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.
![\displaystyle{ C(10, 2)= \frac{10!}{2!8!}= \frac{10\cdot9}{2}=45](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20C%2810%2C%202%29%3D%20%5Cfrac%7B10%21%7D%7B2%218%21%7D%3D%20%5Cfrac%7B10%5Ccdot9%7D%7B2%7D%3D45%20%20)
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](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20C%2810%2C%203%29%3D%20%5Cfrac%7B10%21%7D%7B3%217%21%7D%3D%20%5Cfrac%7B10%20%5Ccdot9%20%5Ccdot8%7D%7B3%5Ccdot2%7D%3D10%5Ccdot3%5Ccdot4%3D120%20%20)
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