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Chapter 6
Decision Making using Probability
In this chapter, we look at more complicated notions of probability and how we can use probability in order to aid in management decision making.
6.1 Conditional Probability
So far we have only considered probabilities of single events or of several independent events, like two rolls of a die. However in reality many events are related. For example the probability of it raining in 5 minutes time is dependent on whether or not it is raining now.
We need a mathematical notation to capture how the probability of one event depends on other events taking place. We do this as follows. Consider two events A and B. We write
P (A|B)
for the probability of A given B has already happened. We describe P (A|B) as the conditional probability of A given B. For example, the probability of it raining in 5 minutes time given that it is raining now would be
P (Rain in 5 minutes|Raining now).
Utility companies need to be able to forecast periods of high demand. They describe their forecasts in terms of probabilities. Gas and electricity suppliers might relate them to air temperature. For example,
P (High demand|air temperature is below normal) = 0. 6 P (High demand|air temperature is normal) = 0. 2 P (High demand|air temperature is above normal) = 0. 05.
We can calculate these conditional probabilities using the formula
P (A|B) =
P (A and B) P (B)
that is, in terms of the probability of both events occurring, P (A and B), and the probability of the event that has already taken place, P (B).
To see how this formula works, let’s consider a simple example based on the class of students in Exercises 5.
Student Height Weight Shoe Student Height Weight Shoe Number Sex (m) (kg) Size Number Sex (m) (kg) Size 1 M 1.91 70 11.0 10 M 1.78 76 8. 2 F 1.73 89 6.5 11 M 1.88 64 9. 3 M 1.73 73 7.0 12 M 1.88 83 9. 4 M 1.63 54 8.0 13 M 1.70 55 8. 5 F 1.73 58 6.5 14 M 1.76 57 8. 6 M 1.70 60 8.0 15 M 1.78 60 8. 7 M 1.82 76 10.0 16 F 1.52 45 3. 8 M 1.67 54 7.5 17 M 1.80 67 7. 9 F 1.55 47 4.0 18 M 1.92 83 12.
Suppose we want the probability that a student chosen at random from this class will be female given that the student’s shoe size is less than 8. We could simply find the proportion of students with shoe sizes less than 8 who are female. There are 7 students with shoe sizes less than 8 and 4 of these are female. So
P (Female|Shoe size less than 8) =
This probability can also be calculated using the above formula as follows:
P (Shoe size less than 8) =
P (Shoe size less than 8 and female) =
and so
P (Female|Shoe size less than 8) =
P (Shoe size less than 8 and female) P (Shoe size less than 8)
6.2 Multiplication of probabilities
We saw in Chapter 5 that, if two events A and B are independent, then P (A andB) = P (A)P (B). Now we know that
P (A|B) =
P (A and B) P (B)
we can easily see that P (A and B) = P (B)P (A|B).
Of course it is also true that P (A and B) = P (A)P (B|A).
and
P (Female) = P (Female and < 30) + P (Female and 30 − 50) + P (Female and 50+) = 0.325 + 0.175 + 0.075 = 0. 575.
Also, the age distribution of the customers is
P (< 30) = P (Male and < 30) + P (Female and < 30) = 0.275 + 0.325 = 0. 6 P (30 − 50) = P (Male and 30 − 50) + P (Female and 30 − 50) = 0.125 + 0.175 = 0. 3 P (50+) = P (Male and 50+) + P (Female and 50+) = 0.025 + 0.075 = 0. 1.
Using this information we can calculate various probabilities such as:
P (Male| 30 − 50) =
P (Male and 30 − 50) P (30 − 50)
P (Female| 30 − 50) = 1 − P (Male| 30 − 50) = 1 − 0 .4167 = 0. 5833
and
P (< 30 |Male) =
P (Male and < 30) P (Male)
P (30 − 50 |Male) =
P (Male and 30 − 50) P (Male)
P (50 + |Male) = 1 − P (< 30 |Male) − P (30 − 50 |Male) = 1 − 0. 6471 − 0 .2941 = 0. 0588.
6.3 Tree Diagrams
Tree diagrams or probability trees are simple clear ways of presenting probabilistic information. Let us first consider a simple example in which a die is rolled twice. Suppose we are interested in the probability that we score a six on both rolls. This probability can be calculated as
P (Six and Six) = P (Six on 1st throw) × P (Six on 2nd throw|Six on 1st throw)
=
×
This example can be represented as a tree diagram in which experiments are represented by circles (called nodes ) and the outcomes of the experiments as branches. The branches are annotated by the probability of the particular outcome.
Here the probability of a six followed by a six is found by tracing the branch corresponding to this outcome through the tree. Note that the ends of the branches of the tree are usually known as terminal nodes.
Consider a more complicated example. A machine is used to produce components. Each time it produces a component there is a chance that the component will be defective. When the machine is working correctly the probability that a component is defective is 0.05. Sometimes, though, the machine requires adjustment and, when this is the case, the probability that a component is defective is 0.2. At the time in question there is a probability of 0.1 that the machine requires adjustment. Components produced by the machine are tested and either accepted or rejected. A component which is not defective is accepted with probability 0.97 and (falsely) rejected with probability 0.03. A defective component is (falsely) accepted with probability 0.15 and rejected with probability 0.85.
We can calculate various probabilities. For example:
P (accepted) = 0 .82935 + 0.00675 + 0.07760 + 0.00300 = 0. 9167 P (defective) = (0. 9 × 0 .05) + (0. 1 × 0 .2) = 0.045 + 0.02 = 0. 065 P (defective and accepted) = 0 .00675 + 0.00300 = 0. 00975
P (accepted | defective) =
P (defective | accepted) =
P (machine OK and accepted) = 0 .82935 + 0.00675 = 0. 8361
P (machine OK | accepted) =
P (machine OK and rejected) = 0 .02565 + 0.03825 = 0. 0639 P (rejected) = 1 − P (accepted) = 0. 0833
P (machine OK | rejected) =
and the managing director believes the probability of the product being successful can be classed into three categories: high, medium or low. She thinks that these categories will occur with prob- abilities 0.2, 0.35 and 0.45 respectively and her thoughts on the likely profits (in £ K) to be earned in each plan are
High Medium Low Direct 100 55 - Internet 46 25 15 Licence 20 20 20
The EMV of each plan can be calculated as follows:
EM V (Direct) = 0. 2 × 100 + 0. 35 × 55 + 0. 45 × (−25) = £ 28 K EM V (Internet) = 0. 2 × 46 + 0. 35 × 25 + 0. 45 × 15 = £ 24. 7 K EM V (Licence) = 0. 2 × 20 + 0. 35 × 20 + 0. 45 × 20 = £ 20 K.
On the basis of expected monetary value, the best choice is the Direct approach.
In this example we have to make a decision. When we include a decision in a probability tree we use a rectangular node, called a decision node to represent the decision. The diagram is then called a decision tree. There are no probabilities at a decision node but we evaluate the expected monetary values of the options. In a decision tree the first node is always a decision node. There may also be other decision nodes. If there is another decision node then we evaluate the options there and choose the best and the expected value of this option becomes the expected value of the branch leading to the decision node.
6.5 Exercises 6
- A company has installed a new computer system and some employees are having difficulty logging on to the system. They have been given training and the problems which arose during training were recorded and their probabilities calculated as follows: - An employee has a probability of 0.9 of logging on successfully on the first attempt. - If the employee logs in successfully then the employee will also be successful on each later attempt with probability 0.9. - If the employee tries to log in and is not successful then the employee loses confidence and the probability of a successful log-in on later occasions drops to 0.5.
Use a tree diagram to find the following probabilities:
(a) An employee successfully logs on in each of the first three attempts. (b) An employee fails in the first attempt but is successful in the next two attempts. (c) An employee logs on successfully only once in three attempts. (d) An employee does not manage to log on successfully in three attempts.
- The owner of a small business has the right to have a retail stall at a large festival to be held during the summer. She judges that this would either be a success or a failure and that the probability that it is a success is 0.4. If the stall was a success, the net income from it would be £ 90,000. If the stall was a failure, there would be a net loss of £ 30,000 from it. To help to make the decision, the owner could pay for market research. This would cost £ 5,000. The market research will either give a positive indication or a negative indication. The conditional probability that it gives a positive indication, given that the stall will actually be a success, is 0.75. The conditional probability that it gives a positive indication, given that the stall will actually be a failure, is 1 / 3. The owner has various options: - Do nothing. - Go ahead without market research. - Pay for the market research. - Sell her right to a stall for £ 10,000.
If she pays for the market reserach then, depending on the outcome, she can:
- Do nothing more.
- Go ahead.
- Sell her right to a stall. If the market research gave a positive indication the price would be £ 35,000. If the market research gave a negative indication the price would be only £ 3,000.
What should she do? This is a fairly complicated question so it is best to tackle it in stages.
