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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?

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opudodennis

Answer:

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;

[tex]p=\frac{5}{40}=0.125[/tex]

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

[tex]E(X)=np\\\\E(F)=np , n=7, p=0.125\\\\E(F)=7\times 0.125\\\\=0.875[/tex]

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;

[tex]P(X\leq 5)=1-P(X=6)-P(X=7)\\\\=1-{7\choose 6}(0.125)^6(1-0.125)^1-{7\choose 7}(0.125)^7(1-0.125)^0\\\\=1-0.000023365-0.000000476\\\\=0.999976158\\\\=99.9976\%[/tex]

Hence, the range of F is 99.9976%

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

[tex]P(X=2)={7\choose 2}(0.125)^2(1-0.125)^5\\\\=0.1683[/tex]

Hence, the probability of F=2 is 0.1683

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:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

  • 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.

In this problem:

  • There are 40 computers, hence [tex]N = 40[/tex].
  • Of those computers, 7 fail, hence [tex]n = 7[/tex].
  • 5 of the computers hold a copy, hence [tex]k = 5[/tex].

Item a:

The expected value of the hypergeometric distribution is:

[tex]E(F) =  \frac{nk}{N}[/tex]

Hence:

[tex]E(F) = \frac{35}{40} = 0.875[/tex]

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.

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 2) = h(2,40,7,5) = \frac{C_{5,2}C_{35, 5}}{C_{40,7}} = 0.1741[/tex]

0.1741 = 17.41% probability that F = 2.

You can learn more about the hypergeometric distribution at https://brainly.com/question/25303388

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