Basic Concepts of Probability

Although typically we expect the conditional probability $P(A \mid B)$ to be different from the probability $P(A)$ of $A$, it does not have to be different from $P(A)$. When $P(A \mid B)=P(A)$, the occurrence of $B$ has no effect on the likelihood of $A$. Whether or not the event $A$ has occurred is independent of the event $B$.

Using algebra it can be shown that the equality $P(A \mid B)=P(A)$ holds if and only if the equality $P(A \cap B)=P(A) \cdot P(B)$ holds, which in turn is true if and only if $P(B \mid A)=P(B)$. This is the basis for the following definition.

Definition

Events $A$ and $B$ are independent if

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

If $A$ and $B$ are not independent then they are dependent.

The formula in the definition has two practical but exactly opposite uses:

1. In a situation in which we can compute all three probabilities $P(A), P(B)$, and $P(A \cap B)$, it is used to check whether or not the events $A$ and $B$ are independent:

• If $P(A \cap B)=P(A) \cdot P(B)$, then $A$ and $B$ are independent.
• If $P(A \cap B) \neq P(A) \cdot P(B)$, then $A$ and $B$ are not independent.

2. In a situation in which each of $P(A)$ and $P(B)$ can be computed and it is known that $A$ and $B$ are independent, then we can compute $P(A \cap B)$ by multiplying together $P(A)$ and $P(B)$ : $P(A \cap B)=P(A) \cdot P(B)$

EXAMPLE 23

A single fair die is rolled. Let $A=\{3\}$ and $B=\{1,3,5\}$. Are $A$ and $B$ independent?

Solution:

In this example we can compute all three probabilities $P(A)=1 / 6, P(B)=1 / 2$, and $P(A \cap B)=P(\{3\})=1 / 6$. Since the product $P(A) \cdot P(B)=(1 / 6)(1 / 2)=1 / 12$ is not the same number as $P(A \cap B)=1 / 6$, the events $A$ and $B$ are not independent.

EXAMPLE 24

The two-way classification of married or previously married adults under 40 according to gender and age at first marriage in Note 3.48 "Example 21" produced the table

Determine whether or not the events F: "female" and E: "was a teenager at first marriage" are independent.

Solution:

The table shows that in the sample of 902 such adults, 452 were female, 125 were teenagers at their first marriage, and 82 were females who were teenagers at their first marriage, so that

$\begin{align*} P(F)=\frac{452}{902} \quad P(E)=\frac{125}{902} \quad P(F \cap E)=\frac{82}{902} \end{align*}$

Since

$\begin{align*} P(F) \cdot P(E)=\frac{452}{902} \cdot \frac{125}{902}=0.069 \end{align*}$

is not the same as

$\begin{align*} P(F \cap E)=\frac{82}{902}=0.091 \end{align*}$

we conclude that the two events are not independent.

EXAMPLE 25

Many diagnostic tests for detecting diseases do not test for the disease directly but for a chemical or biological product of the disease, hence are not perfectly reliable. The sensitivity of a test is the probability that the test will be positive when administered to a person who has the disease. The higher the sensitivity, the greater the detection rate and the lower the false negative rate.

Suppose the sensitivity of a diagnostic procedure to test whether a person has a particular disease is 92%. A person who actually has the disease is tested for it using this procedure by two independent laboratories.

Solution:

EXAMPLE 26

The specificity of a diagnostic test for a disease is the probability that the test will be negative when administered to a person who does not have the disease. The higher the specificity, the lower the false positive rate.

Suppose the specificity of a diagnostic procedure to test whether a person has a particular disease is 89%.

Solution:

EXAMPLE 27

The reliability of a system can be enhanced by redundancy, which means building two or more independent devices to do the same job, such as two independent braking systems in an automobile.

Suppose a particular species of trained dogs has a $90 \%$ chance of detecting contraband in airline luggage. If the luggage is checked three times by three different dogs independently of one another, what is the probability that contraband will be detected?

Solution:

Let $D_{1}$ denote the event that the contraband is detected by the first dog, $D_{2}$ the event that it is detected by the second dog, and $D_{3}$ the event that it is detected by the third. Since each dog has a $90 \%$ of detecting the contraband, by the Probability Rule for Complements it has a $10 \%$ chance of failing. In symbols, $P\left(D_{1}^{c}\right)=0.10, P\left(D_{2}^{c}\right)=0.10$, and $P\left(D_{3}^{c}\right)=0.10$

Let $D$ denote the event that the contraband is detected. We seek $P(D)$. It is easier to find $P\left(D^{c}\right)$, because although there are several ways for the contraband to be detected, there is only one way for it to go undetected: all three dogs must fail. Thus $D^{c}=D_{1}^{c} \cap D_{2}^{c} \cap D_{3}^{c}$, and

$\begin{align*} P(D)=1-P\left(D^{c}\right)=1-P\left(D_{1}^{c} \cap D_{2}^{c} \cap D_{3}^{c}\right) \end{align*}$

But the events $D_{1}, D_{2}$, and $D_{3}$ are independent, which implies that their complements are independent, so

$\begin{align*} P\left(D_{1}^{c} \cap D_{2}^{c} \cap D_{3}^{c}\right)=P\left(D_{1}^{c}\right) \cdot P\left(D_{2}^{c}\right) \cdot P\left(D_{3}^{c}\right)=0.10 \times 0.10 \times 0.10=0.001 \end{align*}$

Using this number in the previous display we obtain

$\begin{align*} P(D)=1-0.001=0.999 \end{align*}$

That is, although any one dog has only a $90 \%$ chance of detecting the contraband, three dogs working independently have a $99.9 \%$ chance of detecting it.

Probabilities on Tree Diagrams

Some probability problems are made much simpler when approached using a tree diagram. The next example illustrates how to place probabilities on a tree diagram and use it to solve a problem.

EXAMPLE 28

A jar contains 10 marbles, 7 black and 3 white. Two marbles are drawn without replacement, which means that the first one is not put back before the second one is drawn.

Solution:

A tree diagram for the situation of drawing one marble after the other without replacement is shown in Figure 3.6 "Tree Diagram for Drawing Two Marbles". The circle and rectangle will be explained later, and should be ignored for now.

Figure 3.6

Tree Diagram for Drawing Two Marbles

The numbers on the two leftmost branches are the probabilities of getting either a black marble, 7 out of 10 , or a white marble, 3 out of 10 , on the first draw. The number on each remaining branch is the probability of the event corresponding to the node on the right end of the branch occurring, given that the event corresponding to the node on the left end of the branch has occurred. Thus for the top branch, connecting the two Bs, it is $P\left(B_{2} \mid B_{1}\right)$, where $B_{1}$ denotes the event "the first marble drawn is black" and $B_{2}$ denotes the event "the second marble drawn is black". Since after drawing a black marble out there are 9 marbles left, of which 6 are black, this probability is $6 / 9$.

The number to the right of each final node is computed as shown, using the principle that if the formula in the Conditional Rule for Probability is multiplied by $P(B)$, then the result is

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

Of course, this answer could have been found more easily using the Probability Law for Complements, simply subtracting the probability of the complementary event, "two white marbles are drawn," from 1 to obtain $1-0.07=0.93$.

As this example shows, finding the probability for each branch is fairly straightforward, since we compute it knowing everything that has happened in the sequence of steps so far. Two principles that are true in general emerge from this example:

Probabilities on Tree Diagrams

1. The probability of the event corresponding to any node on a tree is the product of the numbers on the unique path of branches that leads to that node from the start.
2. If an event corresponds to several final nodes, then its probability is obtained by adding the numbers next to those nodes.