Respuesta :
Answer:
0.1108 = 11.08% probability that there are exactly 3 faulty candy bars among the seven.
Step-by-step explanation:
The bars are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
The probability of x successes 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 successes.
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]
In this question:
60 total candies means that [tex]N = 70[/tex]
12 are faulty, which means that [tex]k = 12[/tex]
Seven are chosen, so [tex]n = 7[/tex]
What is the probability that there are exactly 3 faulty candy bars among the seven?
This is [tex]P(X = 3)[/tex]. So
[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 = 3) = h(3,70,7,12) = \frac{C_{12,3}*C_{48,4}}{C_{60,7}} = 0.1108[/tex]
0.1108 = 11.08% probability that there are exactly 3 faulty candy bars among the seven.
