Understanding Conditional Probability and Independence Through Sample Space and Tree Diagrams

.a /b./.c /d.=a/b*d/c

3/10* 5/2

Multiply fractions

15/20

a/b=.a /5./.b /5.

3/4

Next, find the ratio of P(H⋂ G) to P(G).

P(H⋂ G)/P(G)

▼

Substitute values and simplify

SubstituteValues

Substitute values

310/35

.a /b./.c /d.=a/b*d/c

3/10* 5/3

Multiply fractions

15/30

a/b=.a /15./.b /15.

1/2

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(GgivenH) = 6/8 = 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 34.

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(HgivenG) = 6/12 = 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 12.

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

P(H⋂ G)/P(H) &= 3/4 [0.2cm] P(H⋂ G)/P(G) &= 1/2 [0.2cm] P(GgivenH) &= 3/4 [0.2cm] P(HgivenG) &= 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 P(H⋂ 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 P(H⋂ 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 P(H⋂ 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 P(H⋂ G)P(G).

Example

Finding the Probability of Triggering an Alarm

At Troupial Airport, engineers are testing a prototype of a prohibited substance detector. If there is a forbidden item in a bag, an alarm is supposed to be triggered. To test the detector, 5000 bags will be checked and 7 % of them will randomly contain forbidden items.

If a bag contains a forbidden item, there is a 98 % chance that the alarm is triggered. On the other hand, if a bag does not have a forbidden item, there is a 4 % chance that it triggers the alarm. Calculate the probabilities of the events described below. Write the probabilities as a percent rounded to two decimal places, when necessary.

Find the probability that a bag triggers the alarm.

Find the probability that a bag triggers the alarm and contains a forbidden item. Also, find the probability that a bag does not trigger the alarm and contains a forbidden item.

If Mark's bag triggered the alarm, what is the probability that his bag contains a forbidden item?

If Izabella's bag did not trigger the alarm, what is the probability that her bag contains a forbidden item?

Based on Parts C and D, what can be said about Mark and Izabella situations?

Answer

P(Alarm) = 10.58 %

P(Alarm and Forbidden) = 6.86 % and P(No Alarm and Forbidden) = 0.14 %

The probability that Mark's bag contains a forbidden item given that it triggered the alarm is about 64.84 %.

The probability that Izabella's bag contains a forbidden item given that it did not trigger the alarm is about 0.16 %.

Example Solution:

There is a 64.84 % chance that Mark's bag contains a forbidden item. However, this probability is not big enough to ensure Mark has a forbidden item. Therefore, it is doubtful — but possible — that Mark contains a forbidden item in his bag.
There is a 0.16 % chance that Izabella's bag contains a forbidden item. Since this probability is very small — less than 1 % — it is almost certain that Izabella does not have forbidden items in her bag — but still possible.

Hint

Make a tree diagram and write the number of bags corresponding to every node. The probability of triggering the alarm is the total number of bags that triggered the alarm divided by the total amount of bags checked.

Determine how many bags contain a forbidden item and trigger the alarm. Divide this number by the total number of bags checked. Similar reasoning can be used for the case where the bags do not trigger the alarm.

Considering the diagram drawn in Part A, what is the total number of bags that trigger the alarm? Of all these bags, how many actually had forbidden items?

Of all the bags that did not trigger the alarm, how many had forbidden items? Divide this last number by the total number of bags that did not trigger the alarm.

If the probability of an event is closer to 50 % than 100 %, then it is doubtful that the event actually happened. If the probability of an event is close to 0 %, it is almost certain that the event did not happen.

Solution

To find the probability that a bag triggers the alarm, a tree diagram will be drawn. To do so, first notice that the situation can be divided into three different stages.

All of the previous stages and events can be illustrated using a tree diagram.

Recall that there are 5000 bags, and 7 % of them contain forbidden items. The product of these numbers will give the number of the bags that contain forbidden items. The rest of the bags do not contain forbidden items. Forbidden: & 5000 * 7 % = 350 Not Forbidden: & 5000 - 350 = 4650 Additionally, the following two details about the bags are known.

If a bag contains forbidden items, there is a 98 % chance that it triggers the alarm.
If a bag does not have forbidden items, there is a 4 % chance that it triggers the alarm.

Considering these details, it can be concluded that 2 % of the bags containing forbidden items could trigger the alarm and 96 % of the bags that do not have forbidden items could not trigger the alarm.

Now, using the percentages in the branches, the number of bags for each event can be found.

Forbidden and Alarm	350 * 98 % = 343
Forbidden and No Alarm	350 * 2 % = 7
Not Forbidden and Alarm	4650 * 4 % = 186
Not Forbidden and No Alarm	4650 * 96 % = 4464

Finally, all the information can be shown on the tree diagram.

To find the probability that a bag triggers the alarm, start by finding how many bags triggered the alarm. To do so, add the two corresponding numbers in the tree diagram. Bags that triggered the alarm: 343+ 186 = 529 The probability that a randomly picked bag triggers the alarm P(Alarm) is obtained dividing 529 by the total number of bags.

P(Alarm) = 529/5000

Calculate quotient

P(Alarm) = 0.1058

Convert to percent

P(Alarm) = 10.58 %

Pulling information from the tree diagram, it is seen that the number of bags that triggered the alarm and had forbidden items is 343 and the number of bags that contained forbidden items and did not trigger the alarm is 7.

Since the total number of bags is 5000, the ratio of the number of the favorable outcomes to the total number will give the desired probability.

Probabilities of the Events
P(Alarm and Forbidden)	343/5000 = 6.86 %
P(No Alarm and Forbidden)	7/5000 = 0.14 %

Take note that the sum of the probabilities is equal to 7 %, which is the percentage of the bags that contain forbidden items.

To determine the probability that a bag contains forbidden items — given that it triggered the alarm — divide the total number of bags that both triggered the alarm and had forbidden items by the total number of bags that triggered the alarm.

P = Alarm and Forbidden/Alarm As presented in the tree diagram, the number of bags that trigger the alarm is equal to the sum of the numbers under the word Alarm.

Therefore, there are 529 bags that trigger the alarm. Since Mark's bag triggered the alarm, his bag is one of those 529. Out of this group of bags, 343 had forbidden items.

P = 343/529

▼

Evaluate right-hand side

Calculate quotient

P = 0.648393 ...

RoundDec

Round to 4 decimal place(s)

P ≈ 0.6484

Convert to percent

P ≈ 64.84 %

Therefore, there is about a 64.84 % chance that Mark's bag contains forbidden items given that his bag triggered the alarm.

To determine the probability that a bag contains forbidden items given that it did not trigger the alarm, divide the total number of bags that did not trigger the alarm and had forbidden items by the total number of bags that did not trigger the alarm.

P = No Alarm and Forbidden/No Alarm First, find the number of bags did not trigger the alarm.

Since Izabella's bag did not trigger the alarm, her bag is one of those 4471. Out of these bags, 7 had forbidden items.

P = 7/4471

▼

Evaluate right-hand side

Calculate quotient

P = 0.001565 ...

RoundDec

Round to 4 decimal place(s)

P ≈ 0.0016

Convert to percent

P ≈ 0.16 %

Therefore, there is about a 0.16 % chance that Izabella's bag contains forbidden items given that her bag did not trigger the alarm.

If the probability of an event is closer to 50 % than 100 %, then it is doubtful that the event would actually happen. With that in mind, consider the probability found in Part C.

There is about a 64.84 % chance that Mark's bag contains a forbidden item.

This probability is not close enough to 100 % to ensure that Mark's bag contains a forbidden item. Therefore, it is doubtful — but possible — that Mark's bag contains a forbidden item. Next, recall the answer found in Part D.

There is about a 0.16 % chance that Izabella's bag contains a forbidden item.

Since this probability is very small — less than 1 % — it is almost certain that Izabella does not have forbidden items in her bag — but still possible.

In this example, as well as in the previous one, dependent events and how the occurrence of one event affects the occurrence of the other event were analyzed through counting methods like tables and tree diagrams. Below, a formula for finding the probability of similar situations is shown.

Discussion

Conditional Probability

Conditional probability is the measure of the likelihood of an event B occurring, given that event A has occurred previously. The probability of B given A is written as P(B|A). It can be calculated by dividing the probability of the intersection of A and B by the probability of A.

P(B|A)=P(AandB)/P(A),where P(A)≠0

It is worth noting that usually P(B|A) and P(A|B) are not equal, meaning that conditional probability is not reversible. For example, let A be the event of a prime number and B be the event of an odd number. The probability that a prime number is odd is almost 1, but the reverse, an odd number being prime, is much smaller. P(B|A)≠ P(A|B)

Why

Intuition Behind the Formula

The intuition behind the formula can be visualized by using Venn Diagrams. Consider a sample space S and the events A and B such that P(A)≠0.

Assuming that event A has occurred, the sample space is reduced to A.

This means that the probability that event B can happen is reduced to the outcomes in the intersection of A and B, that is, to those outcomes in A⋂ B.

The possible outcomes are given by P(A) and the favorable outcomes by P(A⋂ B). Therefore, the conditional probability formula can be obtained using the probability formula.

P(B|A)=P(AandB)/P(A)

Now that the formula for finding a conditional probability is known, it can be put to use to solve daily life situations.

Example

Probability of Getting the Correct Size and Color

Diego's generous father has finished doing laundry and put Diego's T-shirts along with those of his big brother into the same ol’ basket. There are orange, blue, and red T-shirts in the basket, of which four are S-sized and eight are M-sized.

Diego is planning to go out with his friends. After taking a shower, he randomly picks a T-shirt from the same ol' laundry basket. Consider the following events. Event S: & Diego picks an S-sized T-shirt. Event O: & Diego picks an orange T-shirt. Event B: & Diego picks a blue T-shirt.

Find and compare P(S|O) and P(O|S). Write the answers as fractions in their simplest form.

Find and compare P(S|B) and P(B|S). Write the answers as fractions in their simplest form.

Hint

Start by finding P(O), P(S), and P(SandO). Then, apply the formula for finding the conditional probability.

Find P(B) and P(SandB). From Part A, P(S) is already known. Apply the formula for finding conditional probability again.

Solution

To start, remember the formula for finding the conditional probability that event B happens given that event A has occurred.

P(B|A) = P(AandB)/P(A) Applying this formula, the required probabilities can be rewritten as follows. P(S| O) &= P(Sand O)/P( O) [0.3cm] P( O|S) &= P(Sand O)/P(S) Consequently, P(Sand O), P( O), and P(S) need to be found. To do so, start by writing the formula to find the probability of an event. P=Number of favorable outcomes/Number of possible outcomes Notice that P(Sand O) represents the probability of picking an S-size orange T-shirt. From the diagram, there are 12 T-shirts in the basket, and there is only one S-size orange T-shirt. P(Sand O) = 1/12 ✓ To find P( O), count how many of the 12 T-shirts are orange. From the diagram, there are only 5. P( O) = 5/12 ✓ To determine P(S), count how many of the 12 T-shirts are S-sized. From the diagram, there are only 4. P(S) = 4/12 = 1/3 ✓

Finding Conditional Probabilities

To find P(S| O), divide P(Sand O)= 112 by P( O) = 512.

P(S| O) = P(Sand O)/P( O)

▼

Substitute values and simplify

P(Sand O)= 1/12, P( O)= 5/12

P(S| O) = 112/512

.a /b./.c /d.=a/b*d/c

P(S| O) = 1/12* 12/5

Multiply fractions

P(S| O) = 12/60

a/b=.a /12./.b /12.

P(S| O) = 1/5

Therefore, the probability that Diego picks an S-size T-shirt given that he picked a orange one is 15, or 20 %. Finally, find P( O|S).

P( O|S) = P(Sand O)/P(S)

▼

Substitute values and simplify

P(Sand O)= 1/12, P(S)= 1/3

P( O|S) = 112/13

.a /b./.c /d.=a/b*d/c

P( O|S) = 1/12* 3/1

Multiply fractions

P( O|S) = 3/12

a/b=.a /3./.b /3.

P( O|S) = 1/4

Consequently, the probability that Diego picks a orange T-shirt given that he picked an S-size one is 14, or 25 %.

Similar to the previous part, start by applying the conditional probability formula to rewrite the given expressions.

P(S| B) &= P(Sand B)/P( B) [0.3cm] P( B|S) &= P(Sand B)/P(S) From Part A, it is known that P(S) equals 13. Thus, only P(Sand B) and P( B) are missing. Note that P(Sand B) is the probability of picking an S-size blue T-shirt. From the 12 T-shirts, only two are blue and S-sized. P(Sand B) = 2/12 = 1/6 ✓ To determine P( B), count how many of the 12 T-shirts are blue. From the diagram, there are 4. P( B) = 4/12 = 1/3 ✓

Finding Conditional Probabilities

Substituting the found probabilities into the initial equations, the required probabilities will be obtained. Start by finding P(S| B).

P(S| B) = P(Sand B)/P( B)

▼

Substitute values and simplify

P(Sand B)= 1/6, P( B)= 1/3

P(S| B) = 16/13

.a /b./.c /d.=a/b*d/c

P(S| B) = 1/6* 3/1

Multiply fractions

P(S| B) = 3/6

a/b=.a /3./.b /3.

P(S| B) = 1/2

Therefore, the probability that Diego picks an S-size T-shirt given that he picked a blue one is 12, or 50 %. Finally, find P( B|S).

P( B|S) = P(S and B)/P(S)

▼

Substitute values and simplify

P(S and B)= 1/6, P(S)= 1/3

P( B|S) = 16/13

.a /b./.c /d.=a/b*d/c

P( B|S) = 1/6* 3/1

Multiply fractions

P( B|S) = 3/6

a/b=.a /3./.b /3.

P( B|S) = 1/2

Consequently, the probability that Diego picks a blue T-shirt given that he picked an S-size one is 12, or 50 %. Time go to go out strutting wearing one of the shirts from the same old laundry basket — hopefully it is not his brothers!

Pop Quiz

Applying the Conditional Probability Formula

Find the required conditional probability and round it to two decimal places.

Random Probabilities of A,B, and 'A and B'

Closure

Destinations, Penguins, and Probabilities

Now that it is known how to compute conditional probabilities, Vincenzo's situation can be better investigated.

1. Tearrik next to the Eiffel Tower. 2. Tearrik in Antartica. 3. Tearrik and a penguin. 4. Frequency table comparing people who go to France/Antarctica and see a penguin with those who do not

External credits: @pikisuperstar, @macrovector, @jcomp

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

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

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.

Hint

Use the formula of conditional probability. Gathering the data from the table, a total of 190 people participated in the survey. Of that total 5, traveled to France and saw penguins.

A total of 35 people saw penguins. How many of these were in France?

Of the 190 people, 155 did not see penguins and 3 of them traveled to Antarctica.

Solution

By the definition of conditional probability, the probability that Vincenzo saw a penguin given that he traveled to France can be expressed in the following manner.

To find the corresponding probabilities, take a look at the table.

A total 190 people participated in the survey and 5 of them traveled to France and saw a penguin. P(Penguin and France) &= 5/190 [0.2cm] &= 1/38 Additionally, 157 people traveled to France. P(France) = 157/190 Next, substitute these two probabilities into the conditional probability formula.

P(Penguin|France) = P(Penguin and France)/P(France)

▼

Substitute values and simplify

P(Penguin and France)= 1/38, P(France)= 157/190

P(Penguin|France) = 138/157190

.a /b./.c /d.=a/b*d/c

P(Penguin|France) = 1/38 * 190/157

Multiply fractions

P(Penguin|France) = 190/5966

Calculate quotient

P(Penguin|France) = 0.031847...

Convert to percent

P(Penguin|France) = 3.1847... %

RoundInt

Round to nearest integer

P(Penguin|France) ≈ 3 %

Consequently, Vincenzo has about a 3 % chance of seeing penguins, given that he traveled to France.

In this case, the situation is opposite to that presented in Part A. Now, it is known that Vincenzo saw penguins, and it is asked to find the probability that Vincenzo is in France. As before, start by applying the conditional probability formula.

From Part A, the numerator is equal to 138. To determine the probability of seeing penguins, determine how many of the 190 people surveyed actually saw penguins. According to the table, 35 people answered that they did. P(Penguins) = 35/190 = 7/38 The next step is to substitute the two probabilities found into the conditional probability formula.

P(France|Penguin) = P(Penguin and France)/P(Penguins)

▼

Substitute values and simplify

P(Penguin and France)= 1/38, P(Penguins)= 7/38

P(France|Penguin) = 138/738

.a /b./.c /d.=a/b*d/c

P(France|Penguin) = 1/38 * 38/7

Multiply fractions

P(France|Penguin) = 38/266

Calculate quotient

P(France|Penguin) = 0.142857...

Convert to percent

P(France|Penguin) = 14.2857... %

RoundInt

Round to nearest integer

P(France|Penguin) ≈ 14 %

In conclusion, there is about a 14 % chance that Vincenzo was in France, given that he saw penguins.

Once more, start by applying the conditional probability formula.

From the second row and second column of the table, 3 of the 190 people traveled to Antarctica and did not see penguins. P(Antarctica and No Penguin) = 3/190 Seen in the third row of the table, 155 people did not see penguins. Knowing this, the probability that a person picked at random did not see penguins can be computed. P(No Penguin) = 155/190 = 31/38 Finally, substitute these values into the conditional probability formula.

P(Antarctica|No Penguin) = P(Antarctica and No Penguin)/P(No Penguin)

▼

Substitute values and simplify

P(Antarctica and No Penguin)= 3/190, P(No Penguin)= 31/38

P(Antarctica|No Penguin) = 3190/3138

.a /b./.c /d.=a/b*d/c

P(Antarctica|No Penguin) = 3/190 * 38/31

Multiply fractions

P(Antarctica|No Penguin) = 114/5890

Calculate quotient

P(Antarctica|No Penguin) = 0.019354...