Respuesta :
Answer:
86.46% probability that at least 2 crashes occur in the next month.
Step-by-step explanation:
For each TIE ghter ights, there are only two possible outcomes. Either it crashes, or it does not. The probability of a TIE ghter ights crashing is independent from other TIE ghter ights. 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.
Every month, there are 1000 independent TIE ghter ights, and each TIE ghter ight crashes with a probability of 0.0035.
This means that [tex]n = 1000, p = 0.0035[/tex]
(a) What is the probability that at least 2 crashes occur in the next month?
Either less than 2 crashes occur, or at least 2 do. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]
So
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = 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_{1000,0}.(0.0035)^{0}.(0.9965)^{1000} = 0.0300[/tex]
[tex]P(X = 1) = C_{1000,1}.(0.0035)^{1}.(0.9965)^{999} = 0.1054[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.0300 + 0.1054 = 0.1354[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1354 = 0.8646[/tex]
86.46% probability that at least 2 crashes occur in the next month.
