### Let’s Think About It

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.

### Guidance

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:

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.

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:

The answer is

.If a student chosen likes fruit juice, then the probability that they also like pizza is

.Notice that you write the conditional probability as:

.

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

or . What is the probability that a student is absent given that today is Friday?First, set up the conditional probability formula.

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

.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.

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

.The probability of choosing a red marble from Jar B is

.### Guided Practice

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*}

### Examples

A jar contains four blue marbles and eight red marbles.

#### Example 1

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:

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.

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

.#### Example 2

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.

The probability of picking two red marbles, without replacement, from this jar is

.#### Example 3

At the local high school, the probability that a student takes math and music is

. The probability that a student is taking math is . 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.

The probability that a student is taking music given that they are taking math is

.### Follow Up

Remember 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

or .Next, multiply the two probabilities together.

The answer is

.There is a

or a 30% chance that the first two appointments booked would be on a Monday, Tuesday or Wednesday morning.### Video Review

https://www.youtube.com/watch?v=u03NipAbyYg

### Explore More

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