Question

Suppose that, next month, the technology group will receive 24 requests for help with computer problems. Among, these, 19 are software related and 5 are hardware problems. If a novice in the group is randomly assign to 3 problems, what are the probabilities that she will address.

Answer 1

See below for the appropriate and correct question.

Suppose that, next month, the technology group will receive 24 requests for help with computer problems. Among, these, 19 are software related and 5 are hardware problems. If a novice in the group is randomly assigned to 3 problems, what are the probabilities that she will address:

a. none of the hardware problems?

b. only one of the hardware problems?

Answer:

a. the probabilities that she will address none of the hardware problems 0.4788

b. the probabilities that she will address only one of the hardware problems 0.4224

Explanation:

From the information given:

Let represent R to be the total number of request received.

i.e. R = 24

Let the number of hardware problems be p = 5

Let the number of the randomly assigned problem be m = 3

Then, we can compute the probability mass function of a hypergeometric distribution by using the formula:

`h(x,m,p,R) = ( (^p_x) (^(R-p)_(m-x)) )/(^R_m)` where x can be 0,1,2,3...

Thus;

`h(x,3,5,24) = ( (^5_x) (^(24-5)_(3-x)) )/((^(24)_3))`

(a). To find the probability that she will address none of the hardware problems; we have:

x = 0

Then:

`h(0;3,5,24) = ( (^5_0) (^(24-5)_(3-0)) )/((^(24)_3))`

`h(0;3,5,24) = ( (^5_0) (^(19)_(3)) )/((^(24)_3))`

`h(0;3,5,24) = ( (5!)/(0!(5-0)!) * (19!)/(3!(19-3)!) )/((24!)/(3!(24-3)!))`

`h(0;3,5,24) = ( 1 * 969 )/(2024)`

`h(0;3,5,24) = 0.4788`

b) To find the probability that she will address only one of the hardware problems; we have:

x = 1

Then:

`h(1;3,5,24) = ( (^5_1) (^(24-5)_(3-1)) )/((^(24)_3))`

`h(1;3,5,24) = ( (^5_1) (^(19)_(2)) )/((^(24)_3))`

`h(1;3,5,24) = ( (5!)/(1!(5-1)!) * (19!)/(2!(19-2)!) )/((24!)/(3!(24-3)!))`

`h(1;3,5,24) = ( 5 * 171 )/(2024)`

`h(1;3,5,24) = 0.4424`