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loris [4]
3 years ago
6

Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

d and destroyed, what is the probability that more than 1 vessel transporting nuclear weapons was destroyed
Mathematics
1 answer:
icang [17]3 years ago
8 0

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:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

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:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question:

Fleet of 17 means that N = 17

4 are carrying nucleas weapons, which means that k = 4

9 are destroyed, which means that n = 9

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

P(X > 1) = 1 - P(X \leq 1)

In which

P(X \leq 1) = P(X = 0) + P(X = 1)

So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294

P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118

Then

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412

P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

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