Question

Naval intelligence reports that 5 enemy vessels in a fleet of 18 are carrying nuclear weapons. If 9 vessels are randomly targeted and destroyed, what is the probability that no more than 1 vessel transporting nuclear weapons was destroyed

Answer 1

0.1471 = 14.71% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

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:

`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 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:

`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)!)`

Naval intelligence reports that 5 enemy vessels in a fleet of 18 are carrying nuclear weapons.

This means, respectively, that
`k = 5, N = 18`.

9 vessels are randomly targeted and destroyed

This means that
`n = 9`

What is the probability that no more than 1 vessel transporting nuclear weapons was destroyed?

This is:

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

In which

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

`P(X = 0) = h(0,18,9,5) = (C_(5,0)*C_(13,9))/(C_(18,9)) = 0.0147`

`P(X = 1) = h(1,18,9,5) = (C_(5,1)*C_(13,8))/(C_(18,9)) = 0.1324`

`P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0147 + 0.1324 = 0.1471`

0.1471 = 14.71% probability that no more than 1 vessel transporting nuclear weapons was destroyed.