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
