A sign on the gas pump of a certain chain of gasoline stations encourages customers to have their oil checked, claiming that one out of every four cars should have its oil topped up. What is the probability that exactly 3 of the next 10 cars entering a station should have their oil topped up?
You can use the binomial probability function for these kinds of problems.
The function is this:
P(x) = (nCx)(p^x)[q^(n-x)]
From your problem, n = 10, x = 3, p = 1/4 or .25, q = .75 (q = 1 - p).
Setting up:
P(3) = (10C3)(.25^3)(.75^7)
Can you take it from here to finish?
is the answer correct?
0.255
Sure, I can help you with that. Let's calculate it step by step.
First, let's calculate the combination using the formula (nCx):
(10C3) = 10! / (3! * (10-3)!) = 120.
Now, let's substitute the values into the binomial probability function:
P(3) = (10C3) * (.25^3) * (.75^7)
P(3) = 120 * (.25^3) * (.75^7)
P(3) ≈ 0.2503
So, the probability that exactly 3 of the next 10 cars entering a station should have their oil topped up is approximately 0.2503.
To calculate the probability that exactly 3 out of the next 10 cars should have their oil topped up, we can use the binomial probability formula.
The formula is as follows:
P(x) = (nCx) * (p^x) * (q^(n-x))
In this formula, P(x) represents the probability of exactly x successes, n represents the number of trials, p represents the probability of success, q represents the probability of failure (which is 1 - p), and (nCx) represents the number of combinations.
In this problem, n = 10 as there are 10 cars, x = 3 as we want to find the probability of exactly 3 cars needing their oil topped up, p = 1/4 or 0.25 (since the sign claims that 1 out of every 4 cars should have their oil topped up), and q = 0.75 (since q = 1 - p).
Substituting these values into the formula, we get:
P(3) = (10C3) * (0.25^3) * (0.75^7)
Now, let's evaluate the individual components of the formula:
(10C3) represents the number of combinations of selecting 3 cars out of 10. It can be calculated as:
(10C3) = 10! / (3! * (10-3)!)
= 10! / (3! * 7!)
= (10 * 9 * 8) / (3 * 2 * 1)
= 120
(0.25^3) represents the probability of 3 out of the 10 cars needing their oil topped up, which is:
(0.25^3) = 0.25 * 0.25 * 0.25
= 0.015625
(0.75^7) represents the probability of the remaining 7 cars not needing their oil topped up, which is:
(0.75^7) = 0.75 * 0.75 * 0.75 * 0.75 * 0.75 * 0.75 * 0.75
≈ 0.1334838867 (approximately)
Now, we can substitute these values into the formula:
P(3) = 120 * 0.015625 * 0.1334838867
Calculating this value, we find that:
P(3) ≈ 0.298845