A trading company has eight computers that it usesto trade on the New York Stock Exchange (NYSE). The probability ofa computer failing in a day is 0.005, and the computers failindependently. Computers are repaired in the evening and each dayis an independent trial.

a) What is the probability that all eight computers fail in aday?

b)What is the mean number of days until a specific computerfails?

c)What is the mean number of days until all eight computers fail inthe same day?

a) 3.906 × 10⁻¹⁹

b) 200 days

c) 2.56 × 10¹⁸ days

Step-by-step explanation:

Probability that a computer fails in a day = 0.005.

This probability is very similar to the number of computers that'll fail in a day, that is, number of computer failures per day = 0.005

a) Probability that all the 8 computers fail in a day.

This probability is setup as a binomial distribution problem, but ⁸C₈ = 1 and q⁰ = 1

So, P(X=8) = (0.005)⁸ = 3.906 × 10¹⁹

b) The mean number of days until a specific computer fails

Number of failure = 0.005 per day

Mean number of days = E(X) = 1/0.005 = 200 days

c) Mean number of days until all the 8 computers fail

E(X) = 1/(3.906 × 10⁻¹⁹)

E(X) = 2.56 × 10¹⁸ days

assuming this is from a math graphing problem.

7 would be on the x-axis which is the horizontal line on the graph

84 would be on the y-axis which is the vertical line on the graph