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Section 5.3 Independence and the Multiplication Rule

Definition 5.3.1.

Two events \(E\) and \(F\) are independent if the occurrence of event \(E\) in a probability experiment does not affect the probability of event \(F\text{.}\) Two events are dependent if the occurrence of event \(E\) in a probability experiment affects the probability of event \(F\text{.}\)

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Reading Questions Reading Questions

Determine if the following pairs of events are independent or dependent.

1.

You are late for work. Your car runs out of gas.

Answer

dependent

2.

Winning a game of Monopoly. Winning the lottery

Answer

independent

3.

You gain weight. You eat fast food for dinner every night.

Answer

dependent

4.

Speeding on the interstate highway. Getting pulled over by a police officer.

Answer

dependent

5.

Owning a dog. Growing an herb garden.

Answer

independent

6.

Robbing a bank. Going to jail.

Answer

dependent

Remark 5.3.2.

Knowing two events are disjoint means that the events are dependent. Why? Consider the example of pulling chips numbered 0-9 out of a bag. Suppose event \(E\) corresponds with pulling a chip labelled with a 1 or a 2 out of the bag, i.e. \(E=\{1,2\}\) and suppose \(F=\{3,4\}\text{,}\) i.e. \(F\) corresponds with pulling a 3 or 4 chip out of the bag. Notice that these events are disjoint because they have no outcomes in common. Finally, notice that if we know that event \(E\) occurred (a 1 or 2 is drawn), then we know for certain that \(F\) (drawing a 3 or 4) did not occur. Thus the occurrence of \(E\) affects the probability of event \(F\) and vice versa.

Example 5.3.4. Flipping Two Heads in a Row.

What is the probability that you flip a fair coin twice and get heads twice?

Solution

Since these events are independent we can use the multiplication rule:

\begin{align*} P(H \text{ and } H) \amp = P(H) \cdot P(H)\\ \amp = \frac{1}{2} \cdot \frac{1}{2}\\ \amp = \frac{1}{4} = 0.25 = 25 \% \end{align*}
Example 5.3.5. Rolling Four Ones in a Row.

What is the probability that you roll 4 ones in a row when you roll a fair, six sided die four times?

Solution

Since these events are independent we can use the multiplication rule:

\begin{align*} P(1 \text{ and } 1 \text{ and } 1 \text{ and } 1) \amp = P(1) \cdot P(1) \cdot P(1) \cdot P(1)\\ \amp = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}\\ \amp = \frac{1}{1296} \approx 0.000716 = .0716 \% \end{align*}
Example 5.3.6.

What is the probability of selecting a two or a club from a deck of cards?

Solution

These events are not disjoint, because there is one card that is both a 2 and a club, thus we must use the general addition rule.

\begin{align*} P(2 \text{ or club} ) \amp = P(2) + P(\text{club}) - P(2\cap \text{club})\\ \amp = \frac{4}{52} + \frac{13}{52} - \frac{1}{52}\\ \amp = \frac{16}{52} = \frac{4}{13} \approx 31\% \end{align*}
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