In a network of 40 computers, 5 hold a copy of a particular file. Suppose that 7 computers at random fail. Let F denote the number of computers that fail and have a copy of the file. A) What is E[F]?B) What is the range of F? C) What is the probability that F = 2?

Using the hypergeometric distribution, it is found that:

a) E(F) = 0.875

b) The range is {0, 1, 2, 3, 4, 5}

c) 0.1741 = 17.41% probability that F = 2.

The computers are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The parameters are:

In this problem:

Item a:

The expected value of the hypergeometric distribution is:

Hence:

Item b:

The range is the possible values that F can assume, which is from 0 to k, hence 0 to 5.

Item c:

The probability is P(X = 2), applying the hypergeometric distribution.

0.1741 = 17.41% probability that F = 2.

You can learn more about the hypergeometric distribution at link

a. E(F)=0.875

b. 99.9976%

c. P(X=2)=0.1683

Step-by-step explanation:

a. We notice that this is a binomial distribution with the probability of success;

#We are given the sample size, n=7. The Expected value is calculated as:

Hence the expectation, E(F)=0.875

b. To calculate the probability of the range of F, we need to calculate all possible outcomes of F in the given sample;

Hence, the range of F is 99.9976%

c. The probability that F=2 is calculated  using the binomial distribution function as:

Hence, the probability of F=2 is 0.1683

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answer: -0.5 would be the most likley outcome

step-by-step explanation: