Understanding Conditional Probability and Independence Through Sample Space and Tree Diagrams

Sometimes, it is useful to identify the connection between different events by analyzing whether they are dependent or not. Another commonly used characteristic is a conditional probability. In this lesson, the conditional probability and its connection with independence will be discovered through the use of real-world examples.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Probability experiment
Probability of an event
Union
Intersection
Complement
Independent and dependent events

Challenge

Analyzing Connections Between Events

Vincenzo is thinking about taking a trip overseas. He is having a tough time deciding between France and Antarctica. In France he can visit the Eiffel Tower — he studies structural engineering — but Antartica has penguins and Vincenzo really loves penguins!

1. Vincenzo next to the Eiffel Tower. 2. Vincenzo in Antartica. 3. Vincenzo and a penguin

External credits: @pikisuperstar, @macrovector, @jcomp

Vincenzo pulls up a search engine and types "France vs. Antartica, I wanna see penguins." He finds an interesting survey that just so happens to contrast France and Antarctica. The survey even asks specifically about penguins!

Of $190$ people surveyed, $157$ people prefer going to France over Antarctica.
Of those $157,$ only five people saw penguins.
Of the people who traveled to Antarctica, only three of them did not see penguins.

The remaining results from the survey are organized in the following table.

Frequency table comparing people who go to France/Antarctica and see a penguin with those who do not

Consider the presented data to find the probabilities of the following scenarios.

a Vincenzo decides to travel to France. What is the probability that he will see a penguin? Round the probability to the nearest percent.

b Yesterday, Vincenzo saw a penguin! What is the probability that he was in France? Round the probability to the nearest percent.

c Vincenzo’s trip just ended. Sadly, he did not see a single penguin. What is the probability that he chose to go to Antarctica? Round the probability to the nearest percent.

Example

Relationship Between Dependent Events

Diego wants to pick two books at random from a pile of five books. Three of them are Geometry books, and the other two are History books. Below, the sample space of this situation is shown, where

G_{1},

G_{2},

G_{3}

represent the Geometry books, and

H_{1}

and

H_{2}

represent the History books.

Three Geometry books and Two History books

Imagine standing next to Diego and calculating the probabilities in real time with him. Let

H

be the event that the first book chosen is a History book, and

G

be the event that the second book chosen is a Geometry book. Now, consider the following scenarios.

a Find

P (H),

P (G),

and

P (H \cap G) .

Express the probabilities as fractions in their simplest form.

b Find

\frac{P ( H \cap G )}{P ( H )}

and

\frac{P ( H \cap G )}{P ( G )} .

Express the probabilities as fractions in their simplest form.

c If the first book Diego picked is a History book, what is the probability that the second book will be a Geometry book? Express the probability as a fraction in its simplest form.

d Diego chose both books before anyone else saw. If the second book Diego picked is a Geometry book, what is the probability that the first book he chose is a History book? Express the probability as a fraction in its simplest form.

e Is there any relationship between the probabilities found in Parts B, C, and D?

Answer

a Probabilities:

$P (H) = \frac{2}{5}$
$P (G) = \frac{3}{5}$
$P (H \cap G) = \frac{3}{1 0}$

\frac{P ( H \cap G )}{P ( H )} = \frac{3}{4}

and

\frac{P ( H \cap G )}{P ( G )} = \frac{1}{2} .

\frac{3}{4}

\frac{1}{2}

e Sample answer:

The probability that event $G$ happens given that event $H$ happened equals $\frac{P ( H \cap G )}{P ( H )} .$
The probability that event $H$ happens, given that event $G$ happened equals $\frac{P ( H \cap G )}{P ( G )} .$

Hint

a Every outcome that has a History book in the first position satisfies

H .

Similarly, every outcome that has a Geometry book in the second position satisfies

G .

Divide the number of favorable outcomes by the total number of outcomes.

b Divide the numbers found in Part A.

c The sample space of the given situation is not the original sample space. Start by finding the new sample space. Later, determine how many of the outcomes have a Geometry book in the second position.

d The sample space of the given situation is not the original sample space. Also, it is different from the one found in Part C. In this new sample space, count the outcomes that have a History book in the first position.

e Compare the probabilities found in Parts B, C, and D. Write a sentence stating the relationship found.

Solution

a By definition, the probability of an event is found by dividing the number of favorable outcomes by the total number of outcomes in the sample space.

P = \frac{Number of favorable outcomes}{Number of possible outcomes}

From the given diagram, the sample space contains

20

different outcomes. The required probabilities can be found one at a time.

To find $P (H),$ divide the number of outcomes in the sample space with a History book in the first position by $20 .$
To find $P (G),$ divide the number of outcomes in the sample space with a Geometry book in the second position by $20 .$
To find $P (H \cap G),$ divide the number of outcomes in the sample space with a History book in the first position and a Geometry book in the second position by $20 .$

The following diagram summarizes all the computations.

Notice that

P (H \cap G) \neq = P (H) \cdot P (G) .

That relationship implies that the given events are dependent.

b In this part, the probabilities found in Part A will be used.

P (H) P (G) P (H \cap G) = \frac{2}{5} = \frac{3}{5} = \frac{3}{1 0}

Start by finding the ratio of

P (H \cap G)

P (H) .

\frac{P ( H \cap G )}{P ( H )}

Substitute values and simplify

SubstituteValues

Substitute values

\frac{\frac{3}{1 0}}{\frac{2}{5}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

\frac{3}{1 0} \cdot \frac{5}{2}

MultFrac

Multiply fractions

\frac{1 5}{2 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 5}{b / 5}$

\frac{3}{4}

Next, find the ratio of

P (H \cap G)

P (G) .

\frac{P ( H \cap G )}{P ( G )}

Substitute values and simplify

SubstituteValues

Substitute values

\frac{\frac{3}{1 0}}{\frac{3}{5}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

\frac{3}{1 0} \cdot \frac{5}{3}

MultFrac

Multiply fractions

\frac{1 5}{3 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 5}{b / 1 5}$

\frac{1}{2}

c For this situation, it is known that the first book chosen is a History book. Knowing this fact reduces the number of possible outcomes because some of the outcomes are ignored in the new situation. Therefore, the new sample space contains fewer outcomes compared to the original one.

As seen, there are

8

outcomes in the new sample space and

6

of them have a Geometry book in the second position. Therefore, the probability that the second book is a Geometry book, given that the first book is a History book, can be found using the following process.

P (G given H) = \frac{6}{8} = \frac{3}{4}

Consequently, if the first book chosen by Diego is a History book, the probability that the second book is a Geometry book is

\frac{3}{4} .

d In this case, it is known that the second book chosen is a Geometry book. This reduces the number of possible outcomes from the original sample space. Therefore, the new sample space contains fewer outcomes and it is different from the one used in Part C.

As seen, there are

12

outcomes in the new sample space and

6

of them have a History book in the first position. Therefore, the probability that the first book is a History book, given that the second book is a Geometry book can be found as follows.

P (H given G) = \frac{6}{1 2} = \frac{1}{2}

Consequently, if the second book chosen by Diego is a Geometry book, the probability that the first book is a History book is

\frac{1}{2} .

e Start by writing the four probabilities found in Parts B, C, and D.

\frac{P ( H \cap G )}{P ( H )} \frac{P ( H \cap G )}{P ( G )} P (G given H) P (H given G) = \frac{3}{4} = \frac{1}{2} = \frac{3}{4} = \frac{1}{2}

Comparing the four probabilities, it can be seen that the first probability found in part B equals the probability found in part C.

The probability that the second book is a Geometry book given that the first book chosen is a History book equals $\frac{P ( H \cap G )}{P ( H )} .$

The previous statement can also be rewritten in terms of $H$ and $G$ as follows.

The probability that event $G$ happens given that event $H$ happened equals $\frac{P ( H \cap G )}{P ( H )} .$

Similarly, the second probability found in part B equals the probability found in part D. This leads to write the following relation.

The probability that the first book is a History book, given that the second book is a Geometry book equals $\frac{P ( H \cap G )}{P ( G )} .$

As before, the previous statement can be rewritten in terms of $H$ and $G .$

The probability that event $H$ happens, given that event $G$ happened equals $\frac{P ( H \cap G )}{P ( G )} .$