Tree and Venn Diagrams

Learning Outcomes

  • Construct and interpret Tree Diagrams
  • Construct and interpret 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

In an urn, there are 11 balls. Three balls are red ([latex]R[/latex]) and eight balls are blue ([latex]B[/latex]). 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. The tree diagram using frequencies that show all the possible outcomes follows.

This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 8B and 3R. The second branch has a set of two lines (8B and 3R) for each line of the first branch. Multiply along each line to find 64BB, 24BR, 24RB, and 9RR.
Total = 64 + 24 + 24 + 9 = 121

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 [latex]R1[/latex], [latex]R2[/latex], and [latex]R3[/latex] and each blue ball as [latex]B1[/latex], [latex]B2[/latex], [latex]B3[/latex], [latex]B4[/latex], [latex]B5[/latex], [latex]B6[/latex], [latex]B7[/latex], and [latex]B8[/latex]. Then the nine [latex]RR[/latex] outcomes can be written as:
[latex]R1R1;\,\, R1R2;\,\, R1R3;\,\, R2R1;\,\, R2R2;\,\, R2R3;\,\, R3R1;\,\, R3R2;\,\, R3R3[/latex]
The other outcomes are similar.

There are a total of [latex]11[/latex] balls in the urn. Draw two balls, one at a time, with replacement. There are [latex]11(11) = 121[/latex] outcomes, the size of the sample space.

  1. List the [latex]24[/latex] [latex]BR[/latex] outcomes: [latex]B1R1[/latex], [latex]B1R2[/latex], [latex]B1R3[/latex], …
  2. Using the tree diagram, calculate [latex]P(RR)[/latex].
  3. Using the tree diagram, calculate [latex]P(RB \text{ OR } BR)[/latex].
  4. Using the tree diagram, calculate [latex]P(R \text{ on 1st draw AND } B \text { on 2nd draw })[/latex].
  5. Using the tree diagram, calculate [latex]P(R \text{ on 2nd draw GIVEN } B \text { on 1st draw })[/latex].
  6. Using the tree diagram, calculate [latex]P(BB)[/latex].
  7. Using the tree diagram, calculate [latex]P(B \text{ on the 2nd draw given } R \text { on the first draw })[/latex].

  1. [latex]B1R1;\, B1R2;\,\, B1R3;\, B2R1;\, B2R2;\, B2R3;\, B3R1;\, B3R2;\, B3R3;\, B4R1;\, B4R2;\, B4R3;\, B5R1;\, B5R2;\, B5R3;\, B6R1;\, B6R2;\, B6R3;\, B7R1;\, B7R2;\, B7R3;\, B8R1;\, B8R2;\, B8R3[/latex]
  2. [latex]P(RR) = \frac{3}{11}\frac{3}{11} = \frac{9}{121}[/latex]
  3. [latex]P(RB \text{ OR } BR) = \frac{3}{11}\frac{8}{11} + \frac{8}{11}\frac{3}{11} = \frac{48}{121}[/latex]
  4. [latex]P(R \text{ on 1st draw AND } B \text{ on 2nd draw}) = P(RB) = \frac{3}{11}\frac{8}{11} = \frac{24}{121}[/latex]
  5. [latex]P(R \text{ on 2nd draw GIVEN } B \text{ on 1st draw}) = P(R \text{ on 2nd }|B \text{ on 1st }) = \frac{24}{88} = \frac{3}{11}.[/latex] This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are [latex]24 + 64 = 88[/latex] possible outcomes ([latex]24 BR \text{ and } 64 BB[/latex]). Twenty-four of the [latex]88[/latex] possible outcomes are [latex]BR[/latex]. [latex]\frac{24}{88} = \frac{3}{11}[/latex].
  6. [latex]P(BB) = \frac{64}{121}[/latex]
  7. [latex]P(B \text{ on 2nd draw }|R \text{ on 1st draw }) = \frac{8}{11}[/latex] There are [latex]9 + 24[/latex] outcomes that have [latex]R[/latex] on the first draw ([latex]9 RR \text{ and } 24 RB[/latex]). The sample space is then [latex]9 + 24 = 33[/latex]. [latex]24[/latex] of the [latex]33[/latex] outcomes have [latex]B[/latex] on the second draw. The probability is then [latex]\frac{24}{33}[/latex].

Example

An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. “Without replacement” means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, (311)(210)=6110.

This is a tree diagram with branches showing probabilities of each draw. The first branch shows 2 lines: B 8/11 and R 3/11. The second branch has a set of 2 lines for each first branch line. Below B 8/11 are B 7/10 and R 3/10. Below R 3/11 are B 8/10 and R 2/10. Multiply along each line to find BB 56/110, BR 24/110, RB 24/110, and RR 6/110.
Total = [latex]\displaystyle\frac{{56+24+24+6}}{{110}}=\frac{{110}}{{110}}=1[/latex]
NOTE

If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn.