Respuesta :
Answer:
0.6864 = 68.64% probability out of the 8 new cars it just received that, when each is tested, no more than 2 of the cars have defective radios
Step-by-step explanation:
The cars are chosen without replacement, so the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
The probability of x sucesses is given by the following formula:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
In which:
x is the number of sucesses.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.
Combinations formula:
[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]
48 new cars being shipped to local dealerships, corporate reports indicate that 12 have defective radios installed.
This means that [tex]N = 48, k = 12[/tex]
8 cars received.
This means that [tex]n = 8[/tex]
What is the probability out of the 8 new cars it just received that, when each is tested, no more than 2 of the cars have defective radios
This is
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
In which
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 0) = h(0,48,8,12) = \frac{C_{12,0}*C_{36,8}}{C_{48,8}} = 0.0802[/tex]
[tex]P(X = 1) = h(1,48,8,12) = \frac{C_{12,1}*C_{36,7}}{C_{48,8}} = 0.2655[/tex]
[tex]P(X = 2) = h(2,48,8,12) = \frac{C_{12,2}*C_{36,6}}{C_{48,8}} = 0.3407[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0802 + 0.2655 + 0.3407 = 0.6864[/tex]
0.6864 = 68.64% probability out of the 8 new cars it just received that, when each is tested, no more than 2 of the cars have defective radios
