Tree and Venn Diagrams Quiz

PDF page 206: 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:

PDF page 206: 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. Therefore, you are selecting from exactly the same group each time, so each draw is independent. The tree diagram shows all the possible outcomes.

PDF page 206: 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. Therefore, you are selecting from exactly the same group each time, so each draw is independent. The tree diagram shows all the possible outcomes.

PDF page 207: The first set of branches represents the first draw. There are 8 ways to draw a blue marble and 3 ways to draw a red one. The second set of branches represents the second draw. Regardless of the choice on the first draw, there are again eight ways to draw a blue marble and 3 ways to draw a red one. Read down each branch to see the total number of possible outcomes. For example, there are 8 ways to get a blue marble on the first draw, and eight ways to get one on the second draw, so there are 8 × 8 = 64 different ways to draw two blue marbles in succession. 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 follows: 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. Problem

PDF page 207: Figure 3.10 Total = 64 + 24 + 24 + 9 = 121.

PDF page 207: The first set of branches represents the first draw. There are 8 ways to draw a blue marble and 3 ways to draw a red one. The second set of branches represents the second draw. Regardless of the choice on the first draw, there are again eight ways to draw a blue marble and 3 ways to draw a red one. Read down each branch to see the total number of possible outcomes. For example, there are 8 ways to get a blue marble on the first draw, and eight ways to get one on the second draw, so there are 8 × 8 = 64 different ways to draw two blue marbles in succession. 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 follows: 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. Problem

PDF page 207: a. List the 24 BR outcomes: B1R1, B1R2, B1R3, . . .

PDF page 207: a. We know that there will be 24 different possible outcomes because there are eight ways to draw blue and three ways

PDF page 208: The events RB and BR are mutually exclusive, so P(RB OR BR) = P(RB) + P(BR) = P(R)P(B) + P(B)P(R) =

PDF page 209: In a standard deck, there are 52 cards. Twelve 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 outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P(FF).

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 two corresponding branches, for example, P(RR) = .

PDF page 210: 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 10 marbles left in the urn.

PDF page 210: 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 10 marbles left in the urn.

PDF page 211: • P(B|R) here means P(B on 2nd|R on 1st)