Probability Calculations For A Number Of Events
Appendix E of AGS (2000) details the equations for calculation of the probability of a rock falling onto a moving vehicle. It is useful for those of us with a shaky understanding of the detail of probability calculations to consider how equation E1 of Appendix E can be derived as this fundamental concept can be applied to other cases.
For simplicity, consider the probability of throwing a six with a normal cubic die. Since there are six possible outcomes, the probability is 1 in 6 (1/6).
What then is the probability of throwing a six in any one or more of ten throws of the die?
- It can not be 10 times (1/6) since that is greater than 1.0 which is impossible.
- It is not (1/6)10 since that is the probability of throwing a six on each of ten throws.
- Consider what the possible outcomes for the ten throws would be in terms of the number of sixes:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10 sixes
All except 0 sixes satisfy the requirement of a six on one or more throws.
The probability of not getting a six on a single throw is (5/6). This can also be derived as (1 – Probability of a six).
Then the probability of 0 sixes in ten throws is (5/6)10.
The numerical total of the probability of each of the possible outcomes must be 1. That is (Probability of 0 sixes + Probability of one or more sixes) = 1
Therefore, the Probability of 1 or more sixes = (1- (5/6)10)
Expressing this algebraically, if
P(6) = Probability of throwing a six
N = Number of throws
Then
P(6:10) = 1- (1 – P(6))N
This is in effect Equation E1.