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Probability Topics: Tree Diagrams (optional)

Module by: Barbara Illowsky, Ph.D., Susan Dean. E-mail the authors

Summary: This module introduces tree diagrams as a method for making some probability problems easier to solve. This module is included in the Elementary Statistics textbook/collection as an optional lesson.

Note: You are viewing an old version of this document. The latest version is available here.

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 calculate. The following example illustrates how to use a tree diagram.

Example 1

In an urn, there are 11 balls. Three balls are red (RR) and 8 balls are blue (BB). 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.

Figure 1: Total=64+24+24 +9= 121 Total=64+24+24+9=121
Tree diagram consisting of the first draw for the first branch and the second draw for the second branch. The first branch consists of 2 lines, 3R and 8B, and the second branch consists of 2 sets of 2 lines of 3R and 8B each. The lines produce 9RR, 24RB, 24BR, and 64BB.

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 R1R1, R2R2, and R3R3 and each blue ball as B1B1, B2B2, B3B3, B4B4, B5B5, B6B6, B7B7, and B8B8. Then the 9 RRRR outcomes can be written as:

R1R1R1R1; R1R2R1R2; R1R3R1R3; R2R1R2R1; R2R2R2R2; R2R3R2R3; R3R1R3R1; R3R2R3R2; R3R3R3R3

The other outcomes are similar.

There are a total of 11 balls in the urn. Draw two balls, one at a time, and with replacement. There are 11  ⋅  11  =  121 11 ⋅ 11 = 121 outcomes, the size of the sample space.

Problem 1

List the 24 BRBR outcomes: B1R1B1R1, B1R2B1R2, B1R3B1R3, ...

Solution

B1R1B1R1; B1R2B1R2; B1R3B1R3; B2R1B2R1; B2R2B2R2; B2R3B2R3; B3R1B3R1; B3R2B3R2; B3R3B3R3; B4R1B4R1; B4R2B4R2; B4R3B4R3; B5R1B5R1; B5R2B5R2; B5R3B5R3; B6R1B6R1; B6R2B6R2; B6R3B6R3; B7R1B7R1; B7R2B7R2; B7R3B7R3; B8R1B8R1; B8R2B8R2; B8R3B8R3

Problem 2

Using the tree diagram, calculate P(RR)P(RR).

Solution

P(RR) = 3 11 3 11 = 9 121 P(RR) = 3 11 3 11 = 9 121

Problem 3

Using the tree diagram, calculate P(RB OR BR)P(RB OR BR).

Solution

P(RB OR BR) = 3 11 8 11 + 8 11 3 11 = 48 121 P(RB OR BR) = 3 11 8 11 + 8 11 3 11 = 48 121

Problem 4

Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw)P(R on 1st draw AND B on 2nd draw).

Solution

P(R on 1st draw AND B on 2nd draw) = P(RB) = 3 11 8 11 = 24 121 P(R on 1st draw AND B on 2nd draw) =P(RB) = 3 11 8 11 = 24 121

Problem 5

Using the tree diagram, calculate P(R on 2nd draw given B on 1st draw)P(R on 2nd draw given B on 1st draw).

Solution

P(R on 2nd draw given B on 1st draw) = P(R on 2nd | B on 1st) = 24 88 = 3 11 P(R on 2nd draw given B on 1st draw) =P(R on 2nd | B on 1st)= 24 88 = 3 11

This problem is a conditional. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24  +  64  =  88 24 + 64 = 88 possible outcomes (24 BRBR and 64 BBBB). Twenty-four of the 88 possible outcomes are BRBR. 24 88 = 3 11 24 88 = 3 11 .

Problem 6

Using the tree diagram, calculate P(BB)P(BB).

Solution

P(BB) = 64121P(BB) = 64121

Problem 7

Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw)P(B on the 2nd draw given R on the first draw).

Solution

P(B on 2nd draw | R on 1st draw)  =  8 11 P(B on 2nd draw | R on 1st draw) =  8 11

There are 9  +  249 + 24 outcomes that have R R on the first draw (9 RR RR and 24 RB RB ). The sample space is then 9  +  24  =  33 9 + 24 = 33. Twenty-four of the 33 outcomes have B B on the second draw. The probability is then 24 33 24 33 .

Example 2

An urn has 3 red marbles and 8 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 ball. Below is a tree diagram. 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, 3 11 2 10 = 6 110 3 11 2 10 = 6 110 .

Figure 2: Total = 56 + 24 + 24 + 6 110 = 110 110 = 1 Total = 56 + 24 + 24 + 6 110 = 110 110 = 1
Tree diagram consisting of the first draw for the first branch and the second draw for the second branch. The first branch consists of 2 lines, B 8/11 and R 3/11, and the second branch consists of 2 sets of 2 lines with B 7/10 and R 3/10 extending from line B 8/11 and B 8/10 and R 2/10 coming from line R 3/11. These 4 lines produce BB 56/110, BR 24/110, RB 24/110, and RR 6/10.

Note:

If you draw a red on the first draw from the 3 red possibilities, there are 2 red left to draw on the second draw. You do not put back or replace the first ball after you have drawn it.

Calculate the following probabilities using the tree diagram. You draw without replacement, so that on the second draw there are 10 marbles left in the urn.

Problem 1

P(RR)P(RR) =

Solution

P(RR) = 3 11 2 10 = 6 110 P(RR)= 3 11 2 10 = 6 110

Problem 2

Fill in the blanks:

P(RB OR BR) = 3 11 8 10 + (___)(___) = 48 110 P(RB OR BR)= 3 11 8 10 +(___)(___)= 48 110

Solution

P(RB or BR) = 3 11 8 10 + P(RB or BR)= 3 11 8 10 + ( 811 ) ( 310 ) (811)(310) = 48 110 = 48 110

Problem 3

P(R on 2d | B on 1st)P(R on 2d | B on 1st) =

Solution

P(R on 2d | B on 1st) =  3 10 P(R on 2d | B on 1st) =  3 10

Problem 4

Fill in the blanks:

P(R on 1st and B on 2nd)  =  P(RB)  =  (___)(___)  =  24 110 P(R on 1st and B on 2nd) = P(RB) = (___)(___) =  24 110

Solution

P(R on 1st and B on 2nd)  =  P(RB)  =  P(R on 1st and B on 2nd) = P(RB) =  ( 3 11 ) ( 8 10 ) ( 3 11 )( 8 10 )  =  24 110  =  24 110

Problem 5

P(BB)P(BB) =

Solution

P(BB)  =  8 11  ⋅  7 10 P(BB) =  8 11  ⋅  7 10

Problem 6

P(B on 2nd | R on 1st) P(B on 2nd | R on 1st) =

Solution

There are 6  +  246 + 24 outcomes that have RR on the first draw (6 RRRR and 24 RBRB). The 6 and the 24 are frequencies. They are also the numerators of the fractions 6 110 6 110 and 24 110 24 110 . The sample space is no longer 110 but 6  +  24  =  30 6 + 24 = 30. Twenty-four of the 30 outcomes have BB on the second draw. The probability is then 24 30 24 30 . Did you get this answer?

If we are using probabilities, we can label the tree in the following general way.

Tree diagram consisting of a first branch and a second branch. The first branch consists of 2 lines, P(R) and P(B), and the second branch consists of 2 sets of 2 lines with one set of P(B)(B) and P(R)(B) from line P(B) and one set of P(B)(R) and P(R)(R) from line P(R). P(B)(B) and P(R)(B) produce P(B and B)=P(BB) and P(B and R)=P(BR) and P(B)(R) and P(R)(R) produce P(R and B)=P(RB) and P(R and R)=P(RR).
  • P(R|R)P(R|R) here means P(R on 2nd | R on 1st)P(R on 2nd | R on 1st)
  • P(B|R)P(B|R) here means P(B on 2nd | R on 1st)P(B on 2nd | R on 1st)
  • P(R|B)P(R|B) here means P(R on 2nd | B on 1st)P(R on 2nd | B on 1st)
  • P(B|B)P(B|B) here means P(B on 2nd | B on 1st)P(B on 2nd | B on 1st)

Glossary

Sample Space:
The set of all possible outcomes of an experiment.
Tree Diagram:
The useful visual representation of a sample space and events in the form of “tree” with branches marked by possible outcomes simultaneously with associated probabilities (frequencies, relative frequencies).

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