Assume that 12 people, including the husband and wife pair, apply for 4 sales positions. people are hired at random. (1) what is the probability that both the husband and wife are hired? (2) what is the probability that one is hired and one is not?

The formula , where r! is 1*2*3*...r

is the formula which gives us the total number of ways of forming groups of r objects, out of n objects.

for example, given 10 objects, there are C(10,6) ways of forming groups of 6, out of the 10 objects.

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Selecting 4 people out of 12 can be done in :

      many ways.

All the possible groups of 4 people, where the husband and wife are included, can be done in C(10, 2) many ways, since we only calculate the possible choices of 2 out of 10 people, to complete the groups of 4.

Thus, the 

thus, 

P(husband hired, wife not hired)=120/495=0.24

thus, 

The probability that one is hired and the other is not = 

P(husband hired, wife not hired) + P(wife hired, husband not hired) =

0.24+0.24=0.48

A) 0.09

B) 0.48

1. sometimes

2. never true

3. sometimes

step-by-step explanation:

these are pretty self explanatory if you know what supp right and com angles are

1st part is 64,000,000

2nd part 500,000

step-by-step explanation: