Respuesta :
Answer:
a) 4.55% probability that both adults dine out more than once per week.
b) 61.80% probability that neither adult dines out more than once per week.
c) 38.20% probability that at least one of the two adults dines out more than once per week.
d) An event is considered unusual if it has a less than 5% probability of happening. Following this rule, in this problem, both adults dining out more than once per week can be considered unusual.
Step-by-step explanation:
The adults are selected without replacement, so we use the hypergeometric distribution 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]
In this question:
800 adults, so [tex]N = 800[/tex]
171 dine out at a resaurant more than once per week, so [tex]k = 171[/tex]
2 are chosen, so [tex]n = 2[/tex]
a. Find the probability that both adults dine out more than once per week.
This is P(X = 2).
[tex]P(X = 2) = h(2,800,2,171) = \frac{C_{171,2}*C_{629,0}}{C_{800,2}} = 0.0455[/tex]
4.55% probability that both adults dine out more than once per week.
b. Find the probability that neither adult dines out more than once per week.
This is P(X = 0).
[tex]P(X = 0) = h(0,800,2,171) = \frac{C_{171,0}*C_{629,2}}{C_{800,2}} = 0.6180[/tex]
61.80% probability that neither adult dines out more than once per week.
c. Find the probability that at least one of the two adults dines out more than once per week.
Either none dines out more than once per week, or at least one does. The sum of the probabilities of these events is 100%. So
p + 61.80 = 100
p = 38.20
38.20% probability that at least one of the two adults dines out more than once per week.
d. Which of the events can be considered unusual? Explain.
An event is considered unusual if it has a less than 5% probability of happening. Following this rule, in this problem, both adults dining out more than once per week can be considered unusual.
