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3 months ago

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

Mathematics

1 answer:

Inessa [12.5K]3 months ago

4 0

The formula $C(n, r)= \frac{n!}{r!(n-r)!}$ , where r! is defined as 1*2*3*...r

provides the total number of combinations for forming groups of r items from a collection of n items.

For instance, with 10 items, there are C(10,6) possible ways to create groups of 6 from these 10 objects.

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Choosing 4 individuals from a total of 12 can be accomplished 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 different ways.

All unique groupings of 4 individuals, including the husband and wife pair, can be computed as C(10, 2) ways, since we only consider the potential selections of 2 from 10 individuals to form a group of 4.

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

Consequently, the probability that both the husband and wife are selected is 45/495=0.09

Part 2)

The chance that one gets selected while the other does not =

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

These two scenarios are precisely equal, so it suffices to compute one.

Let's analyze the scenario: husband chosen, wife not selected.

Assuming the husband is selected, we need to determine the possible formations of 3 from the 11 total excluding the wife=10 individuals.

This results in:

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

Hence,

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

Thus, the overall probability that one is picked while the other is not =

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

0.24+0.24=0.48

Result:

A) 0.09

B) 0.48

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