Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities.
Tree Diagrams
A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.
In a patient registry, there are 11 medical files. Three files indicate high health Risk (R) and eight files indicate a healthy Baseline (B). Select two files, one at a time, with replacement. "With replacement" means that you digitally return the first file to the registry before you select the second file.
The tree diagram using frequencies that shows all the possible outcomes follows.
The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each health risk file drawn as \(R 1, R 2\), and \(R 3\) and each baseline health file drawn as \(B 1, B 2, B 3, B 4, B 5, B 6, B 7\), and \(B 8\). Then the nine \(R R\) outcomes can be written as:
R1R1; R1R2; R1R3; R2R1; R2R2; R2R3; R3R1; R3R2; R3R3
The other outcomes are similar.
There are a total of 11 medical files. Draw two files, one at a time, with replacement. There are \(11(11)=\) 121 outcomes, the size of the sample space.
Problem
a. List the \(24 B R\) outcomes: \(B 1 R 1, B 1 R 2, B 1 R 3, \ldots\)
b. Using the tree diagram, calculate \(P(R R)\).
c. Using the tree diagram, calculate \(P(R B\) OR \(B R)\).
d. Using the tree diagram, calculate \(P(R\) on 1st draw AND \(B\) on 2nd draw).
e. Using the tree diagram, calculate \(P(R\) on 2nd draw GIVEN \(B\) on 1st draw).
f. Using the tree diagram, calculate \(P(B B)\).
- Answer
-
a. \(B 1 R 1 ; B 1 R 2 ; B 1 R 3 ; B 2 R 1 ; B 2 R 2 ; B 2 R 3 ; B 3 R 1 ; B 3 R 2 ; B 3 R 3 ; B 4 R 1 ; B 4 R 2 ; B 4 R 3 ; B 5 R 1 ; B 5 R 2 ;\) B5R3; B6R1; B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B8R2; B8R3
b. \(P(R R)=\left(\frac{3}{11}\right)\left(\frac{3}{11}\right)=\frac{9}{121}\)
c. \(P(R B\) OR \(B R)=\left(\frac{3}{11}\right)\left(\frac{8}{11}\right)+\left(\frac{8}{11}\right)\left(\frac{3}{11}\right)=\frac{48}{121}\)
d. \(P(R\) on 1st draw AND \(B\) on 2nd draw \()=P(R B)=\left(\frac{3}{11}\right)\left(\frac{8}{11}\right)=\frac{24}{121}\)
e. \(P(R\) on 2 nd draw GIVEN \(B\) on 1 st draw \()=P(R\) on 2 nd \(\mid B\) on 1 st \()=\frac{24}{88}=\frac{3}{11}\)
This problem is a conditional one. The sample space has been reduced to those outcomes that already have a healthy Baseline drawn on the first draw. There are \(24+64=88\) possible outcomes ( \(24 B R\) and \(64 B B\) ). Twenty-four of the 88 possible outcomes are \(B R . \frac{24}{88}=\frac{3}{11}\).
f. \(P(B B)=\frac{64}{121}\)
g. \(P(B\) on 2 nd draw \(\mid R\) on 1st draw \()=\frac{8}{11}\)
There are \(9+24\) outcomes that have \(R\) on the first draw (9 \(R R\) and \(24 R B\) ). The sample space is then \(9+24=33.24\) of the 33 outcomes have \(B\) on the second draw. The probability is then \(\frac{24}{33}\).
A patient registry has 11 medical files: three indicate a high health Risk (R) and eight indicate a healthy Baseline (B). Draw two files, one at a time, this time without replacement, from the registry. "Without replacement" means that you do not return the first file to the registry before you select the second file. Because the first file is kept out, the total number of files available for the second draw decreases from 11 to 10, changing the probabilities for the second stage.
The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, \(\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}\).
Calculate the following probabilities using the tree diagram.
Problem
a. \(P(R R)=\) \(\_\_\_\_\)
b. Fill in the blanks:
\[P(R B \text { OR } B R)=\left(\frac{3}{11}\right)\left(\frac{8}{10}\right)+\left(\_\right)\left(\_\right)=\frac{48}{110}\]
c. \(P(R\) on \(2 \mathrm{nd} \mid B\) on 1 st\()=\)
d. Fill in the blanks.
\[P(R \text { on 1st AND } B \text { on 2nd })=P(R B)=(\sqsubseteq)(\square)=\frac{24}{110}\]
e. Find \(P(B B)\).
f. Find \(P(B\) on \(2 \mathrm{nd} \mid R\) on 1 st\()\).
- Answer
-
a. \(P(R R)=\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}\)
b. \(P(R B\) OR \(B R)=\left(\frac{3}{11}\right)\left(\frac{8}{10}\right)+\left(\frac{8}{11}\right)\left(\frac{3}{10}\right)=\frac{48}{110}\)
c. \(P(R\) on \(2 \mathrm{nd} \mid B\) on 1 st\()=\frac{3}{10}\)
d. \(P(R\) on 1 st \(\mathrm{AND} B\) on 2nd \()=P(R B)=\left(\frac{3}{11}\right)\left(\frac{8}{10}\right)=\frac{24}{110}\)
e. \(P(B B)=\left(\frac{8}{11}\right)\left(\frac{7}{10}\right)\)
f. Using the tree diagram, \(P(B\) on \(2 \mathrm{nd} \mid R\) on 1 st\()=P(R \mid B)=\frac{8}{10}\).
If we are using probabilities, we can label the tree in the following general way.
Figure \(\PageIndex{4}\): - \(P(R \mid R)\) here means \(P(R\) on \(2 \mathrm{nd} \mid R\) on 1 st\()\)
- \(P(B \mid R)\) here means \(P(B\) on \(2 \mathrm{nd} \mid R\) on 1 st\()\)
- \(P(R \mid B)\) here means \(P(R\) on \(2 \mathrm{nd} \mid B\) on 1 st\()\)
- \(P(B \mid B)\) here means \(P(B\) on \(2 \mathrm{nd} \mid B\) on 1 st\()\)
Venn Diagrams
A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. Venn diagrams also help us to convert common English words into mathematical terms that help add precision.
Suppose an experiment has the outcomes \(1,2,3, \ldots, 12\) where each outcome has an equal chance of occurring. Let event \(A=\{1,2,3,4,5,6\}\) and event \(B=\{6,7,8,9\}\). Then \(A\) intersect \(B=A \cap B=\{6\}\) and \(A\) union \(B=A \cup B=\{1,2,3,4,5,6,7,8,9\}\). The Venn diagram is as follows:
- Answer
-
Figure \(\PageIndex{7}\)
Flip two fair coins. Let \(A=\) tails on the first coin. Let \(B=\) tails on the second coin. Then \(A=\{T T, T H\}\) and \(B\) \(=\{T T, H T\}\). Therefore, \(A \cap B=\{\mathrm{TT}\} . A \cup B=\{\mathrm{TH}, \mathrm{TT}, \mathrm{HT}\}\).
The sample space when you flip two fair coins is \(X=\{H H, H T, T H, T T\}\). The outcome \(H H\) is in NEITHER \(A\) NOR \(B\). The Venn diagram is as follows:
- Answer
-
Figure \(\PageIndex{8}\)
Solution
a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor.
Forty percent of the students at a local college belong to a health club and \(\mathbf{5 0} \boldsymbol{\%}\) work part time. Five percent of the students work part time and belong to a health club. Draw a Venn diagram showing the relationships. Let \(C=\) student belongs to a club and \(P T=\) student works part time.
If a student is selected at random, find
- the probability that the student belongs to a health club. \(P(C)=0.40\)
- the probability that the student works part time. \(P(P T)=0.50\)
- the probability that the student belongs to a health club AND works part time. \(P(C\) AND \(P T)=0.05\)
- the probability that the student belongs to a health club given that the student works part time.
\[P(C \mid P T)=\frac{P(C \text { AND } P T)}{P(P T)}=\frac{0.05}{0.50}=0.1\]
- the probability that the student belongs to a health club OR works part time. \(P(C\) OR \(P T)=P(C)+P(P T)\) \(P(C\) AND \(P T)=0.40+0.50-0.05=0.85\)
A person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative RH factor, 5−10% of African Americans have the Rh- factor, and 51% have type O blood.
-
Figure \(\PageIndex{9}\)
The “O” circle represents the African Americans with type O blood. The “Rh-“ oval represents the African Americans with the Rh- factor.
We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rh- factor. Let O = African American with Type O blood and R = African American with Rh- factor.
a. \(P(O)=\) \(\qquad\)
b. \(P(R)=\) \(\qquad\)
c. \(P(O \cap R)=\) \(\qquad\)
d. \(P(O \cup R)=\) \(\qquad\)
e. In the Venn Diagram, describe the overlapping area using a complete sentence.
f. In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence.
- Answer
-
a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor.
Solution
a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor.