22 3.5 Tree and Venn Diagrams
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.
Example 1
In an urn, there are 11 balls. Draw two balls, one at a time, with replacement. “With replacement” means that you put the first ball back in the urn before you select the second ball.
a. Use tree diagram to show all possible outcomes.
Show Answer
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 red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8.
Then the nine RR outcomes can be written as:
R1R1 R1R2 R1R3 R2R1 R2R2 R2R3 R3R1 R3R2 R3R3
The other outcomes are similar.
Draw two balls, one at a time, with replacement.
There are 11(11) = 121 outcomes, the size of the sample space.
b. List the 24 BR outcomes.
B1R1, B1R2, B1R3, B2R1, B2R2, B2R3, B3R1, B3R2, B3R3, B4R1, B4R2, B4R3,Show Answer
B5R1, B5R2, B5R3, B6R1, B6R2, B6R3, B7R1, B7R2, B7R3, B8R1, B8R2, B8R3
c. Using the tree diagram, calculate P(RR).
Show Answer
P(RR) = [latex](\frac{3}{11})(\frac{3}{11})[/latex] = [latex]\frac{9}{121}[/latex]
d. Using the tree diagram, calculate P(RB OR BR).
Show Answer
P(RB OR BR) = ([latex]\frac{3}{11}[/latex])([latex]\frac{8}{11}[/latex]) + ([latex]\frac{8}{11}[/latex])([latex]\frac{3}{11}[/latex]) = [latex]\frac{48}{121}[/latex]
e. Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw).
Show Answer
P(R on 1st draw AND B on 2nd draw) =([latex]\frac{3}{11}[/latex])([latex]\frac{8}{11}[/latex]) = [latex]\frac{24}{121}[/latex]
f. Using the tree diagram, calculate P(R on 2nd draw GIVEN B on 1st draw).
Show Answer
P(R on 2nd [latex]\mid[/latex] B on 1st) = [latex]\frac{24}{88}[/latex] = [latex]\frac{3}{11}[/latex]
g. Using the tree diagram, calculate P(BB).
Show Answer
P(BB) = [latex]\frac{64}{121}[/latex]
h. Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw).
Show Answer
]P(B on 2nd [latex]\mid[/latex] R on 1st) = [latex]\frac{8}{11}[/latex]
There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB).
The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw.
Therefore, P(B on the 2nd draw given R on the first draw) = [latex]\frac{24}{33}[/latex] = [latex]\frac{8}{11}[/latex]
Try It
In a standard deck, there are 52 cards. 12 cards are face cards (event F) and 40 cards are not face cards (event N). Draw two cards, one at a time, with replacement. All possible outco