A sign on the gas pumps of a chain of gasoline stations encourages customers to have their oil checked with the claim that one out of four cars needs to have oil added. If this is true, what is the probability of the following events? a. One out of the next four cars needs oil.
b. Two out of the next eight cars needs oil.
c. Three out of the next 12 cars need oil.

a) 0.4219 = 42.19% probability that one out of the next four cars needs oil.

b) 0.3115 = 31.15% probability that two out of the next eight cars needs oil.

c) 0.2581 = 25.81% probability that three out of the next 12 cars need oil.

Step-by-step explanation:

For each car, there are only two possible outcomes. Either they need oil, or they do not need it. The probability of a car needing oil is independent of any other car, which 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.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

One out of four cars needs to have oil added.

This means that

a. One out of the next four cars needs oil.

This is P(X = 1) when n = 4. So

0.4219 = 42.19% probability that one out of the next four cars needs oil.

b. Two out of the next eight cars needs oil.

This is P(X = 2) when n = 8. So

0.3115 = 31.15% probability that two out of the next eight cars needs oil.

c.Three out of the next 12 cars need oil.

This is P(X = 3) when n = 12. So

0.2581 = 25.81% probability that three out of the next 12 cars need oil.

answer:   y=2x-3

step-by-step explanation:

it would take 5 hours i think idk i'm not that good at math.can you me with my questions i need asap.