Homework SourseA trading company has ten computers that it uses to trade on
A trading company has ten computers that it uses to trade on the New York Stock Exchange. The probability of a computer failing in a day is 0.003 and the computers fail independently. Computers are repaired in the evening and each day is an independent trial.
b) What is the probability that a week passes till exactly 3 computers fail? [Hint: consider the probability of exactly n computers failing in a day as a single probability of success for the number of days until n failures occur]
c) What is the mean number of days until all ten computers fail in the same day ?
please answer both part b and c with the method used.
A trading company has ten computers that it uses to trade on the New York Stock Exchange. The probability of a computer failing in a day is 0.003 and the computers fail independently. Computers are repaired in the evening and each day is an independent trial.
b) What is the probability that a week passes till exactly 3 computers fail? [Hint: consider the probability of exactly n computers failing in a day as a single probability of success for the number of days until n failures occur]
c) What is the mean number of days until all ten computers fail in the same day ?
please answer both part b and c with the method used.
b) What is the probability that a week passes till exactly 3 computers fail? [Hint: consider the probability of exactly n computers failing in a day as a single probability of success for the number of days until n failures occur]
c) What is the mean number of days until all ten computers fail in the same day ?
please answer both part b and c with the method used.
Solution
Probability of a computer failing in a day =0.03
number of computers =10;
Probability of one computer fail in a day = 10C1 * (0.003)*(1-0.003)^9 = 0.0292
in a week there are 7 days. Probability of a computer fail in a day =0.0292
n = 7 ; p = 0.228069; three days of a one computer fail from 7 days
= 7C3 * (0.092)^3 * (1-0.029)^4 = 0.000774
b)Probability of all ten computers fail in the same day = (8C8)*(0.003)^10 =5.9049*10^(-26)
Mean number of days until all ten computers fail in the same day = 1/ 5.9049*10^(-26) = 1.69351*(10^25)
