Quantitative evaluation of system reliability and integrity often involves probability related calculations. Basic probability theory, in general, is sufficient to cover these calculations. Here, I will describe basic probability theory.
Simple probability (1 of 2)
What is the probability that a card drawn at random from a deck of cards will be an ace? Since of the 52 cards in the deck, 4 are aces, the probability is 4/52. In general, the probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. (This assumes the outcomes are all equally likely.) In this case there are four favorable outcomes: (1) the ace of spades, (2) the ace of hearts, (3) the ace of diamonds, and (4) the ace of clubs. Since each of the 52 cards in the deck represents a possible outcome, there are 52 possible outcomes.
Simple probability (2 of 2)
The same principle can be applied to the problem of determining the probability of obtaining different totals from a pair of dice. As shown below, there are 36 possible outcomes when a pair of dice is thrown.
|
Die 1 |
Die 2 |
Total |
|
1 |
1 |
2 |
|
1 |
2 |
3 |
|
1 |
3 |
4 |
|
1 |
4 |
5 |
|
1 |
5 |
6 |
|
1 |
6 |
7 |
|
2 |
1 |
3 |
|
.. |
.. |
.. |
To calculate the probability that the sum of the two dice will equal 5, calculate the number of outcomes that sum to 5 and divide by the total number of outcomes (36). Since four of the outcomes have a total of 5 (1,4; 2,3; 3,2; 4,1), the probability of the two dice adding up to 5 is 4/36 = 1/9 . In like manner, the probability of obtaining a sum of 12 is computed by dividing the number of favorable outcomes (there is only one) by the total number of outcomes (36). The probability is therefore 1/36 .
Conditional probability
A conditional probability is the probability of an event given that another event has occurred. For example, what is the probability that the total of two dice will be greater than 8 given that the first die is a 6? This can be computed by considering only outcomes for which the first die is a 6. Then, determine the proportion of these outcomes that total more than 8. All the possible outcomes for two dice are shown in the section on simple probability. There are 6 outcomes for which the first die is a 6, and of these, there are four that total more than 8 (6,3; 6,4; 6,5; 6,6). The probability of a total greater than 8 given that the first die is 6 is therefore 4/6 = 2/3 . More formally, this probability can be written as: p(total>8 | Die 1 = 6) = 2/3 . In this equation, the expression to the left of the vertical bar represents the event and the expression to the right of the vertical bar represents the condition. Thus it would be read as "The probability that the total is greater than 8 given that Die 1 is 6 is 2/3." In more abstract form, p(A|B) is the probability of event A given that event B occurred.
Probability of A and B (1 of 2)
If A and B are Independent
A and B are two events. If A and B are independent, then the probability that events A and B both occur is: p(A and B) = p(A) x p(B). In other words, the probability of A and B both occurring is the product of the probability of A and the probability of B. What is the probability that a fair coin will come up with heads twice in a row? Two events must occur: a head on the first toss and a head on the second toss. Since the probability of each event is 1/2, the probability of both events is: 1/2 x 1/2 = 1/4. Now consider a similar problem: Someone draws a card at random out of a deck, replaces it, and then draws another card at random. What is the probability that the first card is the ace of clubs and the second card is a club (any club). Since there is only one ace of clubs in the deck, the probability of the first event is 1/52. Since 13/52 = 1/4 of the deck is composed of clubs, the probability of the second event is 1/4. Therefore, the probability of both events is: 1/52 x 1/4 = 1/208 .
Probability of A and B (2 of 2)
If A and B are not Independent
If A and B are not independent, then the probability of A and B is p(A and B) = p(A) x p(B|A) where p(B|A) is the conditional probability of B given A. If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces? Event A is that the first card is an ace. Since 4 of the 52 cards are aces, p(A) = 4/52 = 1/13. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A) = 3/51 = 1/17 and the probability of A and B is: 1/13 x 1/17 = 1/221.
Probability of A or B (1 of 3)
If events A and B are mutually exclusive, then the probability of A or B is simply:
p(A or B) = p(A) + p(B).
What is the probability of rolling a die and getting either a 1 or a 6? Since it is impossible to get both a 1 and a 6, these two events are mutually exclusive. Therefore,
p(1 or 6) = p(1) + p(6) = 1/6 + 1/6 = 1/3
If the events A and B are not mutually exclusive, then
p(A or B) = p(A) + p(B) - p(A and B).
The logic behind this formula is that when p(A) and p(B) are added, the occasions on which A and B both occur are counted twice. To adjust for this, p(A and B) is subtracted. What is the probability that a card selected from a deck will be either an ace or a spade? The relevant probabilities are:
p(ace) = 4/52
p(spade) = 13/52
Probability of A or B (2 of 3)
The only way in which an ace and a spade can both be drawn is to draw the ace of spades. There is only one ace of spades, so:
p(ace and spade) = 1/52 .
The probability of an ace or a spade can be computed as:
p(ace)+p(spade)-p(ace and spade) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.
Consider the probability of rolling a die twice and getting a 6 on at least one of the rolls. The events are defined in the following way:
Event A: 6 on the first roll: p(A) = 1/6
Event B: 6 on the second roll: p(B) = 1/6
p(A and B) = 1/6 x 1/6
p(A or B) = 1/6 + 1/6 - 1/6 x 1/6 = 11/36
The same answer can be computed using the following admittedly convoluted approach:
Getting a 6 on either roll is the same thing as not getting a number from 1 to 5 on both rolls. This is equal to: 1 - p(1 to 5 on both rolls). The probability of getting a number from 1 to 5 on the first roll is 5/6. Likewise, the probability of getting a number from 1 to 5 on the second roll is 5/6 . Therefore, the probability of getting a number from 1 to 5 on both rolls is: 5/6 x 5/6 = 25/36. This means that the probability of not getting a 1 to 5 on both rolls (getting a 6 on at least one roll) is:
1 - 25/36 = 11/36.
Despite the convoluted nature of this method, it has the advantage of being easy to generalize to three or more events. For example, the probability of rolling a die three times and getting a six on at least one of the three rolls is:
1 - 5/6 x 5/6 x 5/6 = .421
In general, the probability that at least one of k independent events will occur is:
1 - (1 - a)k
where each of the events has probability a of occurring.
