Respuesta :
Answer:
[tex]3.9 \times 10^{-11}[/tex] probability that both vehicles will have all defective fuses
Step-by-step explanation:
For each fuse, there are only two possible outcomes. Either they are defective, or they are not. The probability of a fuse being defective is independent of any other fuse. This means that the binomial probability distribution is used 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.
Probability of a vehicle having all defective fuses:
5% of fuses are defective, which means that [tex]p = 0.05[/tex]
A vehicle has 4 fuses, which means that [tex]n = 4[/tex].
This probability is P(X = 4). So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{4,4}.(0.05)^{4}.(0.95)^{0} = 0.00000625[/tex]
If 2 assembled vehicles are selected at random and independently of one another, what is the probability that both vehicles will have all defective fuses?
For each vehicle, 0.00000625 probability of having all defective fuses. For 2:
[tex](0.00000625)^2 = 3.9 \times 10^{-11}[/tex]
[tex]3.9 \times 10^{-11}[/tex] probability that both vehicles will have all defective fuses
