# 11.6: Conditional Probability

**At Grade**Created by: CK-12

## Introduction

*Bike Shop Repairs*

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 lesson 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 lesson.**

*What You Will Learn*

By the end of this lesson, you will be able to complete the following skills.

- Recognize and distinguish among independent and dependent events.
- Recognize and apply the definition of conditional probability to find probabilities in finite sample spaces.
- Calculate the probabilities of a series of independent events, mutually exclusive events and events that are not mutually exclusive.
- Make predictions involving conditional probability.

*Teaching Time*

I. **Recognize and Distinguish among Independent and Dependent Events**

In the last few sections, you have been learning all about probability. Now we can think about different events and how these events impact each other. Take a look at this example.

Suppose you have two events:

Event A: Toss 5 on the number cube

Event B: Spin blue on the spinner

The probability of each of these events by itself is easy enough to compute. In general:

\begin{align*}P \text{(event)} = \frac{favorable \ outcomes}{total \ outcomes}\end{align*}

If this is the case, then we can write the following ratios for rolling a 5.

\begin{align*}P (5) &= \frac{favorable \ outcomes}{total \ outcomes} = \frac{1}{6}\\ P(\text{blue}) &= \frac{favorable \ outcomes}{total \ outcomes} = \frac{1}{4}\end{align*}

These two events were performed with a spinner and a number cube.

**Now a question arises.**

*Does event A affect the probability of event B in any way?*

That is, does the number cube landing on 5 affect where the arrow lands in the spinner? If not, then the two events are said to be ** independent events**.

**Definition:** If the outcome of one event has no effect on the outcome of a second event, then the two events are *independent events.*

**Events A and B above are independent events. No matter how the number cube turns up, its outcome does not affect the outcome of spinning the spinner.**

**Now let’s think about a different kind of example, one where the outcome of one event does impact the outcome of another event.**

Example

A bag has 3 red marbles, 4 blue marbles, and 3 green marbles. Irina pulls 1 green marble out of the bag. Does this change the probability that the next marble Irina pulls out of the bag will be green?

**Solution:** Here, the act of taking a marble out of the bag changes the situation. For the first marble, the probability of pulling out a green marble was:

\begin{align*}P \text {(green)} = \frac{favorable \ outcomes}{total \ outcomes} = \frac{3}{10}\end{align*}

For the second marble, there are now only 9 marbles left in the bag and only 2 of them are green. So the probability of pulling out a green marble for the second marble is now:

\begin{align*}P \text {(green)} = \frac{favorable \ outcomes}{total \ outcomes} = \frac{2}{9}\end{align*}

Clearly, the first event affected the outcome of the second event in this situation. So the two events are NOT independent. In other words, they are ** dependent events**.

**Definition:** If the outcome of one event has an effect on the outcome of a second event, then the two events are *dependent events.*

II. **Recognize and Apply the Definition of Conditional Probability to Find Probabilities in Finite Sample Spaces**

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. Let’s look at an example.

Example

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.**

Example

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: 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*}

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 example, 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 examples of how 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.**

III. **Calculate Probabilities of a Series of Independent Events, Mutually Exclusive Events and Events that are not Mutually Exclusive**

Sometimes, we have mutually exclusive events and we have events that overlap and are not mutually exclusive.

In a previous section, you saw that event \begin{align*}R(\text{red})\end{align*} and event \begin{align*}T(\text{top})\end{align*} are ** overlapping events** because both events share one outcome – red-top. The Venn diagram for overlapping events shows that the two events overlap, or share 1 or more outcomes.

**To calculate the probability of overlapping events, list the sample space and find the favorable events.**

\begin{align*} & \mathbf{red - top} \qquad \ \text{blue}-\text{top}\\ & \text{red}-\text{bottom} \quad \text{blue}-\text{bottom}\end{align*}

The probability of red-top is:

\begin{align*}P(\text{red}-\text{top}) = \frac{favorable \ outcomes} {total \ outcomes} = \frac{1}{4} \end{align*}

You can compute the probabilities of other overlapping events in a similar way.

Example

For a single toss of a number cube, what is the probability of event \begin{align*}E(\text{even})\end{align*} and event \begin{align*}S(4)\end{align*} both occurring?

**Step 1:** Identify the overlapping outcomes of both events.

\begin{align*}E (\text{even}) &= 2, \mathbf{4}, 6 \\ S(4) &= \mathbf{4} \end{align*}

**Step 2:** Find the total number of outcomes.

\begin{align*}\text{total outcomes} &= 1, 2, 3, 4, 5, 6 \\ &= 6 \ \text{total outcomes}\end{align*}

**Step 3:** Find the probability of the overlapping events.

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

IV. **Make Predictions Involving Conditional Probability**

**Think back for a minute and remember what we mean when we talk about conditional probability.**

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

We can use conditional probability to determine probabilities, but also to make predictions. Look at this example.

Example

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.**

## Real-Life Example Completed

*Bike Shop Repairs*

**Here is the original problem from the introduction. Reread it and then solve it for the correct 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 the these appointments were booked on a Monday, Tuesday or Wednesday morning?

*Now answer the question at the end of the problem.*

*Solution to Real – Life Example*

**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.**

## Vocabulary

Here are the vocabulary words that are found in this lesson.

- Independent Events
- The outcome of one event has no effect on the outcome of a second event.

- Dependent Events
- If the outcome of one event has an effect on the outcome of another event they are dependent events.

- Overlapping Events
- Events that share one outcome

- Conditional Probability
- Probability that is predicted based on specific criteria or conditions.

## Time to Practice

Directions: Write whether events A and B are dependent or independent.

- A: Doug flips a coin. B: Marlene chooses a card out of a deck.
- A: In a bag with 5 white marbles and 5 black marbles, Sanjay pulls out a white marble. B: Without returning the marble to the bag, Sanjay pulls out a second marble.
- A: Eddie chooses the color blue for his new bike. B: Eddie chooses lasagne from the dinner menu.
- A: The probability that it will rain tomorrow. B: The probability that the Red Wings hockey team will win their game tomorrow.
- A: From a deck of cards, the probability of one player drawing a heart from the deck. B: On the next player’s turn, the probability of drawing another heart.
- A: The probability of a spinner landing on blue 6 times in a row. B: The probability of the spinner landing on blue on the next spin.
- A: The probability that the Rockies will be in the playoffs. B: The probability that the Rockies will win the World Series.
- A: The probability that tomorrow will be sunny. B: The probability that tomorrow will be a full moon.
- A The probability that tomorrow will be sunny. B: The probability that tomorrow will be cloudy.
- A: The probability that it will be a half-moon today. B: The probability that it will be a full moon tomorrow.

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 2 number cubes, predict how likely they are to match.
- If you toss 2 number cubes, predict how likely they are NOT to match.

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