Respuesta :
Answer:
a) 50.41% probability that neither will need repair.
b) 8.41% probability that both will need repair.
c) 49.59% probability that at least one car will need repair.
Step-by-step explanation:
For each car, there are only two possible outcomes. Either it will need repairs, or it will not need repairs. The probability of a car needing repairs is independent of other cars. 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.
17% of cars will need to be repaired once, 10% will need repairs twice, and 2% will require three or more repairs.
This means that [tex]p = 0.17+0.1+0.02 = 0.29[/tex]
Two cars:
This means that [tex]n = 2[/tex]
a) neither will need repair?
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{2,0}.(0.29)^{0}.(0.71)^{2} = 0.5041[/tex]
50.41% probability that neither will need repair.
b) both will need repair?
This is P(X = 2).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{2,2}.(0.29)^{2}.(0.71)^{0} = 0.0841[/tex]
8.41% probability that both will need repair.
c) at least one car will need repair?
Either none will need repair, or at least one will. The sum of these probabilities is 100%.
From a)
50.41% probability that neither will need repair.
p = 100 - 50.41 = 49.59%
49.59% probability that at least one car will need repair.
