Helpppppppppppppppp please

Mathematics

1 answer:

laila [671]3 years ago

6 0

I think it’s A. Here is my sloppy writing.

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Materials

xeze [42]

Answer:

Step-by-step explanation:

3 0

3 years ago

Assume that 12 people, including the husband and wife pair, apply for 4 sales positions. People are hired at random.

konstantin123 [22]

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.

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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

<span>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
</span>

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