Respuesta :
Answer:
0.1471 = 14.71% probability that no more than 1 vessel transporting nuclear weapons was destroyed.
Step-by-step explanation:
Vessels are chosen without replacement, which means that 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]
Naval intelligence reports that 5 enemy vessels in a fleet of 18 are carrying nuclear weapons.
This means, respectively, that [tex]k = 5, N = 18[/tex].
9 vessels are randomly targeted and destroyed
This means that [tex]n = 9[/tex]
What is the probability that no more than 1 vessel transporting nuclear weapons was destroyed?
This is:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/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(0,18,9,5) = \frac{C_{5,0}*C_{13,9}}{C_{18,9}} = 0.0147[/tex]
[tex]P(X = 1) = h(1,18,9,5) = \frac{C_{5,1}*C_{13,8}}{C_{18,9}} = 0.1324[/tex]
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0147 + 0.1324 = 0.1471[/tex]
0.1471 = 14.71% probability that no more than 1 vessel transporting nuclear weapons was destroyed.
