Conditional Probability and the General Multiplication Rule

Section 5.4 Conditional Probability and the General Multiplication Rule

Definition 5.4.1.

The notation \(P(F|E)\) is read “the probability of event \(F\) given event \(E\text{.}\)” It is the probability that event \(F\) occurs, given that event \(E\) occurred.

Fact 5.4.2. Conditional Probability Rule.

\begin{gather*} P(F|E) = \frac{P(E \text{ and } F)}{P(E)} \end{gather*}

Fact 5.4.3. General Multiplication Rule.

\begin{gather*} P(E \text{ and } F) = P(E) \cdot P(F|E) \end{gather*}

Notice that the general multiplication rule follows from the conditional probability rule.

Two events E and F are independent if \(P(E|F) = P(E)\) or, equivalently, if \(P(F|E) = P(F)\text{.}\) This is the same thing as saying that knowing that one event occurred does not affect the probability of the other event occurring (this is the definition of independence from Section 5.3).

Reading Questions Colored Marbles in a Bag

A bag contains 6 blue marbles, 5 red marbles, and 2 green marbles. If two marbles are randomly selected from the bag, what is the probability that:

1.

both are blue?

Solution

\begin{align*} P(B_1 \text{ and } B_2) \amp = P(B_1)\cdot P(B_2|B_1)\\ \amp = \frac{6}{13} \cdot \frac{5}{12} = \frac{5}{26}\\ \amp \approx 0.192 \end{align*}

2.

the first is blue and the second is green?

Solution

\begin{align*} P(B_1 \text{ and } G_2) \amp = P(B_1)\cdot P(G_2|B_1)\\ \amp = \frac{6}{13} \cdot \frac{2}{12} = \frac{1}{13}\\ \amp \approx 0.077 \end{align*}

3.

the first is green and the second is blue?

Solution

\begin{align*} P(G_1 \text{ and } B_2) \amp = P(G_1)\cdot P(B_2 | G_1)\\ \amp = \frac{2}{13} \cdot \frac{6}{12} = \frac{1}{13}\\ \amp \approx 0.077 \end{align*}

4.

one is red, one is blue (order is not specified)?

Solution

\begin{align*} P(R_1 \text{ and } B_2) + P(B_1 \text{ and } R_2) \amp = P(R_1)\cdot P(B_2 | R_1) + P(B_1)\cdot P(R_2 | B_1)\\ \amp = \frac{5}{13} \cdot \frac{6}{12} + \frac{6}{13} \cdot \frac{5}{12}\\ \amp = \frac{5}{26} + \frac{5}{26}\\ \amp = \frac{5}{13} \approx 0.385 \end{align*}

Reading Questions Guessing on a Multiple Choice Quiz

Suppose a student guesses every answer on a multiple choice quiz. There are four questions and each question has four choices: A, B, C or D.

1.

Find the probability that all four answers are correct.

Solution

Multiple choice questions are independent because the answer you give on one does not affect your probability of getting any other question correct.

\begin{align*} P(C \text{ and } C \text{ and } C \text{ and } C) \amp = P(C) \cdot P(C) \cdot P(C) \cdot P(C)\\ \amp = \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}\\ \amp = \left( \frac{1}{4} \right)^4 = \frac{1}{256} \approx 0.0039 \end{align*}

2.

Find the probability that all four answers are incorrect.

Solution

\begin{align*} P(C^c \text{ and } C^c \text{ and } C^c \text{ and } C^c) \amp = P(C^c) \cdot P(C^c) \cdot P(C^c) \cdot P(C^c)\\ \amp = \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4}\\ \amp = \left( \frac{3}{4} \right)^4 = \frac{81}{256} \approx 0.3164 \end{align*}

3.

Find the probability that at least one answer is correct.

Solution

This first solution will take advantage of our solution to the last problem.

\begin{align*} P(\text{at least one correct}) \amp = 1-P(\text{none correct})\\ \amp = 1 - \frac{81}{256}\\ \amp = \frac{175}{256} \approx 0.6836 \end{align*}

We can also approach this problem by summing the probabilities of disjoint events. This solution won't make sense until you read Section 5.5.

\begin{align*} P(\text{at least one correct}) = \amp P(\text{1}) + P(\text{2}) + P(\text{3}) + P(\text{4})\\ = \amp 4 \left( \frac{1}{4}\cdot \frac{3}{4}\cdot \frac{3}{4}\cdot \frac{3}{4} \right) +\\ \amp 6 \left( \frac{1}{4}\cdot \frac{1}{4}\cdot \frac{3}{4}\cdot \frac{3}{4} \right) +\\ \amp 4 \left( \frac{1}{4}\cdot \frac{1}{4}\cdot \frac{1}{4}\cdot \frac{3}{4} \right) +\\ \amp 1 \left( \frac{1}{4}\cdot \frac{1}{4}\cdot \frac{1}{4}\cdot \frac{1}{4} \right)\\ = \amp \frac{108}{256} + \frac{54}{256} + \frac{12}{256} + \frac{1}{256}\\ = \amp \frac{175}{256} \approx .6836 \end{align*}

4.

Find the probability that one is wrong and 3 are correct.

Solution

\begin{align*} P(\text{1 wrong, 3 correct}) = \amp P(C^cCCC \text{ or } CC^cCC \text{ or } CCC^cC \text{ or } CCCC^c)\\ = \amp \frac{3}{4}\cdot \frac{1}{4}\cdot \frac{1}{4}\cdot \frac{1}{4} +\\ \amp \frac{1}{4}\cdot \frac{3}{4}\cdot \frac{1}{4}\cdot \frac{1}{4} +\\ \amp \frac{1}{4}\cdot \frac{1}{4}\cdot \frac{3}{4}\cdot \frac{1}{4} +\\ \amp \frac{1}{4}\cdot \frac{1}{4}\cdot \frac{1}{4}\cdot \frac{3}{4}\\ = \amp 4 \left( \frac{3}{256} \right)\\ = \amp \frac{12}{256} \approx 0.0469 \end{align*}