So, we had to create and prove our own formula for a given event. I think it looks good, but since I just spent a huge amount of time working on it, I'm obviously very biased. Math isn't my strongest suit, and I'd really appreciate a set of fresh eyes.

Probability of being in a car accident


A How many times out of five do you drive above the posted speed limit?
B How many times out of five do you break a common traffic rule?
C How many times out of five do you drive while you have some sort of distraction (kids, cell phone, etc.)?
D How many times out of five do you drive at night?
E How many times out of five do you drive on a busy street?
F How many times out of five do you drive with some sort of impairment (tiredness, after drinking, on some drug, etc.)?


My research shows that your chances of being involved in a car accident are about 1 in 5000, or .02%.





Trial #
Possible
Equation Test
Values Equation
With Test Values Answer
1 6(A)+5(B)+4(C)+3(D)+2(E)+1(F)
5 A-1
B-1.75
C-4
D-2
E-2.5
F-3 6(1)+5(1.75)+4(4)+3(2)+2(2.5)+1(3)
5 895%. This answer is obviously much too high. We don’t all get into car accidents every single time that we drive.
2 1(A)+.5(B)+3(C)+1.5(D)+2(E)+2(F)
10 A-1
B-1.75
C-4
D-2
E-2.5
F-3 1(1)+.5(1.75)+3(4)+1.5(2)+2(2.25)+2(3)
10 33.3%. This answer is much closer, but still significantly high.
3 .2(A)+.1(B)+.75(C)
.37(D)+.4(E)+.4(F) A-1
B-1.75
C-4
D-2
E-2.5
F-3 .2(1)+.1(1.75)+.75(4)
.37(2)+.4(2.5)+.4(3) 114.7%. This answer is higher, which I didn’t expect. Apparently dividing it this way won’t work. Since none of my factors would really increase as the formula increases, it wasn’t the best of ideas in hindsight.
4 1(A)+.5(B)+3(C)+1.5(D)+2(E)+2(F)
1000 A-1
B-1.75
C-4
D-2
E-2.5
F-3 1(1)+.5(1.75)+3(4)+1.5(2)+2(2.25)+2(3)
1000 3.34%. This is much closer to my research number. Just a little farther to go.
5 1(A)+.25(B)+1.5(C)+.75(D)+1(E)+1.5(F)
1000 A-1
B-1.75
C-4
D-2
E-2.5
F-3 1(1)+.25(1.75)+1.5(4)+.75(2)+1(2.25)+1.5(3)
1000 1.57%. This number is getting closer to my research number.
6 .05(A)+.02(B)+.3(C)+.07(D)+.08(E)+.3(F)
2000 A-1
B-1.75
C-4
D-2
E-2.5
F-3 .05(1)+.02(1.75)+.3(4)+.07(2)+.08(2.5)+.3(3)
2000 .12% This number is really close to my research number and just needs a little bit more tweaking.
7 .05(A)+.02(B)+.3(C)+.07(D)+.08(E)+.3(F)
12500 A-1
B-1.75
C-4
D-2
E-2.5
F-3 .05(1)+.02(1.75)+.3(4)+.07(2)+.08(2.8)+.3(3)
12500 .0202 This number is really close to my expected number.


Trial # Formula New Test Set Equation with Numbers Result
1 .05(A)+.02(B)+.3(C)+.07(D)+.08(E)+.3(F)
12500 A-2
B-1
C-3
D-5
E-1
F-2 .05(2)+.02(1)+.3(3)+.07(5)+.08(1)+.3(2)
12500 .016%. This number is very close to my research number of .02%.
2 .05(A)+.02(B)+.3(C)+.07(D)+.08(E)+.3(F)
12500 A-5
B-3
C-.5
D-4
E-2
F-3 .05(5)+.02(3)+.3(.5)+.07(4)+.08(2)+.3(3)
12500 .015%. This number is very close to my test number of .02%.
3 .05(A)+.02(B)+.3(C)+.07(D)+.08(E)+.3(F)
12500 A-5
B-1.5
C-2.75
D-4
E-2
F-3 .05(5)+.02(1.5)+.3(2.75)+.07(4)+.08(2)+.3(3)
12500 .0195%. This number is extremely close to my test number of .02%.
.05(A)+.02(B)+.3(C)+.07(D)+.08(E)+.3(F)
12500 A-5
B-2
C-4.3
D-4
E-1
F-3 .05(5)+.02(2)+.3(4.3)+.07(4)+.08(1)+.3(3)
12500 .022% This is very close to my research number of .02%.


Based on this, I declare my formula to be:

.05(A)+.02(B)+.3(C)+.07(D)+.08(E)+.3(F)
12500

So, the attachment is MS Word. I couldn't get the formatting to work well just copying the post. It's much nicer in the actual document.