MA121: Introduction to Statistics

LEARNING OBJECTIVES

1. To learn the concept of a conditional probability and how to compute it.
2. To learn the concept of independence of events, and how to apply it.

Conditional Probability

Suppose a fair die has been rolled and you are asked to give the probability that it was a five. There are six equally likely outcomes, so your answer is $1 / 6 .$ But suppose that before you give your answer you are given the extra information that the number rolled was odd. Since there are only three odd numbers that are possible, one of which is five, you would certainly revise your estimate of the likelihood that a five was rolled from $1 / 6$ to $1 / 3$. In general, the revised probability that an event $A$ has occurred, taking into account the additional information that another event $B$ has definitely occurred on this trial of the experiment, is called the conditional probability of $A$ given $B$ and is denoted by $P(A \mid B)$. The reasoning employed in this example can be generalized to yield the computational formula in the following definition.

Definition

The conditional probability of $A$ given $B$, denoted $P(A \mid B)$, is the probability that event $A$ has occurred in a trial of a random experiment for which it is known that event $B$ has definitely occurred. It may be computed by means of the following formula:

Rule for Conditional Probability

$\begin{align*} P(A \mid B)=\frac{P(A \cap B)}{P(B)} \end{align*}$

EXAMPLE 20

A fair die is rolled.

a. Find the probability that the number rolled is a five, given that it is odd.

b. Find the probability that the number rolled is odd, given that it is a five.

Solution:

The sample space for this experiment is the set $S=\{1,2,3,4,5,6\}$ consisting of six equally likely outcomes. Let $F$ denote the event "a five is rolled" and let $O$ denote the event "an odd number is rolled, "so that

$\begin{align*} F=\{5\} \quad \text { and } \quad O=\{1,3,5\} \end{align*}$

a. This is the introductory example, so we already know that the answer is $1 / 3$. To use the formula in the definition to confirm this we must replace $A$ in the formula (the event whose likelihood we seek to estimate) by $F$ and replace $B$ (the event we know for certain has occurred) by $O$ :

$\begin{align*} P(F \mid O)=\frac{P(F \cap O)}{P(O)} \end{align*}$

Since $F \cap O=\{5\} \cap\{1,3,5\}=\{5\}, P(F \cap O)=1 / 6$

Since $O=\{1,3,5\}, P(O)=3 / 6$

Thus

$\begin{align*} P(F \mid O)=\frac{P(F \cap O)}{P(O)}=\frac{1 / 6}{3 / 6}=\frac{1}{3} \end{align*}$

b. This is the same problem, but with the roles of $F$ and $O$ reversed. Since we are given that the number that was rolled is five, which is odd, the probability in question must be 1. To apply the formula to this case we must now replace $A$ (the event whose likelihood we seek to estimate) by $O$ and $B$ (the event we know for certain has occurred) by $F$ :

$\begin{align*} P(O \mid F)=\frac{P(O \cap F)}{P(F)} \end{align*}$

Obviously $P(F)=1 / 6$. In part (a) we found that $P(F \cap O)=1 / 6$. Thus

$\begin{align*} P(O \mid F)=\frac{P(O \cap F)}{P(F)}=\frac{1 / 6}{1 / 6}=1 \end{align*}$

Just as we did not need the computational formula in this example, we do not need it when the information is presented in a two-way classification table, as in the next example.

EXAMPLE 21

In a sample of 902 individuals under 40 who were or had previously been married, each person was classified according to gender and age at first marriage. The results are summarized in the following two-way classification table, where the meaning of the labels is:

• M: male
• F: female
• E: a teenager when first married
• W: in one’s twenties when first married
• H: in one’s thirties when first married

The numbers in the first row mean that 43 people in the sample were men who were first married in their teens, 293 were men who were first married in their twenties, 114 men who were first married in their thirties, and a total of 450 people in the sample were men. Similarly for the numbers in the second row. The numbers in the last row mean that, irrespective of gender, 125 people in the sample were married in their teens, 592 in their twenties, 185 in their thirties, and that there were 902 people in the sample in all. Suppose that the proportions in the sample accurately reflect those in the population of all individuals in the population who are under 40 and who are or have previously been married. Suppose such a person is selected at random.

Solution:

It is natural to let $E$ also denote the event that the person selected was a teenager at first marriage and to let $M$ denote the event that the person selected is male.

The proportion of males in the sample who were in their teens at their first marriage is $43 / 450$. This is the relative frequency of such people in the population of males, hence $P(E \mid M)=43 / 450 \approx 0.096$ or about 10%.

In the next example, the computational formula in the definition must be used.

EXAMPLE 22

Suppose that in an adult population the proportion of people who are both overweight and suffer hypertension is 0.09; the proportion of people who are not overweight but suffer hypertension is 0.11; the proportion of people who are overweight but do not suffer hypertension is 0.02; and the proportion of people who are neither overweight nor suffer hypertension is 0.78. An adult is randomly selected from this population.

Solution:

Let $H$ denote the event "the person selected suffers hypertension". Let $O$ denote the event "the person selected is overweight". The probability information given in the problem may be organized into the following contingency table: