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# Conditional Probability

## P(B|A) = P(A and B)/ P(A)

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Understanding Conditional Probability

On Thursday, Carey was in charge of answering the phones and booking appointments for bike repairs. The bike shop repairs bikes on Monday, Tuesday and Wednesday mornings and on Thursday and Friday afternoons. All appointments are booked randomly. The person making the appointment can choose or the person answering the phone can choose. Carey booked two appointments right away.

What are the chances that both of these appointments were booked for a Monday, Tuesday or Wednesday morning?

In this concept, you will learn to understand conditional probability.

### Conditional Probability

Sometimes the outcome that you get when figuring out a probability is conditional. A conditional outcome means that you will only get an outcome if the conditions are designed to cause a specific result.

Let’s look at an example.

Consider a jar with 4 black marbles and 6 white marbles. If you randomly pull out 2 marbles from the jar, one at a time, without replacing the first marble, what is the probability that both marbles will be white?

First, start by approaching the problem the same as you would with dependent events. There are 6 white marbles and there are 10 marbles in the jar. The probability of the first marble being white is:

\begin{align*}\begin{array}{rcl} P( 1^{st} \ \text{white marble}) &=& \frac{6}{10} \\ &=& \frac{3}{5} \end{array}\end{align*}

Next, having removed the first marble from the jar, now instead of 6 white marbles out of 10 total marbles, there are only 5 white marbles left out of 9 total marbles. The probability of the second marble being white is:

\begin{align*}P( 2^{nd} \ \text{white marble}) = \frac{5}{9}\end{align*}

Then, calculate the probability of both events occurring.

\begin{align*}\begin{array}{rcl} P(\text{white then white}) &=& P( 1^{st} \ \text{white marble}) \times P( 2^{nd} \ \text{white marble}) \\ &=& \frac{3}{5} \times \frac{5}{9} \\ &=& \frac{15}{45} \\ &=& \frac{1}{3} \end{array}\end{align*}
The answer is \begin{align*}\frac{1}{3}\end{align*}.

The probability of picking two white marbles, without replacement, from this jar is \begin{align*}\frac{1}{3}\end{align*}.

The same general method works for calculating any two (or more) dependent events.

Now let’s look at conditional probability and outcomes.

Conditional probability involves situations in which you determine the probability of an event based on another event having occurred.

Let’s look at an example.

In a class of 25 students, 14 like pizza and 16 like fruit juice. One student likes neither and 6 students like both. If one student is chosen at random, what is the probability that this student likes pizza given that he/she likes fruit juice?

First, to solve this problem, you could draw a Venn diagram to represent the data.

Next, of the 16 who like fruit juice, 6 like pizza. Therefore the probability is:

\begin{align*}\begin{array}{rcl} P(\text{pizza } \big | \text{ fruit juice}) &=& \frac{6}{16} \\ &=& \frac{3}{8} \end{array} \end{align*}

The answer is \begin{align*}\frac{3}{8}\end{align*}.

If a student chosen likes fruit juice, then the probability that they also like pizza is \begin{align*}\frac{3}{8}\end{align*}.

Notice that you write the conditional probability as:

\begin{align*}P\text{(second choice} \ \big | \ \text{first choice)}\end{align*}.

You read this as the probability of the second choice given the first choice.

Another way to look at the conditional probability formula is:

\begin{align*}P(\text{second} \ \big| \ \text{first} )= \frac{P(\text{first choice and second choice})}{P \text{(first choice)}}\end{align*}

Let’s look at an example.

The probability that it is Friday and that a student is absent is 0.05. Since there are 5 school days in a week, the probability that it is Friday is \begin{align*}\frac{1}{5}\end{align*} or \begin{align*}0.2\end{align*}. What is the probability that a student is absent given that today is Friday?

First, set up the conditional probability formula.

\begin{align*}P(\text{absent} \ \big| \ \text{Friday}) = \frac{P(\text{Friday and absent})}{P \text{(Friday)}}\end{align*}

Next, fill in the probabilities and solve.

\begin{align*}\begin{array}{rcl} P(\text{absent} \ \big| \ \text{Friday}) &=& \frac{P(\text{Friday and absent})}{P \text{(Friday)}} \\ &=& \frac{0.05}{0.2} \\ &=& \frac{1}{4} \end{array}\end{align*}

The answer is \begin{align*}\frac{1}{4}\end{align*}.

The probability that a student is absent given that it is Friday is \begin{align*}\frac{1}{4}\end{align*}.

Let’s look at another example.

Jar A contains 2 red and 3 blue marbles. Jar B contains 4 red and 1 blue marble. Thomas selects a jar by tossing a coin and takes a marble from that jar. What is the probability that given the marble is red that it came from jar B?

First, let’s draw a tree diagram. Remember there are 5 marbles in each jar. Since Thomas flips a coin he has a 1 in 2 chance of choosing each jar.

Next, calculate the probability that the marble chosen is red. Notice that there are two ways you can choose a red marble, from jar A or from Jar B.

\begin{align*}\begin{array}{rcl} P(\text{red}) &=& P_{\text{Jar A}} + P_{\text{Jar B}} \\ &=& \frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{4}{5} \\ &=& \frac{2}{10} + \frac{4}{10} \\ &=& \frac{6}{10} \\ &=& \frac{3}{5} \end{array}\end{align*}

Then, calculate the probability of choosing a marble from Jar B given that it is red. Use algebra to simplify.

\begin{align*}\begin{array}{rcl} P(\text{Jar B} \ \big| \ \text{red}) &=& \frac{P \text{(red and Jar B)}}{P \text{(red)}} \\ \\ &=& \frac{\frac{1}{2} \times \frac{4}{5}} {\frac{3}{5}} \\ \\ &=& \frac{\frac{4}{10}}{\frac{3}{5}} \\ \\ &=& \frac{4}{10} \times \frac{5}{3} \\ \\ &=& \frac{20}{30} \\ \\ &=& \frac{2}{3} \end{array}\end{align*}

The answer is \begin{align*}\frac{2}{3}\end{align*}.

The probability of choosing a red marble from Jar B is \begin{align*}\frac{2}{3} \end{align*}.

### Examples

#### Example 1

Earlier, you were given a problem about the appointments at the bike shop.

Carey is randomly booking appoints at the bike shop for Monday, Tuesday, and Wednesday mornings as well as Thursday and Friday afternoons. You need to find the probability that the first two appointments booked will be in the morning.

First, to work on this probability, you must determine the probability of the first appointment booked being on a Monday, Tuesday or Wednesday. There are five possible days for appointments, but three favorable outcomes.

Probability of the first appointment being on Mon, Tues, or Wed \begin{align*}= \frac{3}{5} \end{align*}.

Probability of the second appointment being on Mon, Tues, or Wed \begin{align*}= \frac{2}{4} \end{align*} or \begin{align*}\frac{1}{2}\end{align*}.

Next, multiply the two probabilities together.

\begin{align*}\begin{array}{rcl} P(\text{morning then morning}) &=& P(1^{st} \ \text{appointment}) \times P(2^{nd} \ \text{appointment}) \\ &=& \frac{3}{5} \times \frac{1}{2} \\ &=& \frac{3}{10} \end{array}\end{align*}

The answer is \begin{align*}\frac{3}{10}\end{align*}.

There is a \begin{align*}\frac{3}{10}\end{align*} or a 30% chance that the first two appointments booked would be on a Monday, Tuesday or Wednesday morning.

#### Example 2

Jack’s Catering Service is accepting weekday appointments for Monday through Thursday, and weekend appointments for Friday through Sunday. If appointment dates are made randomly, what is the probability that 2 weekdays will be the first 2 days to be booked?

First, calculate the probability that the first day will be a weekday. Jack only books appointments 4 of the seven days in the week.

\begin{align*}P(1^{st}\text{weekday}) = \frac{4}{7}\end{align*}

Next, calculate the probability that the second booked day will also be a weekday. Remember now, there are only six days left to choose from and three of these will be workdays.

\begin{align*}\begin{array}{rcl} P(2^{nd}\ \text{weekday}) &=& \frac{3}{6} \\ P(2^{nd}\ \text{weekday}) &=& \frac{1}{2} \end{array} \end{align*}

Then, calculate the probability that the two days chosen will be weekdays.

\begin{align*}\begin{array}{rcl} P(\text{weekday and weekday}) &=& P( 1^{st} \ \text{weekday}) \times P( 2^{nd} \ \text{weekday}) \\ P(\text{weekday and weekday}) &=& \frac{4}{7} \times \frac{1}{2} \\ P(\text{weekday and weekday}) &=& \frac{4}{14} \\ P(\text{weekday and weekday}) &=& \frac{2}{7} \end{array} \end{align*}

#### Example 3

What is the probability of taking out a blue, then a red marble without replacing the first marble?

First, start by approaching the problem the same as you would with dependent events.

There are 4 blue marbles and 8 red marbles in the jar. The probability of the first marble being blue is:

\begin{align*}\begin{array}{rcl} P(1^{st}\ \text{blue marble}) &=& \frac{4}{12} \\ &=& \frac{1}{3} \end{array}\end{align*}

Next, having removed the first marble from the jar, you now have 11 marbles in the jar. The probability of the second marble being red is:

\begin{align*}P(2^{nd}\ \text{red marble}) = \frac{8}{11}\end{align*}

Then, calculate the probability of both events occurring.

\begin{align*}\begin{array}{rcl} P(\text{blue then red}) &=& P(1^{st}\ \text{blue marble}) \times P(2^{nd}\ \text{red marble}) \\ &=& \frac{1}{3} \times \frac{8}{11} \\ &=& \frac{8}{33} \end{array} \end{align*}

The answer is \begin{align*}\frac{8}{33}\end{align*}.

The probability of picking a blue then a red marble, without replacement, from this jar is \begin{align*}\frac{8}{33}\end{align*}.

#### Example 4

What is the probability of taking out two red marbles without replacement?

First, start by approaching the problem the same as you would with dependent events. There are 4 blue marbles and 8 red marbles in the jar. The probability of the first marble being red is:

\begin{align*}\begin{array}{rcl} P(1^{st}\ \text{red marble}) &=& \frac{8}{12} \\ &=& \frac{2}{3} \end{array} \end{align*}

Next, having removed the first marble from the jar, you now have 11 marbles in the jar and only 7 of them are red. The probability of the second marble being red is:

\begin{align*}P(2^{nd}\ \text{red marble}) = \frac{7}{11}\end{align*}

Then, calculate the probability of both events occurring.

\begin{align*}\begin{array}{rcl} P(\text{red then red}) &=& P(1^{st} \ \text{red marble}) \times P(2^{nd}\ \text{red marble}) \\ &=& \frac{2}{3} \times \frac{7}{11} \\ &=& \frac{14}{3} \end{array}\end{align*}
The answer is \begin{align*}\frac{14}{33}\end{align*}.

The probability of picking two red marbles, without replacement, from this jar is \begin{align*}\frac{14}{33}\end{align*}.

#### Example 5

At the local high school, the probability that a student takes math and music is \begin{align*}\frac{1}{4}\end{align*}. The probability that a student is taking math is \begin{align*} \frac{17}{20} \end{align*}. What is the probability that a student is taking music given that they are taking math?

First, set up the conditional probability formula.

\begin{align*}P(\text{music}\ \big| \ \text{math}) = \frac{P \text{(math and music)}}{P \text{(math)}}\end{align*}

Next, fill in the probabilities and solve.

\begin{align*}\begin{array}{rcl} P(\text{music} \ \big| \ \text{math}) &=& \frac{P \text{(math and music)}}{P \text{(math)}} \\ \\ &=& \frac{\frac{1}{4}} {\frac{17}{20}} \\ \\ &=& \frac{1}{4} \times {\frac{20}{17}} \\ \\ &=& \frac{20}{68} \\ \\ &=& \frac{5}{17} \end{array}\end{align*}
The answer is \begin{align*}\frac{5}{17}\end{align*}.

The probability that a student is taking music given that they are taking math is \begin{align*}\frac{5}{17}\end{align*}.

### Review

Solve the problems.

1. A stack of 12 cards has 4 Aces, 4 Kings, and 4 Queens. What is the probability of picking 2 Aces from the stack at random?
2. What is the probability of picking an Ace then a King from the stack above?
3. What is the probability of picking 3 Queens from the stack above?
4. Stoyko’s shirt drawer has 4 colored t-shirts and 4 white t-shirts. If Stoyko picks out 2 shirts at random, what is the probability that they will both be colored?
5. If Stoyko picks out 2 shirts at random from the drawer above, what is the probability that the first one will be colored and the second one will be white?
6. On a game show, there are 16 questions: 8 easy, 5 medium-hard, and 3 hard. If contestants are given questions randomly, what is the probability that the first two contestants will get easy questions?
7. On the game show above, what is the probability that the first contestant will get an easy question and the second contestant will get a hard question?
8. On the game show above, what is the probability that both of the first two contestants will get hard questions?
9. For a single toss of a number cube, what is the probability that the cube will land on a number that is both odd and greater than 2?
10. For a single toss of a number cube, what is the probability that the cube will land on a number that is greater than 2 and less than 6?
11. For a single toss of a number cube, what is the probability that the cube will land on a number that is greater than 1 and less than 6?
12. What is the probability that a sum of a pair of number cubes will be 11 if the first cube lands on 5?
13. What is the probability that a sum of a pair of number cubes will be odd if the first cube lands on 2?
14. What is the probability that a sum of a pair of number cubes will be even and greater than 6 if the first cube lands on 4?
15. If you toss a number cubes, predict how likely it is to roll a number less than 6?

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Color Highlighted Text Notes

### Vocabulary Language: English

conditional probability

The probability of a particular dependent event  given the outcome of the event on which it occurs.

conditional probability formula

The conditional probability formula is P(A/B) = P(AUB)/P(B)

Favorable Outcome

A favorable outcome is the outcome that you are looking for in an experiment.

Independent Events

Two events are independent if the occurrence of one event does not impact the probability of the other event.

Multiplication Rule

States that for 2 events (A and B), the probability of A and B is given by: P(A and B) = P(A) x P(B).

Mutually Exclusive Events

Mutually exclusive events have no common outcomes.

Overlapping Events

Overlapping events are events that have outcomes in common.

Sample Space

In a probability experiment, the sample space is the set of all the possible outcomes of the experiment.