the binomial distribution and conditional probability.
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.
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
Find the probability that exactly 3 devices broke down if it is known that at least 1 device broke down
So, by the conditional probability:
Event A: At least 1 device broke down.
Event B: Exactly 3 devices broke down.
The solution will be found by the conditional probability formula, and the probabilities will be found using the binomial distribution.
There are 10 devices. Every device independently breaks down with probability 0.15.
This means that [tex]n = 10, p = 0.15[/tex]
Probability of at least 1 device breaking down:
This is:
[tex]P(A) = P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.15)^{0}.(0.85)^{10} = 0.1969[/tex]
So
[tex]P(A) = P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1969 = 0.8031[/tex]
Intersection of events A and B:
The intersection between at least one breaking down, and exactly three is exactly three. So
[tex]P(A \cap B) = P(X = 3) = C_{10,3}.(0.15)^{3}.(0.85)^{7} = 0.1298[/tex]
Probability:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1298}{0.8031} = 0.1616[/tex]
0.1616 = 16.16% probability that exactly 3 devices broke down if it is known that at least 1 device broke down
