Question

Car colors preference change over the years and according to the particular model that the customer selects. In a recent year, 10% of all luxury cars sold were black. If 25 cars of that year and type are randomly selected, find the probabilitiy that at most six cars are black.

Answer 1

Answer:

99.05% probabilitiy that at most six cars are black.

Step-by-step explanation:

For each luxury car, there are only two possible outcomes. Either it is black, or it is not black. The probability of a luxury car being black is independent from other luxury 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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$p = 0.1, n = 25$

If 25 cars of that year and type are randomly selected, find the probabilitiy that at most six cars are black.

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.1)^{0}.(0.9)^{25} = 0.0718$

$P(X = 1) = C_{25,1}.(0.1)^{1}.(0.9)^{24} = 0.1994$

$P(X = 2) = C_{25,2}.(0.1)^{2}.(0.9)^{23} = 0.2659$

$P(X = 3) = C_{25,3}.(0.1)^{3}.(0.9)^{22} = 0.2265$

$P(X = 4) = C_{25,4}.(0.1)^{4}.(0.9)^{21} = 0.1384$

$P(X = 5) = C_{25,5}.(0.1)^{5}.(0.9)^{20} = 0.0646$

$P(X = 6) = C_{25,6}.(0.1)^{6}.(0.9)^{19} = 0.0239$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0718 + 0.1994 + 0.2659 + 0.2265 + 0.1384 + 0.0646 + 0.0239 = 0.9905$

99.05% probabilitiy that at most six cars are black.