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leva [86]
4 years ago
14

Please help! I really don’t want this to be wrong ;-;

Mathematics
2 answers:
solmaris [256]4 years ago
6 0

280 in i did caculations

MissTica4 years ago
4 0
I am pretty sure this is 280
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Assume that 12 people, including the husband and wife pair, apply for 4 sales positions. People are hired at random.
Iteru [2.4K]

brainly.com/question/5218999


The formula C(n, r)= \frac{n!}{r!(n-r)!}, 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.

-----------------------------------------------------------------------------------------------


Selecting 4 people out of 12 can be done in :

\displaystyle{C(12, 4)= \frac{12!}{4!8!}= \frac{12\cdot11\cdot10\cdot9\cdot8!}{4!8!}= \frac{12\cdot11\cdot10\cdot9}{4!}=11\cdot5\cdot9= 495       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.


\displaystyle{ C(10, 2)= \frac{10!}{2!8!}= \frac{10\cdot9}{2}=45


Thus, the 

probability that both the husband and wife are hired is 45/495=0.09


Part 2)

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

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

these 2 are clearly equal, so it is enough to calculate one.


Consider the case : husband hired, wife not hired.

assuming the husband is hired, we have to calculate the possible groups of 3 that can be formed from 11-1 (the wife)=10 people.

this is 

\displaystyle{ C(10, 3)= \frac{10!}{3!7!}= \frac{10 \cdot9 \cdot8}{3\cdot2}=10\cdot3\cdot4=120


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



Answer:


A) 0.09


B) 0.48

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