Question

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?

Answer 1

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;

$p=\frac{5}{40}=0.125$

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

$E(X)=np\\\\E(F)=np , n=7, p=0.125\\\\E(F)=7\times 0.125\\\\=0.875$

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;

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

Hence, the range of F is 99.9976%

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

$P(X=2)={7\choose 2}(0.125)^2(1-0.125)^5\\\\=0.1683$

Hence, the probability of F=2 is 0.1683