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

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:

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:

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

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

This means, respectively, that .

9 vessels are randomly targeted and destroyed

This means that

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

This is:

In which

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

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

the two may have that, even a breakdown is probably offered