Have you ever worked in a bike shop? Take a look at this dilemma.

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 the these appointments were booked on a Monday, Tuesday or Wednesday morning?

**Answering this question will require you to understand conditional probability. This Concept will teach you all that you need to know so that you will be able to figure out the solution to the problem by the end of the Concept.**

### Guidance

Sometimes the outcome that you get when figuring out a probability is what we call “conditional.” This means that we will only an outcome if the conditions designed to cause a specific result.

Take a look at this situation.

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

**Start by approaching the problem the same as you would with independent events.**

The probability of the first marble being white is:

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

**What about the second marble?**

Having removed the first marble from the bag, now instead of 6 white marbles out of 10 total marbles, there are only 5 white marbles left out of 9 total marbles:

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

This gives a probability of both events occurring as:

@$$\begin{align*}P (\text{white then white}) &= P (\text{white} \ 1 \text{st marble}) \cdot P (\text{white} \ 2 \text{nd marble}) \\ &= \frac{3}{5} \cdot \frac{5}{9} \\ &= \frac{1}{3}\end{align*}@$$

**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**.

For instance, suppose you roll two number cubes on a table. The first cube lands face up on 5. The second cube falls off of the table so you can’t see how it landed. Given what you know so far, what is the probability that the sum of the number cubes will be 9?

**To solve this problem, consider the entire sample space for rolling two number cubes.**

@$$\begin{align*} &66 \quad 56 \quad {\color{red}46} \quad 36 \quad 26 \quad 16 \\ &65 \quad 55 \quad {\colorbox{yellow}{\color{red}45}} \ \ 35 \quad 25 \quad 15 \\ &64 \quad 54 \quad {\color{red}44} \quad 34 \quad 24 \quad 14 \\ &63 \quad 53 \quad {\color{red}43} \quad 33 \quad 23 \quad 13 \\ &62 \quad 52 \quad {\color{red}42} \quad 32 \quad 22 \quad 12 \\ &61 \quad 51 \quad {\color{red}41} \quad 31 \quad 21 \quad 11\end{align*}@$$

You already know that the first number cube landed on 4, so now you need to consider only those outcomes marked in red. Only 1 of those 6 results in a 9, so:

@$$\begin{align*}P (9 | 4) = \frac{favorable \ outcomes}{total \ outcomes} = \frac{1}{6} \end{align*}@$$

Notice that we write the conditional probability as @$\begin{align*}P (9 | 4)\end{align*}@$. You can read this as:

@$\begin{align*}P (9|4) \Longleftarrow\end{align*}@$ the probability of 9, given 4

Here are some other ways to read this notation.

@$\begin{align*}P (B|A) \Longleftarrow\end{align*}@$ the probability of @$\begin{align*}B\end{align*}@$, given @$\begin{align*}A\end{align*}@$

@$\begin{align*}P (7|3) \Longleftarrow\end{align*}@$ the probability of 7, given 3

@$\begin{align*}P (\text{heads}| \text{tails}) \Longleftarrow\end{align*}@$ the probability of heads, given tails

@$\begin{align*}P (\text{red}| \text{blue}) \Longleftarrow\end{align*}@$ the probability of red, given blue

**The probability is determined because certain factors are in place.**

We can use conditional probability to determine probabilities, but also to make predictions.

A stack of 12 cards has the Ace, King, and Queen of all 4 suits, spades, hearts, diamonds, and clubs. What is the probability that if you draw 2 cards randomly, they will both be hearts? Make a prediction.

**Step 1:** Draw the first card. The probability of it being a heart is 3 of 12.

**Step 2:** Now draw the second card. Since the first card was a heart, there are only 11 cards left and only 2 of them are hearts.

**Step 3:** Calculate the final probability.

**So @$\begin{align*}P(\text{heart and heart}) = \frac{1}{22}\end{align*}@$. You would predict that both cards would be hearts @$\begin{align*}\frac{1}{22}\end{align*}@$ of the time.**

A jar contains four blue marbles and eight red marbles.

#### Example A

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

**Solution: @$\begin{align*}\frac{8}{33}\end{align*}@$**

#### Example B

What is the probability of taking out two red marbles?

**Solution: @$\begin{align*}\frac{7}{24}\end{align*}@$**

#### Example C

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

**Solution: @$\begin{align*}\frac{8}{33}\end{align*}@$**

Now let's go back to the dilemma from the beginning of the Concept.

**To work on this probability, first we 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 first appointment being Mon, Tues or Weds @$\begin{align*}=\frac{3}{5}\end{align*}@$**

**Probability of second appointment being Mon, Tues or Weds @$\begin{align*}=\frac{2}{4}\end{align*}@$ or @$\begin{align*}\frac{1}{2}\end{align*}@$**

**Now we can multiply them for the conditional probability.**

@$\begin{align*}\frac{3}{5} \cdot \frac{1}{2} = \frac{3}{10}\end{align*}@$ or 30%

**There is a 30% chance that the first two appointments booked would be on a Monday, Tuesday or Wednesday morning.**

### Guided Practice

Here is one for you to try on your own.

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?

**Solution**

Solution: The probability that the first day will be a weekday is:

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

The probability that the second booked day will also be a weekday is:

@$$\begin{align*}P (\text{weekday} \ 2 \text{nd}) = \frac {3}{6} = \frac{1}{2}\end{align*}@$$

@$$\begin{align*} P \text{(weekday and weekday)} &= P \text{(weekday} \ 1 \text{st}) \cdot P \text{(weekday} \ 2 \text{nd}) \\ &= \frac{4}{7} \cdot \frac{1}{2}\\ &= \frac{2}{7}\end{align*}@$$

### Video Review

### Explore More

Directions: Solve the problems.

- 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?
- What is the probability of picking an Ace then a King from the stack above?
- What is the probability of picking 3 Queens from the stack above?
- 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?
- 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?
- 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?
- 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?
- On the game show above, what is the probability that both of the first two contestants will get hard questions?
- 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?
- 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?
- 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?
- What is the probability that a sum of a pair of number cubes will be 11 if the first cube lands on 5?
- What is the probability that a sum of a pair of number cubes will be odd if the first cube lands on 2?
- 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?
- If you toss a number cubes, predict how likely is it to roll a number less than 6?