Three Special Relationships among Events

Independent Events

Two events A and B are said to be independent if the occurrence of one does not influence the probability of the occurrence of the other. The one sure way to check to see if events are independent is to confirm that

P{A} = P{A|B}

orP{B} = P{B|A}.

Note that if one of these is true, the other must be true. If the above expressions do not hold, then the events are said to be dependent.

Example 1: In the merchandise example, are the events A and B independent or dependent?

Example 2: Toss a fair die. Let

A: {even number}

B: {number less than or equal to 4}

Are A and B independent?

Recall that If A and B are independent, P{A|B} = P{A}. By substitution

Thus we have the following special case:

This implies that computing the joint probabilities (intersections) is easy for independent events, since you just multiply their respective probabilities. As a result, there are no conditional probabilities involved.

In Example 2, what is

Mutually Exclusive Events

Two events are mutually exclusive if they cannot happen at the same time. Probabilistically,

Now remember that we established the addition rule

What happens if A and B are mutually exclusive?

This implies that computing the probability of a union is easy to do. In trying to compute the probability of an event, it is frequently advantageous to decompose the event, A, into mutually exclusive events, say A1, A2, …, An, compute P{A1}, P{A2}, …, P{An}, and then compute the probability of A by

P{A} = P{A1} + P{A2} + … + P{An}.

Complementary Events

The set AC (or ) is referred to as the complement of A if it contains all simple events not in A.

Thus

Q1: Are A and AC mutually exclusive?

Q2: Are A and AC independent? (Assume P{A} ≠ 0 and P{AC} ≠ 0.)

Q3: If A and B are mutually exclusive, are they independent? (Assume again P{A} ≠ 0 and P{B} ≠ 0.)

A Serious Example

A test is being introduced to detect a terminal disease in order to enhance early detection. If the person has the disease, the test will show positive 99% of the time. (How do we find this out?) If the person does not have the disease, the test will show negative 95% of the time. One half of one percent (0.5% = .005) of the population has the disease in question. If the test is administered to an individual and is positive, what is the probability that he/she has the disease? (Why isn’t the answer 0.5%?)

Let S:{test is positive}

D:{person has the disease}

Fill in the following in terms of S and D.

______= .005

______= .99

______= .95

What is the probability that we want?

Can we fill in the contingency table?

S (positive) / SC (negative) / Total
D (disease)
DC (no disease)
Total

Note

Thus

P{D|S} =

The problem that we have solved is actually an application of Bayes’ Theorem. In notational form, Bayes’ Theorem may be stated as:

where the Ej are mutually exclusive and collectively exhaustive; i.e.,

P{E1} + P{E2} + … + P{En} = 1.

Now that the revised probability of have the disease is about 9%, it would be natural for the test to be repeated. In this case, a new revised probability could be computed if the test were positive by using P{D} = .0904. Using our same approach with this new P{D}, you can show that this new revised probability is .665.

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