{"id":134,"date":"2016-04-21T22:43:44","date_gmt":"2016-04-21T22:43:44","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstats1xmaster\/?post_type=chapter&p=134"},"modified":"2017-07-11T22:23:35","modified_gmt":"2017-07-11T22:23:35","slug":"tree-and-venn-diagrams","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-fmcc-introstats1\/chapter\/tree-and-venn-diagrams\/","title":{"raw":"Tree and Venn Diagrams","rendered":"Tree and Venn Diagrams"},"content":{"raw":"

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.<\/p>\r\n\r\n

\r\n

Tree Diagrams<\/h2>\r\n

A tree diagram<\/span> 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.<\/p>\r\n\r\n

\r\n

Example<\/h3>\r\n
In an urn, there are 11 balls. Three balls are red (R<\/em>) and eight balls are blue (B<\/em>). Draw two balls, one at a time, with replacement<\/strong>. \"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.\r\n
\"This<\/span>
Total = 64 + 24 + 24 + 9 = 121<\/figcaption><\/figure>\r\nThe 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 R<\/em>1, R<\/em>2, and R<\/em>3 and each blue ball as B<\/em>1, B<\/em>2, B<\/em>3, B<\/em>4, B<\/em>5, B<\/em>6, B<\/em>7, andB<\/em>8. Then the nine RR<\/em> outcomes can be written as:\r\n

R<\/em>1R<\/em>1<\/span> R<\/em>1R<\/em>2<\/span> R<\/em>1R<\/em>3<\/span> R<\/em>2R<\/em>1<\/span> R<\/em>2R<\/em>2<\/span> R<\/em>2R<\/em>3<\/span> R<\/em>3R<\/em>1<\/span> R<\/em>3R<\/em>2<\/span> R<\/em>3R<\/em>3<\/span><\/span><\/p>\r\n

The other outcomes are similar.<\/p>\r\nSolution:\r\n\r\n<\/section>

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<\/span>.\r\n\r\n<\/section><\/div>\r\n<\/section>\r\n
\r\n
\r\n
\r\n

<\/h3>\r\n
\r\n
\r\n
\r\n

Try it<\/h3>\r\n

a. List the 24 BR<\/em> outcomes: B<\/em>1R<\/em>1, B<\/em>1R<\/em>2, B<\/em>1R<\/em>3, ...<\/p>\r\nb. Using the tree diagram, calculate P<\/em>(RR<\/em>).\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n

\r\n
\r\n
\u00a0c. Using the tree diagram, calculate P<\/em>(RB<\/em> OR BR<\/em>).<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
\r\n
\r\n
\u00a0d. Using the tree diagram, calculate P<\/em>(R<\/em> on 1st draw AND B<\/em> on 2nd draw).<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
\r\n
\r\n
\u00a0e. Using the tree diagram, calculate P<\/em>(R<\/em> on 2nd draw GIVEN B<\/em> on 1st draw).<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
\r\n
\r\n
\u00a0f. Using the tree diagram, calculate P<\/em>(BB<\/em>).<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
\r\n
\r\n
\u00a0g. Using the tree diagram, calculate P<\/em>(B<\/em> on the 2nd draw given R<\/em> on the first draw).<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
\r\n
\r\n
\r\n

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\"<\/strong> 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, (<\/span>3<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>2<\/span>10<\/span><\/span><\/span><\/span>)<\/span><\/span>=<\/span>6<\/span>110<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>.<\/p>\r\n\r\n

\"This<\/span>
Total = [latex]\\displaystyle\\frac{{56+24+24+6}}{{110}}=\\frac{{110}}{{110}}=1[\/latex]\r\nNOTE<\/figcaption><\/figure>\r\n<\/div>\r\n
\r\n

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<\/strong>, so that on the second draw there are ten marbles left in the urn.<\/p>\r\n\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n \r\n\r\n<\/div>\r\n<\/section><\/div>\r\n

\r\n
\r\n
\r\n

Try it<\/h3>\r\n
\r\n
\r\n
\r\n

a. P<\/em>(RR<\/em>) = ________<\/p>\r\n\r\n<\/div>\r\n

\r\n
\u00a0b. Fill in the blanks:<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
\r\n
\r\n

P<\/em>(RB<\/em> OR BR<\/em>) = (<\/span>3<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>8<\/span>10<\/span><\/span><\/span><\/span>)<\/span><\/span>\u00a0<\/span>+<\/span>\u00a0(___)(___)\u00a0<\/span>=<\/span>\u00a0<\/span>48<\/span><\/span>110<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n
\u00a0c. P<\/em>(R<\/em> on 2nd|B<\/em> on 1st) =<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
\r\n
\r\n
\u00a0d. Fill in the blanks.<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
\r\n
\r\n

P<\/em>(R<\/em> on 1st AND B<\/em> on 2nd) = P<\/em>(RB<\/em>) = (___)(___)\u00a0= 24<\/span><\/span>100<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n
\u00a0e. Find P<\/em>(BB<\/em>).<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
\r\n
\r\n\r\nf. Find P<\/em>(B<\/em> on 2nd|R<\/em> on 1st).\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n
<\/div>\r\n

If we are using probabilities, we can label the tree in the following general way.<\/p>\r\n\"This<\/span>\r\n

    \r\n \t
  • P<\/em>(R<\/em>|R<\/em>) here means P<\/em>(R<\/em> on 2nd|R<\/em> on 1st)<\/li>\r\n \t
  • P<\/em>(B<\/em>|R<\/em>) here means P<\/em>(B<\/em> on 2nd|R<\/em> on 1st)<\/li>\r\n \t
  • P<\/em>(R<\/em>|B<\/em>) here means P<\/em>(R<\/em> on 2nd|B<\/em> on 1st)<\/li>\r\n \t
  • P<\/em>(B<\/em>|B<\/em>) here means P<\/em>(B<\/em> on 2nd|B<\/em> on 1st)<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n<\/div>\r\n
    <\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\nhttps:\/\/youtu.be\/Zxvc6iPKdec\r\n\r\n
    \r\n

    Venn Diagram<\/h2>\r\n

    A Venn diagram<\/span> is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events.<\/p>\r\n\r\n

    \r\n

    Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an equal chance of occurring. Let event A<\/em> = {1, 2, 3, 4, 5, 6} and event B<\/em> = {6, 7, 8, 9}. Then A<\/em> AND B<\/em> = {6} and A<\/em>\u00a0OR\u00a0B<\/em> = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:<\/p>\r\n\r\n

    \"A<\/span><\/figure>\r\n \r\n\r\n<\/section><\/div>\r\n
    \r\n
    \r\n
    \r\n

    Example<\/h3>\r\nFlip two fair coins. Let A<\/em> = tails on the first coin. Let B<\/em> = tails on the second coin. Then A<\/em> = {TT<\/em>, TH<\/em>} and B<\/em> = {TT<\/em>, HT<\/em>}. Therefore,A<\/em> AND B<\/em> = {TT<\/em>}. A<\/em> OR B<\/em> = {TH<\/em>, TT<\/em>, HT<\/em>}.\r\n

    The sample space when you flip two fair coins is X<\/em> = {HH<\/em>, HT<\/em>, TH<\/em>, TT<\/em>}. The outcome HH<\/em> is in NEITHER A<\/em> NOR B<\/em>. The Venn diagram is as follows:<\/p>\r\n\r\n

    \"This<\/span><\/figure>\r\n<\/div>\r\n \r\n\r\n<\/section><\/div>\r\n
    <\/div>\r\n<\/div>\r\n<\/section>https:\/\/youtu.be\/MassxXy8iko\r\n

    Glossary<\/h2>\r\n
    \r\n \t
    Tree Diagram<\/dt>\r\n \t
    the useful visual representation of a sample space and events in the form of a \u201ctree\u201d with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies)<\/dd>\r\n<\/dl>\r\n
    \r\n \t
    Venn Diagram<\/dt>\r\n \t
    the visual representation of a sample space and events in the form of circles or ovals showing their intersections<\/dd>\r\n<\/dl>\r\n<\/section><\/div>\r\n
    Solutions to Try These 1:<\/div>\r\n
    a. B<\/em>1R<\/em>1<\/span> B<\/em>1R<\/em>2<\/span> B<\/em>1R<\/em>3<\/span> B<\/em>2R<\/em>1<\/span> B<\/em>2R<\/em>2<\/span> B<\/em>2R<\/em>3<\/span> B<\/em>3R<\/em>1<\/span> B<\/em>3R<\/em>2<\/span> B<\/em>3R<\/em>3<\/span> B<\/em>4R<\/em>1<\/span> B<\/em>4R<\/em>2<\/span> B<\/em>4R<\/em>3<\/span> B<\/em>5R<\/em>1<\/span> B<\/em>5R<\/em>2<\/span> B<\/em>5R<\/em>3<\/span> B<\/em>6R<\/em>1<\/span> B<\/em>6R<\/em>2<\/span> B<\/em>6R<\/em>3<\/span> B<\/em>7R<\/em>1<\/span>B<\/em>7R<\/em>2<\/span> B<\/em>7R<\/em>3<\/span> B<\/em>8R<\/em>1<\/span> B<\/em>8R<\/em>2<\/span> B<\/em>8R<\/em>3<\/span><\/span><\/div>\r\n
    b. P<\/em>(RR<\/em>) = (<\/span>3<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>3<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 9<\/span>121<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\r\n
    c. P<\/em>(RB<\/em> OR BR<\/em>) = (<\/span>3<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>8<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> + (<\/span>8<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>3<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 48<\/span><\/span>121<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\r\n
    d. P<\/em>(R<\/em> on 1st draw AND B<\/em> on 2nd draw) = P<\/em>(RB<\/em>) = (<\/span>3<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>8<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 24<\/span><\/span>121<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\r\n
    \r\n

    e. P<\/em>(R<\/em> on 2nd draw GIVEN B<\/em> on 1st draw) = P<\/em>(R<\/em> on 2nd|B<\/em> on 1st) = 24<\/span><\/span>88<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 3<\/span><\/span>11<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\r\n

    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<\/em> and 64 BB<\/em>). Twenty-four of the 88 possible outcomes are BR<\/em>. 24<\/span><\/span>88<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 3<\/span><\/span>11<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>.<\/p>\r\nf. P<\/em>(BB<\/em>)\u00a0=\u00a064<\/span><\/span>121<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\r\n

    \r\n
    \r\n
    \r\n

    g. P<\/em>(B<\/em>\u00a0on\u00a02nd\u00a0draw|R<\/em>\u00a0on\u00a01st\u00a0draw) =\u00a08<\/span>11<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\r\n

    There are 9 + 24 outcomes that have R<\/em> on the first draw (9 RR<\/em> and 24 RB<\/em>). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B<\/em> on the second draw. The probability is then 24<\/span>33<\/span><\/span><\/span><\/span><\/span><\/span><\/span>.<\/p>\r\nSolutions to Try These 2:\r\n\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

    a. P<\/em>(RR<\/em>) = (<\/span>3<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>2<\/span>10<\/span><\/span><\/span><\/span>)<\/span><\/span>=<\/span>6<\/span>110<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\r\n
    b. P<\/em>(RB<\/em> OR BR<\/em>) = (<\/span>3<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>8<\/span>10<\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> + (<\/span>8<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>3<\/span>10<\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 48<\/span><\/span>110<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\r\n
    c. P<\/em>(R<\/em> on 2nd|B<\/em> on 1st) = 3<\/span>10<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\r\n
    d. P<\/em>(R<\/em> on 1st AND B<\/em> on 2nd) = P<\/em>(RB<\/em>) = (<\/span>3<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>8<\/span>10<\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 24<\/span><\/span>100<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\r\n
    e. P<\/em>(BB<\/em>) = (<\/span>8<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>7<\/span>10<\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\r\n
    f. Using the tree diagram, P<\/em>(B<\/em> on 2nd|R<\/em> on 1st) = P<\/em>(R<\/em>|B<\/em>) = 8<\/span>10<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>.<\/div>\r\n
    <\/div>\r\n<\/div>\r\n
    <\/div>","rendered":"

    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.<\/p>\n

    \n

    Tree Diagrams<\/h2>\n

    A tree diagram<\/span> 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.<\/p>\n

    \n

    Example<\/h3>\n
    In an urn, there are 11 balls. Three balls are red (R<\/em>) and eight balls are blue (B<\/em>). Draw two balls, one at a time, with replacement<\/strong>. “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.<\/p>\n
    \"This<\/span>
    Total = 64 + 24 + 24 + 9 = 121<\/figcaption><\/figure>\n

    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 R<\/em>1, R<\/em>2, and R<\/em>3 and each blue ball as B<\/em>1, B<\/em>2, B<\/em>3, B<\/em>4, B<\/em>5, B<\/em>6, B<\/em>7, andB<\/em>8. Then the nine RR<\/em> outcomes can be written as:<\/p>\n

    R<\/em>1R<\/em>1<\/span> R<\/em>1R<\/em>2<\/span> R<\/em>1R<\/em>3<\/span> R<\/em>2R<\/em>1<\/span> R<\/em>2R<\/em>2<\/span> R<\/em>2R<\/em>3<\/span> R<\/em>3R<\/em>1<\/span> R<\/em>3R<\/em>2<\/span> R<\/em>3R<\/em>3<\/span><\/span><\/p>\n

    The other outcomes are similar.<\/p>\n

    Solution:<\/p>\n<\/section>\n

    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<\/span>.<\/p>\n<\/section>\n<\/div>\n<\/section>\n
    \n
    \n
    \n
    \n
    \n

    <\/h3>\n
    \n
    \n
    \n
    \n

    Try it<\/h3>\n

    a. List the 24 BR<\/em> outcomes: B<\/em>1R<\/em>1, B<\/em>1R<\/em>2, B<\/em>1R<\/em>3, …<\/p>\n

    b. Using the tree diagram, calculate P<\/em>(RR<\/em>).<\/p>\n<\/div>\n<\/section>\n<\/div>\n

    \n
    \n
    \n
    \u00a0c. Using the tree diagram, calculate P<\/em>(RB<\/em> OR BR<\/em>).<\/div>\n<\/div>\n<\/section>\n<\/div>\n
    \n
    \n
    \n
    \u00a0d. Using the tree diagram, calculate P<\/em>(R<\/em> on 1st draw AND B<\/em> on 2nd draw).<\/div>\n<\/div>\n<\/section>\n<\/div>\n
    \n
    \n
    \n
    \u00a0e. Using the tree diagram, calculate P<\/em>(R<\/em> on 2nd draw GIVEN B<\/em> on 1st draw).<\/div>\n<\/div>\n<\/section>\n<\/div>\n
    \n
    \n
    \n
    \u00a0f. Using the tree diagram, calculate P<\/em>(BB<\/em>).<\/div>\n<\/div>\n<\/section>\n<\/div>\n
    \n
    \n
    \n
    \u00a0g. Using the tree diagram, calculate P<\/em>(B<\/em> on the 2nd draw given R<\/em> on the first draw).<\/div>\n<\/div>\n<\/section>\n<\/div>\n
    \n
    \n
    \n
    \n

    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”<\/strong> 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, (<\/span>3<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>2<\/span>10<\/span><\/span><\/span><\/span>)<\/span><\/span>=<\/span>6<\/span>110<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>.<\/p>\n

    \"This<\/span>
    Total = [latex]\\displaystyle\\frac{{56+24+24+6}}{{110}}=\\frac{{110}}{{110}}=1[\/latex]
    \nNOTE<\/figcaption><\/figure>\n<\/div>\n
    \n

    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<\/strong>, so that on the second draw there are ten marbles left in the urn.<\/p>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n

     <\/p>\n<\/div>\n<\/section>\n<\/div>\n

    \n
    \n
    \n
    \n
    \n

    Try it<\/h3>\n
    \n
    \n
    \n
    \n
    \n

    a. P<\/em>(RR<\/em>) = ________<\/p>\n<\/div>\n

    \n
    \u00a0b. Fill in the blanks:<\/div>\n<\/div>\n<\/section>\n<\/div>\n
    \n
    \n
    \n

    P<\/em>(RB<\/em> OR BR<\/em>) = (<\/span>3<\/span>11<\/span><\/span><\/span><\/span>)<\/span><\/span>(<\/span>8<\/span>10<\/span><\/span><\/span><\/span>)<\/span><\/span>\u00a0<\/span>+<\/span>\u00a0(___)(___)\u00a0<\/span>=<\/span>\u00a0<\/span>48<\/span><\/span>110<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/div>\n

    \n
    \u00a0c. P<\/em>(R<\/em> on 2nd|B<\/em> on 1st) =<\/div>\n<\/div>\n<\/section>\n<\/div>\n
    \n
    \n
    \n
    \u00a0d. Fill in the blanks.<\/div>\n<\/div>\n<\/section>\n<\/div>\n
    \n
    \n
    \n

    P<\/em>(R<\/em> on 1st AND B<\/em> on 2nd) = P<\/em>(RB<\/em>) = (___)(___)\u00a0= 24<\/span><\/span>100<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/div>\n

    \n
    \u00a0e. Find P<\/em>(BB<\/em>).<\/div>\n<\/div>\n<\/section>\n<\/div>\n
    \n
    \n
    \n

    f. Find P<\/em>(B<\/em> on 2nd|R<\/em> on 1st).<\/p>\n<\/div>\n<\/section>\n<\/div>\n

    <\/div>\n

    If we are using probabilities, we can label the tree in the following general way.<\/p>\n

    \"This<\/span><\/p>\n

      \n
    • P<\/em>(R<\/em>|R<\/em>) here means P<\/em>(R<\/em> on 2nd|R<\/em> on 1st)<\/li>\n
    • P<\/em>(B<\/em>|R<\/em>) here means P<\/em>(B<\/em> on 2nd|R<\/em> on 1st)<\/li>\n
    • P<\/em>(R<\/em>|B<\/em>) here means P<\/em>(R<\/em> on 2nd|B<\/em> on 1st)<\/li>\n
    • P<\/em>(B<\/em>|B<\/em>) here means P<\/em>(B<\/em> on 2nd|B<\/em> on 1st)<\/li>\n<\/ul>\n<\/section>\n<\/div>\n<\/div>\n
      <\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n