Question

The sales department at a certain company consists of four​people, the manufacturing department consists of seven ​people, and the accounting department consists of five people.
Three people will be selected at random from these people and will be given gift certificates to a local restaurant.

Determine the probability that two of those selected will be from the accounting department and one will be from the sales department.

Assume that the selection is done without replacement.

Answer 1

Answer:

$P=\frac{40}{560}=0.0714$

Step-by-step explanation:

Notation

$n_{sales}=4, n_{manufacturing}=7, n_{accounting}=5$

Total = n= 4+7+5=16 people

We are going to select 3 people and will be given gift certificates to a local restaurant so then r =3.

Determine the probability that two of those selected will be from the accounting department and one will be from the sales department.

For this case we can use combinatory nCx, since the selection is without replacment.  

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$  

So then the definition of probability is given by :

$P=\frac{Possible outcomes}{Total outcomes}$

Let's begin with the total outcomes, we have a total of n=16 people and we wan't to select 3 of them, so the possible outcomes are:

$16C3= \frac{16!}{(16-3)! 3!}=560$

And now let's analyze the possible outcomes, we need that the group of 3 would be conformed by two people from the accounting department and one from the sales deparment. So then the possible outcomes are:

$(5C2)*(4C1)= \frac{5!}{(5-2)! 2!} \frac{4!}{(4-1)! 1!}=10*4=40$

And the reason is because we have a total of 5 people at the accounting and we want to select 2. And we have a total of 4 people at the sales department and we want to select just 1. And the multiplication it's because the order on the selection no matter (we assume this).

So then replacing into our formula of probability we got:

$P=\frac{40}{560}=0.0714$