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Sagot :
Answer:
0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed
Step-by-step explanation:
The vessels are destroyed 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:
Fleet of 17 means that [tex]N = 17[/tex]
4 are carrying nucleas weapons, which means that [tex]k = 4[/tex]
9 are destroyed, which means that [tex]n = 9[/tex]
What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?
This is:
[tex]P(X > 1) = 1 - P(X \leq 1)[/tex]
In which
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/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 = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294[/tex]
[tex]P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118[/tex]
Then
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412[/tex]
[tex]P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588[/tex]
0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed
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