Answer:
a)
{FVVV,FFVV,VFVV,VVVV,FVFV,FFFV,VFFV,VVFV,FVVF,FFVF,VFVF,VVVF,FVFF,FFFF,VFFF,VVFF}
b)
{FFFV,FFVF,FVFF,VFFF}
c)
{FFFF,VVVV}
d)
{FFFV,FFVF,FVFF,FFFF,VFFF}
e)
{FFFV,FFVF,FVFF,FFFF,VFFF,VVVV}
{FFFF}
f)
{FFFV,FFVF,FVFF,VFFF,FFFF,VVVV}
{ }
Step-by-step explanation:
a)
The 16 outcomes for a sample of four home mortgages classified as fixed and variable rate are
Sr.No Outcomes
1 FVVV
2 FFVV
3 VFVV
4 VVVV
5 FVFV
6 FFFV
7 VFFV
8 VVFV
9 FVVF
10 FFVF
11 VFVF
12 VVVF
13 FVFF
14 FFFF
15 VFFF
16 VVFF
b)
Let A be the event that exactly three are fixed rate. Thus, event A consists of following outcomes
A={FFFV,FFVF,FVFF,VFFF}
c)
Let B be the event that all four are of same type. The event B consists of either all fixed rate or all variable rate. Thus, event B consists of following outcomes
B={FFFF,VVVV}
d)
Let C be the event that at most one of four is variable rate. The event C consists of less than and equal to one variable rate. Thus, event C consists of following outcomes
C={FFFV,FFVF,FVFF,FFFF,VFFF}
e)
Let union of part(c) and part(d) can be represented as BUC. The union represents all the outcomes in event B and event C.
B∪C={FFFF,VVVV}∪{FFFV,FFVF,FVFF,FFFF,VFFF}
B∪C={FFFV,FFVF,FVFF,FFFF,VFFF,VVVV}
Let intersection of part(c) and part(d) can be represented as B∩C. The intersection represents the common outcomes in event B and event C.
B∩C={FFFF,VVVV}∩{FFFV,FFVF,FVFF,FFFF,VFFF}
B∩C={FFFF}
f)
Let union of part(b) and part(c) can be represented as AUB. The union represents all the outcomes in event A and event B.
A∪B={FFFV,FFVF,FVFF,VFFF}∪{FFFF,VVVV}
A∪B={FFFV,FFVF,FVFF,VFFF,FFFF,VVVV}
Let intersection of part(b) and part(c) can be represented as A∩B. The intersection represents the common outcomes in event A and event B.
A∩B={FFFV,FFVF,FVFF,VFFF}∩{FFFF,VVVV}
A∩B={ }
So intersection of part(b) and part(c) is an empty set as there is no common outcome in these two sets.
![\bf sec(s)=cos(s)\qquad but\implies sec(\theta)=\cfrac{1}{cos(\theta)} \\\\\\ thus\cfrac{1}{cos(s)}=cos(s)\implies 1=cos^2(s)\implies \pm \sqrt{1}=cos(s) \\\\\\ \pm 1=cos(s)\impliedby \textit{now taking }cos^{-1}\textit{ to both sides} \\\\\\ cos^{-1}(\pm 1)=cos^{-1}[cos(s)]\implies cos^{-1}(\pm 1)=\measuredangle s](https://tex.z-dn.net/?f=%5Cbf%20sec%28s%29%3Dcos%28s%29%5Cqquad%20but%5Cimplies%20sec%28%5Ctheta%29%3D%5Ccfrac%7B1%7D%7Bcos%28%5Ctheta%29%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%5Ccfrac%7B1%7D%7Bcos%28s%29%7D%3Dcos%28s%29%5Cimplies%201%3Dcos%5E2%28s%29%5Cimplies%20%5Cpm%20%5Csqrt%7B1%7D%3Dcos%28s%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cpm%201%3Dcos%28s%29%5Cimpliedby%20%5Ctextit%7Bnow%20taking%20%7Dcos%5E%7B-1%7D%5Ctextit%7B%20to%20both%20sides%7D%0A%5C%5C%5C%5C%5C%5C%0Acos%5E%7B-1%7D%28%5Cpm%201%29%3Dcos%5E%7B-1%7D%5Bcos%28s%29%5D%5Cimplies%20cos%5E%7B-1%7D%28%5Cpm%201%29%3D%5Cmeasuredangle%20s)