Question

An office has 30 computers. Seventeen of the 30 are Macintosh, and the remaining thirteen are windows. Two computers are randomly selected without replacement. What is the probability that the sample contains exactly one windows machine and exactly one Macintosh? If needed, round to FOUR decimal places. Pr(One Widows and One Macintosh) = ___________

Answer 1

To find the probability, you need to calculate the number of ways to choose one Macintosh and one Windows computer from the given options and divide it by the total number of ways to choose two computers. The probability is 0.507.

Step-by-step explanation:

To find the probability that the sample contains exactly one Windows machine and exactly one Macintosh machine, we can use the concept of combinations. There are a total of 30 computers, out of which 17 are Macintosh and 13 are Windows. The number of ways to choose one Macintosh and one Windows computer can be calculated by multiplying the number of ways to choose one Macintosh from 17 and the number of ways to choose one Windows from 13.

The number of ways to choose one Macintosh from 17 is C(17, 1) = 17 and the number of ways to choose one Windows from 13 is C(13, 1) = 13. Therefore, the total number of ways to choose one Macintosh and one Windows computer is 17 * 13 = 221.

The sample space is the total number of ways to choose two computers from 30, which is C(30, 2) = 435. So the probability of selecting exactly one Windows machine and exactly one Macintosh machine is 221 / 435 = 0.507.

Answer 2

0.5085

Step-by-step explanation:

Given that an office has 30 computers. Seventeen of the 30 are Macintosh, and the remaining thirteen are windows.

When two computers are randomly selected without replacement

no of ways of selection =
`60C2 = 1770`

No of ways of selecting one window and one Mac =
`30C1(30C1) = 900`

Prob that the sample contains exactly one windows machine and exactly one Macintosh=
`(900)/(1770) =0.5085`