In a start-up company which has 20 computers, some of the computers are infected with virus. The probability that a computer is infected with the virus is 0.4, independently of other computers. A technician tests the computers, one after another, to see if they are infected.1. What is the probability that she has to test at least 5 computers to find the first (if any) defective one?2. Find the probability that on this day at least 5 computers are infected.

(1) The probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2) The probability at least 5 computers are infected is 0.949.

Step-by-step explanation:

The probability that a computer is defective is, p = 0.40.

(1)

Let X = number of computers to be tested before the 1st defect is found.

Then the random variable .

The probability function of a Geometric distribution for k failures before the 1st success is:

Compute the probability that the technician tests at least 5 computers before the 1st defective computer is found as follows:

P (X ≥ 5) = 1 - P (X < 5)

              = 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]

Thus, the probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2)

Let Y = number of computers infected.

The number of computers in the company is, n = 20.

Then the random variable .

The probability function of a binomial distribution is:

Compute the probability at least 5 computers are infected as follows:

P (Y ≥ 5) = 1 - P (Y < 5)

             = 1 - [P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3) + P (Y = 4)]              

Thus, the probability at least 5 computers are infected is 0.949.

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1

i think it would be 42cm