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Sagot :
Answer:
0.2348 = 23.48% probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.
Step-by-step explanation:
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:
Population of 280, which means that [tex]N = 280[/tex]
112 use e-ticket check-in kiosk, which means that [tex]k = 112[/tex]
Sample of eight passengers means that [tex]n = 8[/tex]
Find the approximate hypergeometric probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.
This is P(X = 4). 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 = 4) = h(4,280,8,112) = \frac{C_{112,4}*C_{168,4}}{C_{280,4}} = 0.2348[/tex]
0.2348 = 23.48% probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.
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