Question

Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targeted and destroyed, what is the probability that more than 1 vessel transporting nuclear weapons was destroyed

Answer 1

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

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) = (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) = (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) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 0) = h(0,17,9,4) = (C_(4,0)*C_(13,9))/(C_(17,9)) = 0.0294`

`P(X = 1) = h(1,17,9,4) = (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