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maksim [4K]
3 years ago
11

Any home mortgage is classified to be of fixed-rate (F) or variable rate (V). Consider an experiment of randomly selecting a sam

ple of four mortgages.
a. Find S, that is the sample space of all possible outcomes.
b. Which outcomes are in the event that exactly three of the
selected mortgages are fixed rate?
c. Which outcomes are in the event that all four mortgages
are of the same type?
d. Which outcomes are in the event that at most one of the
four is a variable-rate mortgage?
e. What is the union of the events in parts (c) and (d), and
what is the intersection of these two events?
f. What are the union and intersection of the two events in
parts (b) and (c)?
Business
1 answer:
Lisa [10]3 years ago
4 0

Answer:

A) we would have 16 outcomes in the sample space.

B) when exactly 3 are fixed rate, then it would be 3 out of the 16 outcomes

C) when we have Event (all being the same), then we would have it to be 2 out of the 16 overall outcomes

D) for Event (almost 1 being variable rate), then the answer would be 3 out of the 16 overall

E) for the union of Event( C) and Event (D), then the answer would be 5 out of the 16 overall outcomes

F)1. for the union of Event (B) and Event (C) , we would have it to be 5 out of the 16 overall outcomes

F) 2. For the intersection of Event (B) and Event (C), then the result would be 3/16*2/16 =6/256

Explanation:

Note : by using the tree diagram,and taking f=fixed rate v=variable rate

A) Sample space ={ffff, fffv, ffvf, ffvv, fvff, fvfv, fvvf, fvvv, vfff, vffv, vfvf, vfvv, vvff, vvfv, vvvf, vvvv} =16 possible outcomes in all

B) Event (Exactly 3 are fixed) = {ffvf, fvff, vfff} =3/16

C) Event (The Same) ={ffff, vvvv} =2/16

D) Event(having at most 1 variable rate) ={ffvf, fvff, vfff) = 3/16

E) for the union of Event C and D above, we must recall that union is considered as addition in probability theory. So we just do, 2/16 + 3/16 =5/16

F)1. For the Union of Event B and C above, we would just do our usual addition, ie, 3/16 + 2/16 =5/16

F) 2. For the intersection of the same as question F) 1 above, it would be required that we multiple, since intersection is handled as multiplication in probability theory.

Hence, we'd have, 3/16 * 2/16 =6/256

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