Question

10% of the cars in my neighborhood are red, and the rest of the cars in the neighborhood are silver. We'll call "seeing a red car" a success, and "seeing a silver car" a failure for the purposes of this problem.

Suppose that I watch 3 cars pass my house and that I become interested in the probability that exactly one of the three cars is red.

Apply the binomial formula to find the probability that exactly one of the three cars is red. Be sure to clearly state the values of n, x, and p in this case.

Answer 1

In this problem, we want to find the probability that exactly one of the three cars is red, given that 10% of the cars in the neighborhood are red and we watch 3 cars pass by.

Let's define the following variables:

n = 3 (the number of trials or cars we observe)
x = 1 (the number of successes we want, i.e. seeing one red car)
p = 0.1 (the probability of success, i.e. seeing a red car)

Using the binomial formula, the probability of getting exactly one red car in three cars passing by is:

P(x = 1) = (n choose x) * p^x * (1 - p)^(n - x)

where (n choose x) is the binomial coefficient, which represents the number of ways to choose x objects from a set of n objects, and is calculated as:

(n choose x) = n! / (x! * (n - x)!)

Plugging in the values for n, x, and p, we get:

P(x = 1) = (3 choose 1) * 0.1^1 * (1 - 0.1)^(3 - 1)
= 3 * 0.1 * 0.81
= 0.243

Therefore, the probability that exactly one of the three cars is red is 0.243 or approximately 24.3%.

Answer 2

In this scenario, we have:

n = 3 (since we are watching 3 cars)

x = 1 (since we are interested in the probability of exactly one car being red)

p = 0.1 (since the probability of a car being red is 10%, or 0.1)

The binomial formula for calculating the probability of exactly x successes in n independent trials with a probability of success p is:

P(x) = (nCx) * p^x * (1-p)^(n-x)

where nCx is the binomial coefficient, which can be calculated as:

nCx = n! / (x! * (n-x)!)

Using these values and the binomial formula, we can calculate the probability of exactly one of the three cars being red as:

P(1) = (3C1) * 0.1^1 * (1-0.1)^(3-1)

= (3) * 0.1 * 0.81

= 0.243

Therefore, the probability of exactly one of the three cars being red is 0.243.