If the sample probability of an individual traveling to the statue is 25%, the number of trials is and the number of successful trials is . The probability that exactly 7 out of 10 individuals will visit the Statue of Liberty is .

0.0031 = 0.31% probability that exactly 7 out of 10 individuals will visit the Statue of Liberty

Step-by-step explanation:

For each individual, there are only two possible outcomes. Either they visit the Statue of the Liberty, or they do not. The probability of an individual visiting the statue is independent of any other individual. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

And p is the probability of X happening.

The sample probability of an individual traveling to the statue is 25%

This means that

The probability that exactly 7 out of 10 individuals will visit the Statue of Liberty is

This is when . So

0.0031 = 0.31% probability that exactly 7 out of 10 individuals will visit the Statue of Liberty

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,