Tree and Venn Diagrams Quiz
The question sheet
Reveal any answer as you study-
What are the two tools mentioned for visualizing and solving conditional probabilities?
- Bar and pie charts
- Line and scatter plots
- Tree and Venn diagrams
- Box and stem plots
Reveal answer
Answer: Tree and Venn diagrams
Source evidence
PDF page 207: 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.
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A tree diagram consists of branches labeled with either frequencies or what?
- Names
- Probabilities
- Colors
- Averages
Reveal answer
Answer: Probabilities
Source evidence
PDF page 207: 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.
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In the urn with 11 balls, how many balls are red?
- Five
- Eight
- Three
- Eleven
Reveal answer
Answer: Three
Source evidence
PDF page 207: In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). 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.
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When drawing two balls 'with replacement,' what do you do before the second draw?
- Mix in new balls
- Add a red ball
- Put the first ball back
- Remove a blue ball
Reveal answer
Answer: Put the first ball back
Source evidence
PDF page 207: In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). 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.
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In the tree diagram for 11 balls with replacement, what is the total number of outcomes?
- 121
- 24
- 110
- 64
Reveal answer
Answer: 121
Source evidence
PDF page 207: Figure 3.2 Total = 64 + 24 + 24 + 9 = 121
PDF page 207: 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. There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.
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The size of the sample space (121) comes from which calculation?
- 11/11
- 11(10)
- 11(11)
- 11+11
Reveal answer
Answer: 11(11)
Source evidence
PDF page 207: 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. There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.
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How many BR outcomes are listed in the tree diagram example?
- 24
- 48
- 9
- 64
Reveal answer
Answer: 24
Source evidence
PDF page 207: a. List the 24 BR outcomes: B1R1, B1R2, B1R3, ...
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In the conditional problem P(R on 2nd|B on 1st), how many possible outcomes remain?
- 121
- 88
- 33
- 64
Reveal answer
Answer: 88
Source evidence
PDF page 208: 88 11 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 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible 24 3 outcomes are BR. = . 88 11
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The 88 outcomes with blue on first draw consist of 24 BR and how many BB?
- 48
- 9
- 24
- 64
Reveal answer
Answer: 64
Source evidence
PDF page 208: 88 11 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 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible 24 3 outcomes are BR. = . 88 11
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For P(B on 2nd|R on 1st), the reduced sample space has how many outcomes?
- 24
- 121
- 33
- 88
Reveal answer
Answer: 33
Source evidence
PDF page 208: 11 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 24 of the 33 outcomes have B on the second draw. The probability is then . 33
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In the marble urn drawn without replacement, how many red marbles are there?
- Eight
- Three
- Eleven
- Two
Reveal answer
Answer: Three
Source evidence
PDF page 209: 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 6 ⎛ 3 ⎞⎛ 2 ⎞ two corresponding branches, for example, . = ⎝11⎠⎝10⎠ 110
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'Without replacement' means you do not do what before the second draw?
- Put the first ball back
- Shuffle the urn
- Remove a marble
- Count the balls
Reveal answer
Answer: Put the first ball back
Source evidence
PDF page 209: 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 6 ⎛ 3 ⎞⎛ 2 ⎞ two corresponding branches, for example, . = ⎝11⎠⎝10⎠ 110
Introductory Statistics
Introductory Statistics by OpenStax, used under CC BY 4.0. Changes made by Stratacademy.
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