*Theorem of Total Probability*

*Theorem of Total Probability*

*If E _{i}, i = 1, 2, . . . , n are disjoint events, with P(E_{1}) + P(E_{2}) + . . . + P(E_{n}) = 1, then for any event F: P(F) = P(F and E_{1}) + P(F and E_{2}) + . . . + P(F and E_{n}) = P(F|E_{1})P(E_{1}) + P(F|E_{2})P(E_{2}) + . . . +P(F|E_{n})P(E_{n})*

**Example**

Suppose that a company uses a specialized part in its manufacturing process, and the part can be purchased from only three suppliers. The company consistently purchases 60%, 30%, and 10% of its stock of this part from suppliers *A*, *B*, and *C*, respectively. A portion of the parts from each supplier is defective. The company has determined that 1%, 3%, and 8% of the parts purchased from suppliers *A*, *B*, and *C*, respectively, are defective. What is the probability that a randomly selected part from the company's stock is defective?

**Solution**

When working probability problems, it is best to start by presenting what is known using symbols. Notice that we know the probability that a randomly selected part comes from each of the suppliers. Let *E _{i}* be the event that a randomly selected part from the company's stock is from supplier

*i*,

*i*=

*A*,

*B*, or

*C*. Then, for a randomly selected part,

*P*(

*E*) = 0.6,

_{A}*P*(

*E*) = 0.3, and

_{B}*P*(

*E*) = 0.1.Notice that

_{C}*E*,

_{A}*E*, and

_{B}*E*are disjoint (a randomly selected part cannot come from two different suppliers) and exhaustive (any randomly selected part must have come from one of these three suppliers). Consequently,

_{C}*P*(

*E*and

_{A}*E*) =

_{B}*P*(

*E*and

_{A}*E*) =

_{C}*P*(

*E*and

_{B}*E*) = 0 (because any two events are disjoint, the probability of both occurring is zero), and

_{C}*P*(

*E*) +

_{A}*P*(

*E*) +

_{B}*P*(

*E*) = 1.

_{C}Let *D* be the event that a randomly selected part is defective. Notice that *D* overlaps with *E _{A}*,

*E*, and

_{B}*E*because it is possible for a defective part to come from any supplier. Further, although we do not know the probability of

_{C}*D*,we do know the probability of

*D*if we are given the supplier. That is, we know

*P*(

*D*|

*E*) = 0.01,

_{A}*P*(

*D*|

*E*) = 0.03, and

_{B}*P*(

*D*|

*E*) = 0.08. Therefore, if we add together the probabilities that a part was defective and from supplier

_{C}*A*, defective and from supplier

*B*, and defective and from supplier

*C*, we would have the probability of a defective part.We can represent this in symbols as

*P*(

*D*) =

*P*(

*D*and

*E*) +

_{A}*P*(

*D*and

*E*) +

_{B}*P*(

*D*and

*E*) =

_{C}*P*(

*D*|

*E*)

_{A}*P*(

*E*) +

_{A}*P*(

*D*|

*E*)

_{B}*P*(

*E*) +

_{B}*P*(

*D*|

*E*)

_{C}*P*(

*E*) = 0.01(0.6) + 0.03(0.3) + 0.08(0.1) = 0.023 Thus, the probability that a randomly selected part is defective is 0.023.

_{C}**Bayes' Theorem**

As with the Theorem of Total Probability, suppose that the sample space is partitioned into *n* mutually exclusive and exhaustive events, *E _{i}*,

*i*= 1, 2, . . . ,

*n*. Also, assume that for an event

*F*on the same sample space

*S*,

*P*(

*F*|

*E*) is known for all

_{i}*i*. However, instead of finding

*P*(

*F*),we want to find

*P*(

*E*|

_{i}*F*). Bayes' Theorem says that, if

*E*,

_{i}*i*= 1, 2, . . . ,

*n*are disjoint events with

*P*(

*E*

_{1}) +

*P*(

*E*

_{2}) + . . . +

*P*(

*E*) = 1, then for any event

_{n}*F*,

*Bayes' Theorem*

*Bayes' Theorem*

*If E _{i}, i = 1, 2, . . . , n are disjoint events with P(E_{1}) + P(E_{2}) + . . . + P(E_{n}) = 1, then for any event F, *.

**Example**

Suppose that, for the same manufacturer and parts suppliers in the last example, the manufacturer selects a part at random and determines it to be defective. What is the probability that it came from supplier *A*? Notice that we want to find *P*(*E _{A}*|

*D*). If we begin by using the definition of conditional probability,we have

Using the properties of conditional probability and the Theorem of Total Probability, we can rewrite the conditional probability of *E _{A}* given

*D*as

*P*(

*E*|

_{A}*D*)

Of course, because we had already found the probability that a randomly selected part is defective, it was not necessary to expand the denominator using the Theorem of Total Probability. Finding the probability that the supplier was *E _{A}* given that the part was defective is an example of the application of Bayes' Theorem.

**Basic Ideas in Probability In Short**

Probability is the foundation of statistics. In this lesson, we learned that the sample space is the set of all possible outcomes of an experiment, and an event is any subset of the sample space. The classical definition and the relative frequency definition of probability were compared. Conditional probabilities and independent events were discussed. Finally, the Theorem of Total Probability and Bayes' Theorem were presented as useful methods of finding probabilities of interest.

Find practice problems and solutions for these concepts at Basic Ideas in Probability Practice Questions.

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