Suppose a network has 40 computers of which 5 fail. (a) How many possibilities are there for the five that fail? (b) Suppose that 3 of the computers in the network have a copy of a particular file. How many sets of failures wipe out all the copies of the file? That is, how many 5-subsets contain the three computers that have the file?

Respuesta :

Answer: a) 658008, b) 666.

Step-by-step explanation:

Since we have given that

Total number of computers = 40

Number of failed computers = 5

(a) How many possibilities are there for the five that fail?

We will use "Combination" to find all 5 failed.

[tex]^{40}C_5=\dfrac{40!}{5!\times 35!}=658008[/tex]

(b) Suppose that 3 of the computers in the network have a copy of a particular file. How many sets of failures wipe out all the copies of the file?

So, number of computers failed left = 5-3 =2

Total number of computers = 40-3=37

So, the number of set of failures wipe out would be

[tex]^{37}C_2=\dfrac{37!}{2!\times 35!}=666[/tex]

Hence, a) 658008, b) 666.

The number of possibilities that exists for the five that fail is; 658008

How to solve probability combination?

We are given;

Total number of computers = 40

Number of failed computers = 5

(A) To find the number of possibilities that are there for the five that fail, we will use combination method. Thus;

40C5 = 40!/(5! * (40 - 5)!)

>> 40!/(5! * 35!)

>> 658008

B) If we assume that 3 of the computers in the network have a copy of a particular file. Then;

number of computers failed that are left = 5 - 3 = 2

Thus;

Total number of computers = 40-3=37

Finally, the number of set of failures wipe out would be; 37C2 = 40!/(5! * (40 - 5)!)

>> 40!/(5! * 35!)

>> 658008

Read more about Probability Combination at;https://brainly.com/question/4658834