A stock brokerage has four computers that are used for making trades on the New York Stock Exchange. The probability that a computer fails on any single day is 0.005. Failures occur independently. Any failed computers are repaired after the exchange closes, so each day can be considered an independent trial.(a) What is the probability that all four computers fail on one day?(b) What is the probability that at least one computer fails on a day?(c) What is the mean number of days until a specific computer fails?

A) 0 , B)  0.0199 , C) 200

Step-by-step explanation:

Let X denotes no. of computers fail on day

X ~ Bin (n = 4 , p = 0.05)

A] P (X = x) = (n c x) p^x q^n-x

P (X = 4) = (4 c 4) (0.005)^4 (0.995)^0

(0.05)^4 ~ 0

B] P (X > 1) = 1 - P (X < 1) = 1 - P (X = 0)

1 - { ( 4 c 0) (0.005)^0 (0.995)^ 4 }

1 -  (0.995)^ 4 ~ 0.0199

C] Let Y depict mean number of days until a specific computer fails

Y ~ Geom (p = 0.005)

Mean = E (Y) = 1/p = 1/0.05 = 200

we need more info to solve this

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.