AP Statistics Lectures
by Arnold Kling

Independence and Failure Models

We use a failure model when we look at how the probability of the success of a system depends on the success of independent individual events.

For example, suppose that we think of a motorboat as the system. The system fails only if every individual engine on the boat fails. If the boat has three engines, all three have to fail in order for the boat to fail.

We assume that each individual event is independent of the others. The first engine failing is one event. The second engine failing is another event. Independence means that the probability of both engines failing is equal to the product of the probability of the first engine failing times the probability of the second engine failing.

There are two common forms of failure models.

1. In the single-failure model, one individual failure causes the entire system to fail. For example, in a single-elimination play-off tournament (such as the NFL play-offs), one loss means that you are out of the tournament.

2. In the compound-failure model, every component must fail in order for the entire system to fail. For example, if the boat will operate as long as at least one engine is running, then every engine must fail in order for the boat to fail.

The single-failure and the compound-failure models are mirror images of one another. A single-failure model is the same as a compound-success model: if a single component failure causes the system to fail, then every componenet must succeed for the system to succed. A compound failure model is the same as a single-success model: if every component must fail to cause the system to fail, then only one component must succeed to cause the system to succeed.

Here is a mathematical description of the failure model:

P(s) = the probability that one component (one engine) is successful
P(f) = 1 - P(s) is the probability that one component fails.
P(S) = the probability that the system (the entire motorboat) is successful
P(F) = 1 - P(S) = the probability that the system fails.
n = the number of components (engines)

For the single-failure model, P(S) = P(s)n.
For the compound-failure model, P(F) = P(f)n.

For example, suppose that we have to win three play-off games to make it to the finals, and that our chance of winning any one game is 0.6, or 60 percent. This is a single-failure model, and the probability of making the playoffs, P(S), equals (0.6)3.

On the other hand, suppose that we have a motorboat where every engine must fail in order for the boat to fail. This is a compound-failure model. Suppose that the probability that an engine will succeed, P(s) = 0.6, or 60 percent. That means that P(f) = 0.4, or 40 percent. With three engines, the probabiliby that the boat will fail, P(F) = (0.4)3. The probability that the boat will succeed, P(S) = 1 - P(F).

You can use the failure model to solve problems where the probability of a systemic success or failure depends on the probability of individually independent successes or failures. To use a failure model,

• Determine from the context whether it is a single-failure or a compound-failure model.
• Determine what information is given and what information is to be solved for. For example, sometimes you might be given P(f), the probability of an individual failure, and a desired value for P(S), the probability of systemic success. You might then be asked to solve for n, the number of components needed to achieve the desired value of P(S).
• If necessary, use the facts that P(s) + P(f) = 1 and P(S) + P(F) = 1 to help solve for missing information.
• Use either the single-failure formula or the compound-failure formula, depending on which fits the problem.

Some sample problems:

1. The probability of getting an odd number less than 4 when you roll one die is 1/3. What is the probability of doing this four times in a row?
2. Joanna is practicing her basketball shots. She usually makes 55 percent of her shots. How many shots will she have to try before she has at least a 95 percent change of having made at least one basket?

Problem one is a single-failure model. Therefore, P(S) = p(s)4 = (1/3)4 = 1/81.

Problem two is a compound-failure model. We want P(S) to be 0.95, which means that P(F) = 0.05, or 5 percent. We know that P(f) = 1 - P(s) = 1 - .55 = .45, or 45 percent. P(F) = P(f)n, and we are solving for n. It turns out that if n=3, P(F) is above .05, but if n=4, then P(F) is below .05. So n = 4.