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== 7.9#h(0.6em)Conditional Probability and the Multiplication Rule


  #figure(figph[Two red dice are shown. The first die shows 5 and the second die shows 1.], alt: "Two red dice are shown. The first die shows 5 and the second die shows 1.", caption: [If you roll two dice by throwing them one at a time, the face showing on the first die will affect the possible outcomes for the sum of the two dice.])


  
    
=== Learning Objectives


    #strong[After completing this section, you should be able to:]


    + Calculate conditional probabilities.
+ Apply the Multiplication Rule for Probability to compute probabilities.


  
  Back in Example 3 in Section 7.5, we constructed the following table  to help us find the probabilities associated with rolling two standard 6-sided dice:


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  For example, 3 of these 36 equally likely outcomes correspond to rolling a sum of 10, so the probability of rolling a 10 is #math.equation(block: false, alt: "the fraction 3 over 36 equals the fraction 1 over 12")[$frac(3, 36) = frac(1, 12)$]. However, if you choose to roll the dice one at a time, the probability of rolling a 10 will change after the first die comes to rest. For example, if the first die shows a 5, then the probability of rolling a sum of 10 has jumped to #math.equation(block: false, alt: "the fraction 1 over 6")[$frac(1, 6)$]—the event will occur if the second die also shows a 5, which is 1 of 6 equally likely outcomes for the second die. If instead the first die shows a 3, then the probability of rolling a sum of 10 drops to 0—there are no outcomes for the second die that will give us a sum of 10.


  Understanding how probabilities can shift as we learn new information is critical in the analysis of our second type of compound events: those built with “and.” This section will explain how to compute probabilities of those compound events.


  
    
=== Conditional Probabilities


    When we analyze experiments with multiple stages, we often update the probabilities of the possible final outcomes or the later stages of the experiment based on the results of one or more of the initial stages. These updated probabilities are called #strong[conditional probabilities].


    In other words, if #math.equation(block: false, alt: "O")[$O$] is a possible outcome of the first stage in a multistage experiment, then the probability of an event #math.equation(block: false, alt: "E")[$E$] conditional on #math.equation(block: false, alt: "O")[$O$] (denoted #math.equation(block: false, alt: "P open parenthesis E | O close parenthesis")[$P ( E | O )$], read “the probability of #math.equation(block: false, alt: "E")[$E$] given #math.equation(block: false, alt: "O")[$O$]”) is the updated probability of #math.equation(block: false, alt: "E")[$E$] under the assumption that #math.equation(block: false, alt: "O")[$O$] occurred.


    In the example that opened this section, we might consider rolling two dice as a multistage experiment: rolling one, then the other. If we define #math.equation(block: false, alt: "E")[$E$] to be the event “roll a sum of 10,” #math.equation(block: false, alt: "O")[$O$] to be the event “first die shows 5,” and #math.equation(block: false, alt: "Q")[$Q$] to be the event “first die shows 3,” then we computed #math.equation(block: false, alt: "P open parenthesis E close parenthesis equals the fraction 1 over 12")[$P ( E ) = frac(1, 12)$], #math.equation(block: false, alt: "P open parenthesis E | O close parenthesis equals the fraction 1 over 6")[$P ( E | O ) = frac(1, 6)$], and #math.equation(block: false, alt: "P open parenthesis E | Q close parenthesis equals 0")[$P ( E | Q ) = 0$].


    #examplebox("Example 1")[Computing Conditional Probabilities][
+ April is playing a coin-flipping game with Ben. She will flip a coin 3 times. If the coin lands on heads more than tails, April wins; if it lands on tails more than heads, Ben wins. Let #math.equation(block: false, alt: "A")[$A$] be the event “April wins,” #math.equation(block: false, alt: "H")[$H$] be “first flip is heads,” and #math.equation(block: false, alt: "T")[$T$] be “first flip is tails.” Compute #math.equation(block: false, alt: "P open parenthesis A close parenthesis")[$P ( A )$], #math.equation(block: false, alt: "P open parenthesis A | H close parenthesis")[$P ( A | H )$], and #math.equation(block: false, alt: "P open parenthesis A | T close parenthesis")[$P ( A | T )$].
+ You are about to draw 2 cards without replacement from a deck containing only these 10 cards: #math.equation(block: false, alt: "A ♡")[$"A" ♡$], #math.equation(block: false, alt: "A ♠")[$"A" ♠$], #math.equation(block: false, alt: "A ♣")[$"A" ♣$], #math.equation(block: false, alt: "A ♢")[$"A" ♢$], #math.equation(block: false, alt: "K ♠")[$"K" ♠$], #math.equation(block: false, alt: "K ♣")[$"K" ♣$], #math.equation(block: false, alt: "Q ♡")[$"Q" ♡$], #math.equation(block: false, alt: "Q ♠")[$"Q" ♠$], #math.equation(block: false, alt: "J ♡")[$"J" ♡$], #math.equation(block: false, alt: "J ♠")[$"J" ♠$]. We’ll define the following events: #math.equation(block: false, alt: "F")[$F$] is “both cards are the same rank,” #math.equation(block: false, alt: "A")[$A$] is “first card is an ace,” and #math.equation(block: false, alt: "K")[$K$] is “first card is a king.” Compute #math.equation(block: false, alt: "P open parenthesis F | A close parenthesis")[$P ( F | A )$] and #math.equation(block: false, alt: "P open parenthesis F | K close parenthesis")[$P ( F | K )$].
+ Jim’s sock drawer contains 5 black socks and 3 blue socks. To avoid waking his partner, Jim doesn’t want to turn the lights on, so he puts on 2 socks at random. Let #math.equation(block: false, alt: "M")[$M$] be the event “Jim’s 2 socks match,” let #math.equation(block: false, alt: "K")[$K$] be the event “the sock on Jim’s left foot is black,” and let #math.equation(block: false, alt: "L")[$L$] be the event “the sock on Jim’s left foot is blue.” Compute #math.equation(block: false, alt: "P open parenthesis M close parenthesis")[$P ( M )$], #math.equation(block: false, alt: "P open parenthesis M | K close parenthesis")[$P ( M | K )$], and #math.equation(block: false, alt: "P open parenthesis M | L close parenthesis")[$P ( M | L )$].


      
      #solutionbox[
+ #strong[Step 1.] The sample space is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. The event #math.equation(block: false, alt: "A")[$A$] consists of the first 4 of those outcomes: HHH, HHT, HTH, and THH. Thus, #math.equation(block: false, alt: "P open parenthesis A close parenthesis equals the fraction 4 over 8 equals the fraction 1 over 2")[$P ( A ) = frac(4, 8) = frac(1, 2)$].

#strong[Step 2.] Now, let’s compute #math.equation(block: false, alt: "P open parenthesis A | H close parenthesis")[$P ( A | H )$]. We are assuming the result of the first flip is heads. That leaves us with 4 possible outcomes: HHH, HHT, HTH, and HTT. Of those, April wins 3 (HHH, HHT, HTH) and loses one (HTT). So, #math.equation(block: false, alt: "P open parenthesis A | H close parenthesis equals the fraction 3 over 4")[$P ( A | H ) = frac(3, 4)$].

#strong[Step 3.] If the result of the first flip is instead tails, the 4 possible outcomes are THH, THT, TTH, and TTT. Of those, April wins 1 (THH) and loses 3 (THT, TTH, TTT). So, #math.equation(block: false, alt: "P open parenthesis A | T close parenthesis equals the fraction 1 over 4")[$P ( A | T ) = frac(1, 4)$].
+ #strong[Step 1.] If the event #math.equation(block: false, alt: "A")[$A$] happens, then 1 of the 4 aces is drawn first; the remaining cards in the deck are 3 aces, 2 kings, 2 queens, and 2 jacks. In order for the event #math.equation(block: false, alt: "F")[$F$] to occur, the second card drawn has to be an ace. Since there are 3 aces among the remaining 9 cards, #math.equation(block: false, alt: "P open parenthesis F | A close parenthesis equals the fraction 3 over 9 equals the fraction 1 over 3")[$P ( F | A ) = frac(3, 9) = frac(1, 3)$].

#strong[Step 2.] If the event #math.equation(block: false, alt: "K")[$K$] happens instead, then the first card drawn is a king. That leaves 4 aces, 1 king, 2 queens, and 2 jacks in the deck. Under the assumption that the first card is a king, the event #math.equation(block: false, alt: "F")[$F$] will occur only if the second card is also a king. Since only one of the remaining 9 cards is a king, we have #math.equation(block: false, alt: "P open parenthesis F | K close parenthesis equals the fraction 1 over 9")[$P ( F | K ) = frac(1, 9)$].
+ #strong[Step 1.] We can view the event #math.equation(block: false, alt: "M")[$M$] as a compound event using “or”: both socks are blue or both socks are black. Let’s compute the probability that both socks are blue using combinations. We’re choosing 2 socks from a group of 8; 3 of the 8 are blue. So, #math.equation(block: false, alt: "P open parenthesis both socks blue close parenthesis equals the fraction sub 3 C sub 2 over sub 8 C sub 2 equals the fraction 3 over 28")[$P ( "both socks blue" ) = frac(3 C_(2), 8 C_(2)) = frac(3, 28)$]. Similarly, #math.equation(block: false, alt: "P open parenthesis both socks black close parenthesis equals the fraction sub 5 C sub 2 over sub 8 C sub 2 equals the fraction 10 over 28")[$P ( "both socks black" ) = frac(5 C_(2), 8 C_(2)) = frac(10, 28)$]. Therefore, since these events are mutually exclusive, we can use the Addition Rule: #math.equation(block: false, alt: "P open parenthesis M close parenthesis equals P open parenthesis both socks blue close parenthesis plus P open parenthesis both socks black close parenthesis equals the fraction 3 over 28 plus the fraction 10 over 28 equals the fraction 13 over 28")[$P ( M ) = P ( "both socks blue" ) + P ( "both socks black" ) = frac(3, 28) + frac(10, 28) = frac(13, 28)$].

#strong[Step 2.] If the sock on Jim’s left foot is black (i.e., #math.equation(block: false, alt: "K")[$K$] occurred), then there are 4 remaining black socks of the 7 in the drawer. So, #math.equation(block: false, alt: "P open parenthesis M | K close parenthesis equals the fraction 4 over 7")[$P ( M | K ) = frac(4, 7)$].

#strong[Step 3.] If the sock on Jim’s left foot is blue (#math.equation(block: false, alt: "L")[$L$] occurred), then there are 2 blue socks among the 7 remaining in the drawer. So, #math.equation(block: false, alt: "P open parenthesis M | L close parenthesis equals the fraction 2 over 7")[$P ( M | L ) = frac(2, 7)$].
]
]


    
    #notebox("Checkpoint", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
#emph[In Tree Diagrams, Tables, and Outcomes, we introduced the concept of dependence between stages of a multistage experiment. We stated at the time that two stages were dependent if the result of one stage affects the other stage. We explained that dependence in terms of the sample space, but sometimes that dependence can be a little more subtle; it’s more properly understood in terms of conditional probabilities. Two stages of an experiment are dependent if] #math.equation(block: false, alt: "P open parenthesis E | F close parenthesis not equal to P open parenthesis E | F ′ close parenthesis")[$P ( E | F ) ≠ P ( E | F ′ )$] #emph[for some outcome of the second stage] #math.equation(block: false, alt: "E")[$E$] #emph[and outcome of the first stage] #math.equation(block: false, alt: "F")[$F$].
]


    #notebox("Who Knew?", rgb("#183B6F"), rgb("#183B6F"), rgb("#EFF1F5"))[
#emph[Protecting Bombers in World War II]


      In his book #emph[How Not to Be Wrong], Jordan Ellenberg recounts this anecdote: During World War II, the American military wanted to add additional armor plating to bomber aircraft, in order to reduce the chances that they get shot down. So, they collected data on planes after returning from missions. The data showed that the fuselage, wings, and fuel system had many more bullet holes (per unit area) than the engine compartments, so the military brass wanted to add additional armor to the parts of the plane that were hit most often. Luckily, before they added the armor to the planes, they asked for a second opinion. Abraham Wald, a Jewish mathematician who had fled the rising Nazi regime, pointed out that it was far more important that the armor plating be added to areas where there were #emph[fewer] bullet holes. Why? The planes they were studying had already completed their missions, so the military was essentially looking at conditional probabilities: the probability of suffering a bullet strike, #emph[given that] the plane made it back safely. More bullet holes in an area on the plane indicated that was a region that wasn’t as important for the plane’s survival!
]


  
  
    
=== Compound Events Using “And” and the Multiplication Rule


    For multistage experiments, the outcomes of the experiment as a whole are often stated in terms of the outcomes of the individual stages. Commonly, those statements are joined with “and.” For example, in the sock drawer example just above, one outcome might be “the left sock is black and the right sock is blue.” As with “or” compound events, these probabilities can be computed with basic arithmetic.


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
Multiplication Rule for Probability: If #math.equation(block: false, alt: "E")[$E$] and #math.equation(block: false, alt: "F")[$F$] are events associated with the first and second stages of an experiment, then #math.equation(block: false, alt: "P open parenthesis E and ⁢ F close parenthesis equals P open parenthesis E close parenthesis times P open parenthesis F | E close parenthesis")[$P ( E #h(0.28em) "and" F ) = P ( E ) × P ( F | E )$].
]


    #notebox("Checkpoint", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
#emph[In The Addition Rule for Probability, we considered probabilities of events connected with “and” in the statement of the Inclusion/Exclusion Principle. These two scenarios are different; in the statement of the Inclusion/Exclusion Principle, the events connected with “and” are both events associated with the #strong[same] single-stage experiment (or the same stage of a multistage experiment). In the Multiplication Rule, we’re looking at events associated with #strong[different] stages of a multistage experiment.]
]


    #examplebox("Example 2")[Using the Multiplication Rule for Probability][
You are president of a club with 10 members: 4 seniors, 3 juniors, 2 sophomores, and 1 first-year. You need to choose 2 members to represent the club on 2 college committees. The first person selected will be on the Club Awards Committee and the second will be on the New Club Orientation Committee. The same person cannot be selected for both. You decide to select these representatives at random.


        + What is the probability that a senior is chosen for both positions?
+ What is the probability that a junior is chosen first and a sophomore is chosen second?
+ What is the probability that a sophomore is chosen first and a senior is chosen second?


      
      #solutionbox[
+ We need the probability that a senior is chosen first and a senior is chosen second. These are two stages of a multistage experiment, so we’ll apply the Multiplication Rule for Probability: #math.equation(block: false, alt: "P open parenthesis senior chosen first and senior chosen second close parenthesis equals P open parenthesis senior chosen first close parenthesis times P open parenthesis senior chosen second | senior chosen first close parenthesis .")[$P ( " senior chosen first and senior chosen second" ) = P ( "senior chosen first" ) × P ( "senior chosen second" | "senior chosen first" ) "."$] Since there are 4 seniors among the 10 members, #math.equation(block: false, alt: "P open parenthesis senior chosen first close parenthesis equals the fraction 4 over 10 equals the fraction 2 over 5")[$P ( "senior chosen first" ) = frac(4, 10) = frac(2, 5)$]. Next, assuming a senior is chosen first, there are 3 seniors among the 9 remaining members. So, #math.equation(block: false, alt: "P open parenthesis senior chosen second | senior chosen first close parenthesis equals the fraction 3 over 9 equals the fraction 1 over 3")[$P ( "senior chosen second" | "senior chosen first" ) = frac(3, 9) = frac(1, 3)$]. Putting this all together, we get #math.equation(block: false, alt: "P open parenthesis senior chosen first and senior chosen second close parenthesis equals the fraction 2 over 5 times the fraction 1 over 3 equals the fraction 2 over 15")[$P ( "senior chosen first and senior chosen second" ) = frac(2, 5) × frac(1, 3) = frac(2, 15)$].
+ There are 3 juniors among the 10 members, so #math.equation(block: false, alt: "P open parenthesis junior chosen first close parenthesis equals the fraction 3 over 10")[$P ( "junior chosen first" ) = frac(3, 10)$]. Assuming a junior is chosen first, there are 2 sophomores among the remaining 9 members, so #math.equation(block: false, alt: "P open parenthesis sophomore chosen second | junior chosen first close parenthesis equals the fraction 2 over 9")[$P ( "sophomore chosen second" | "junior chosen first" ) = frac(2, 9)$]. Thus, using the Multiplication Rule for Probability, we have #math.equation(block: false, alt: "P open parenthesis junior chosen first and sophomore chosen second close parenthesis equals the fraction 3 over 10 times the fraction 2 over 9 equals the fraction 1 over 15")[$P ( "junior chosen first and sophomore chosen second" ) = frac(3, 10) × frac(2, 9) = frac(1, 15)$].
+ The probability that a sophomore is chosen first is #math.equation(block: false, alt: "the fraction 2 over 10 equals the fraction 1 over 5")[$frac(2, 10) = frac(1, 5)$], and the probability that a senior is chosen second given that a sophomore was chosen first is #math.equation(block: false, alt: "the fraction 4 over 9")[$frac(4, 9)$]. Thus, using the Multiplication Rule for Probability, we have: #math.equation(block: false, alt: "P open parenthesis sophomore chosen first and senior chosen second close parenthesis equals the fraction 1 over 5 times the fraction 4 over 9 equals the fraction 4 over 45")[$P ( "sophomore chosen first and senior chosen second" ) = frac(1, 5) × frac(4, 9) = frac(4, 45)$].
]
]


    
    #notebox("Work It Out", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
#emph[The Birthday Problem]


      One of the most famous problems in probability theory is the Birthday Problem, which has to do with shared birthdays in a large group. To make the analysis easier, we’ll ignore leap days, and assume that the probability of being born on any given date is #math.equation(block: false, alt: "the fraction 1 over 365")[$frac(1, 365)$]. Now, if you have 366 people in a room, we’re guaranteed to have at least one pair of people who share a single birthday. Imagine filling the room by first admitting someone born on January 1, then someone born on January 2, and so on… The 365th person admitted would be born on December 31. If you add one more person to the room, that person’s birthday would #emph[have] to match someone else’s.


      Let’s look at the other end of the spectrum. If you choose two people at random, what is the probability that they share a birthday? As with many probability questions, this is best addressed by find out the probability that they do #emph[not] share a birthday. The first person’s birthday can be anything (probability 1), and the second person’s birthday can be anything other than the first person’s birthday (probability #math.equation(block: false, alt: "the fraction 364 over 365")[$frac(364, 365)$]). The probability that they have #emph[different] birthdays is #math.equation(block: false, alt: "1 times the fraction 364 over 365 equals the fraction 364 over 365")[$1 × frac(364, 365) = frac(364, 365)$]. So, the probability that they share a birthday is #math.equation(block: false, alt: "1 minus the fraction 364 over 365 equals the fraction 1 over 365")[$1 − frac(364, 365) = frac(1, 365)$].


      What if we have three people? The probability that they all have different birthdays can be obtained by extending our previous calculation: The probability that two people have different birthdays is #math.equation(block: false, alt: "the fraction 364 over 365")[$frac(364, 365)$], so if we add a third to the mix, the probability that they have a different birthday from the other two is #math.equation(block: false, alt: "the fraction 363 over 365")[$frac(363, 365)$]. So, the probability that all three have different birthdays is #math.equation(block: false, alt: "the fraction 364 over 365 times the fraction 363 over 365 approximately equals 0.9918")[$frac(364, 365) × frac(363, 365) ≈ 0.9918$], and thus the probability that there’s a shared birthday in the group is #math.equation(block: false, alt: "1 minus 0.9918 approximately equals 0.0082")[$1 − 0.9918 ≈ 0.0082$].


      The big question is this: How many people do we need in the room to have the probability of a shared birthday greater than #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]? Make a guess, then with a partner keep adding hypothetical people to the group and computing probabilities until you get there!
]


    It is often useful to combine the rules we’ve seen so far with the techniques we used for finding sample spaces. In particular, trees can be helpful when we want to identify the probabilities of every possible outcome in a multistage experiment. The next example will illustrate this.


    #examplebox("Example 3")[Using Tree Diagrams to Help Find Probabilities][
The board game #emph[Clue] uses a deck of 21 cards: 6 suspects, 6 weapons, and 9 rooms. Suppose you are about to draw 2 cards from this deck. There are 6 possible outcomes for the draw: 2 suspects, 2 weapons, 2 rooms, 1 suspect and 1 weapon, 1 suspect and 1 room, or 1 weapon and 1 room. What are the probabilities for each of these outcomes?


      
      #solutionbox[
#strong[Step 1:] Let’s start by building a tree diagram that illustrates both stages of this experiment. Let’s use S, W, and R to indicate drawing a suspect, weapon, and room, respectively .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((-0.00pt, 0.00pt), (-129.60pt, -54.00pt), stroke: 0.7pt + black)
  line((-0.00pt, 0.00pt), (0.00pt, -54.00pt), stroke: 0.7pt + black)
  line((-0.00pt, 0.00pt), (129.60pt, -54.00pt), stroke: 0.7pt + black)
  line((-129.60pt, -54.00pt), (-172.80pt, -108.00pt), stroke: 0.7pt + black)
  line((-129.60pt, -54.00pt), (-129.60pt, -108.00pt), stroke: 0.7pt + black)
  line((-129.60pt, -54.00pt), (-86.40pt, -108.00pt), stroke: 0.7pt + black)
  line((0.00pt, -54.00pt), (-43.20pt, -108.00pt), stroke: 0.7pt + black)
  line((0.00pt, -54.00pt), (0.00pt, -108.00pt), stroke: 0.7pt + black)
  line((0.00pt, -54.00pt), (43.20pt, -108.00pt), stroke: 0.7pt + black)
  line((129.60pt, -54.00pt), (86.40pt, -108.00pt), stroke: 0.7pt + black)
  line((129.60pt, -54.00pt), (129.60pt, -108.00pt), stroke: 0.7pt + black)
  line((129.60pt, -54.00pt), (172.80pt, -108.00pt), stroke: 0.7pt + black)
  circle((-0.00pt, 0.00pt), radius: 3.00pt, stroke: 0.7pt + black, fill: black)
  circle((-129.60pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-129.60pt, -54.00pt), text(size: 9pt)[S])
  circle((0.00pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((0.00pt, -54.00pt), text(size: 9pt)[W])
  circle((129.60pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((129.60pt, -54.00pt), text(size: 9pt)[R])
  circle((-172.80pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-172.80pt, -108.00pt), text(size: 9pt)[S])
  circle((-129.60pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-129.60pt, -108.00pt), text(size: 9pt)[W])
  circle((-86.40pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-86.40pt, -108.00pt), text(size: 9pt)[R])
  circle((-43.20pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-43.20pt, -108.00pt), text(size: 9pt)[S])
  circle((0.00pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((0.00pt, -108.00pt), text(size: 9pt)[W])
  circle((43.20pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((43.20pt, -108.00pt), text(size: 9pt)[R])
  circle((86.40pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((86.40pt, -108.00pt), text(size: 9pt)[S])
  circle((129.60pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((129.60pt, -108.00pt), text(size: 9pt)[W])
  circle((172.80pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((172.80pt, -108.00pt), text(size: 9pt)[R])
  content((-172.80pt, -127.80pt), anchor: "north", text(size: 9pt)[S S])
  content((-129.60pt, -127.80pt), anchor: "north", text(size: 9pt)[S W])
  content((-86.40pt, -127.80pt), anchor: "north", text(size: 9pt)[S R])
  content((-43.20pt, -127.80pt), anchor: "north", text(size: 9pt)[W S])
  content((0.00pt, -127.80pt), anchor: "north", text(size: 9pt)[W W])
  content((43.20pt, -127.80pt), anchor: "north", text(size: 9pt)[W R])
  content((86.40pt, -127.80pt), anchor: "north", text(size: 9pt)[R S])
  content((129.60pt, -127.80pt), anchor: "north", text(size: 9pt)[R W])
  content((172.80pt, -127.80pt), anchor: "north", text(size: 9pt)[R R])
})]),
  alt: "A tree diagram with three stages. The diagram shows a node in the first stage branching into three nodes labeled S, W, and R in the second stage. The second stage represents the first card. The third stage representing the second card is as follows. The node, S branches into three nodes labeled S, W, and R. The node, W branches into three nodes labeled S, W, and R. The node, R branches into three nodes labeled S, W, and R. The possible outcomes are as follows: S S, S W, S R, W S, W W, W R, R S, R W, and R R.", caption: none,
)


        #strong[Step 2:] We want to start computing probabilities, starting with the first stage. The probability that the first card is a suspect is #math.equation(block: false, alt: "the fraction 6 over 21 equals the fraction 2 over 7")[$frac(6, 21) = frac(2, 7)$]. The probability that the first card is a weapon is the same: #math.equation(block: false, alt: "the fraction 2 over 7")[$frac(2, 7)$]. Finally, the probability that the first card is a room is #math.equation(block: false, alt: "the fraction 9 over 21 equals the fraction 3 over 7")[$frac(9, 21) = frac(3, 7)$].


        #strong[Step 3:] Let’s incorporate those probabilities into our tree: label the edges going into each of the nodes representing the first-stage outcomes with the corresponding probabilities .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((-0.00pt, 0.00pt), (-129.60pt, -54.00pt), stroke: 0.7pt + black)
  content((-71.28pt, -27.00pt), anchor: "east", text(size: 8pt)[2/7])
  line((-0.00pt, 0.00pt), (0.00pt, -54.00pt), stroke: 0.7pt + black)
  content((6.48pt, -27.00pt), anchor: "west", text(size: 8pt)[2/7])
  line((-0.00pt, 0.00pt), (129.60pt, -54.00pt), stroke: 0.7pt + black)
  content((71.28pt, -27.00pt), anchor: "west", text(size: 8pt)[3/7])
  line((-129.60pt, -54.00pt), (-172.80pt, -108.00pt), stroke: 0.7pt + black)
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  line((-129.60pt, -54.00pt), (-129.60pt, -108.00pt), stroke: 0.7pt + black)
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  line((0.00pt, -54.00pt), (-43.20pt, -108.00pt), stroke: 0.7pt + black)
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  line((0.00pt, -54.00pt), (0.00pt, -108.00pt), stroke: 0.7pt + black)
  content((0.00pt, -81.00pt), anchor: "south", text(size: 8pt)[1/3])
  line((0.00pt, -54.00pt), (43.20pt, -108.00pt), stroke: 0.7pt + black)
  content((28.08pt, -81.00pt), anchor: "west", text(size: 8pt)[1/3])
  line((129.60pt, -54.00pt), (86.40pt, -108.00pt), stroke: 0.7pt + black)
  content((101.52pt, -81.00pt), anchor: "east", text(size: 8pt)[1/3])
  line((129.60pt, -54.00pt), (129.60pt, -108.00pt), stroke: 0.7pt + black)
  content((136.08pt, -81.00pt), anchor: "west", text(size: 8pt)[1/3])
  line((129.60pt, -54.00pt), (172.80pt, -108.00pt), stroke: 0.7pt + black)
  content((157.68pt, -81.00pt), anchor: "west", text(size: 8pt)[1/3])
  circle((-0.00pt, 0.00pt), radius: 3.00pt, stroke: 0.7pt + black, fill: black)
  circle((-129.60pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-129.60pt, -54.00pt), text(size: 9pt)[S])
  circle((0.00pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((0.00pt, -54.00pt), text(size: 9pt)[W])
  circle((129.60pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((129.60pt, -54.00pt), text(size: 9pt)[R])
  circle((-172.80pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-172.80pt, -108.00pt), text(size: 9pt)[S])
  circle((-129.60pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-129.60pt, -108.00pt), text(size: 9pt)[W])
  circle((-86.40pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-86.40pt, -108.00pt), text(size: 9pt)[R])
  circle((-43.20pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-43.20pt, -108.00pt), text(size: 9pt)[S])
  circle((0.00pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((0.00pt, -108.00pt), text(size: 9pt)[W])
  circle((43.20pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((43.20pt, -108.00pt), text(size: 9pt)[R])
  circle((86.40pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((86.40pt, -108.00pt), text(size: 9pt)[S])
  circle((129.60pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((129.60pt, -108.00pt), text(size: 9pt)[W])
  circle((172.80pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((172.80pt, -108.00pt), text(size: 9pt)[R])
  content((-172.80pt, -127.80pt), anchor: "north", text(size: 9pt)[SS])
  content((-129.60pt, -127.80pt), anchor: "north", text(size: 9pt)[SW])
  content((-86.40pt, -127.80pt), anchor: "north", text(size: 9pt)[SR])
  content((-43.20pt, -127.80pt), anchor: "north", text(size: 9pt)[WS])
  content((0.00pt, -127.80pt), anchor: "north", text(size: 9pt)[WW])
  content((43.20pt, -127.80pt), anchor: "north", text(size: 9pt)[WR])
  content((86.40pt, -127.80pt), anchor: "north", text(size: 9pt)[RS])
  content((129.60pt, -127.80pt), anchor: "north", text(size: 9pt)[RW])
  content((172.80pt, -127.80pt), anchor: "north", text(size: 9pt)[RR])
})]),
  alt: "A tree diagram with three stages. The diagram shows a node in the first stage branching into three nodes labeled S, W, and R in the second stage with the probabilities, two-sevenths, two-sevenths, and three-sevenths, respectively. The second stage represents the first card. The third stage representing the second card is as follows. The node, S branches into three nodes labeled S, W, and R. The node, W branches into three nodes labeled S, W, and R. The node, R branches into three nodes labeled S, W, and R. The possible outcomes are as follows: S S, S W, S R, W S, W W, W R, R S, R W, and R R.", caption: none,
)


        Note that the sum of the probabilities coming out of the initial node is 1; this should always be the case for the probabilities coming out of any node!


        #strong[Step 4:] Let’s look at the case where the first card is a suspect. There are 3 edges emanating from that node (leading to the outcomes SS, SW, and SR). We’ll label those edges with the appropriate #emph[conditional] probabilities, under the assumption that the first card is a suspect. First, there are 5 remaining suspect cards among the 20 left in the deck, so #math.equation(block: false, alt: "P open parenthesis second is suspect | first is suspect close parenthesis equals the fraction 5 over 20 equals the fraction 1 over 4")[$P ( "second is suspect" | "first is suspect" ) = frac(5, 20) = frac(1, 4)$]. Using similar reasoning, we can compute #math.equation(block: false, alt: "P open parenthesis second is weapon | first is suspect close parenthesis equals the fraction 6 over 20 equals the fraction 3 over 10")[$P ( "second is weapon" | "first is suspect" ) = frac(6, 20) = frac(3, 10)$] and #math.equation(block: false, alt: "P open parenthesis second is room | first is suspect close parenthesis equals the fraction 9 over 20")[$P ( "second is room" | "first is suspect" ) = frac(9, 20)$].


        #strong[Step 5:] Checking our work, we see that the sum of these 3 probabilities is again equal to 1. Let’s add those to our tree .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((-0.00pt, 0.00pt), (-129.60pt, -54.00pt), stroke: 0.7pt + black)
  content((-71.28pt, -27.00pt), anchor: "east", text(size: 8pt)[2/7])
  line((-0.00pt, 0.00pt), (0.00pt, -54.00pt), stroke: 0.7pt + black)
  content((6.48pt, -27.00pt), anchor: "west", text(size: 8pt)[2/7])
  line((-0.00pt, 0.00pt), (129.60pt, -54.00pt), stroke: 0.7pt + black)
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  line((-129.60pt, -54.00pt), (-172.80pt, -108.00pt), stroke: 0.7pt + black)
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  line((-129.60pt, -54.00pt), (-129.60pt, -108.00pt), stroke: 0.7pt + black)
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  line((-129.60pt, -54.00pt), (-86.40pt, -108.00pt), stroke: 0.7pt + black)
  content((-101.52pt, -81.00pt), anchor: "west", text(size: 8pt)[9/20])
  line((0.00pt, -54.00pt), (-43.20pt, -108.00pt), stroke: 0.7pt + black)
  line((0.00pt, -54.00pt), (0.00pt, -108.00pt), stroke: 0.7pt + black)
  line((0.00pt, -54.00pt), (43.20pt, -108.00pt), stroke: 0.7pt + black)
  line((129.60pt, -54.00pt), (86.40pt, -108.00pt), stroke: 0.7pt + black)
  line((129.60pt, -54.00pt), (129.60pt, -108.00pt), stroke: 0.7pt + black)
  line((129.60pt, -54.00pt), (172.80pt, -108.00pt), stroke: 0.7pt + black)
  circle((-0.00pt, 0.00pt), radius: 3.00pt, stroke: 0.7pt + black, fill: black)
  circle((-129.60pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-129.60pt, -54.00pt), text(size: 9pt)[S])
  circle((0.00pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((0.00pt, -54.00pt), text(size: 9pt)[W])
  circle((129.60pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((129.60pt, -54.00pt), text(size: 9pt)[R])
  circle((-172.80pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-172.80pt, -108.00pt), text(size: 9pt)[S])
  circle((-129.60pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-129.60pt, -108.00pt), text(size: 9pt)[W])
  circle((-86.40pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-86.40pt, -108.00pt), text(size: 9pt)[R])
  circle((-43.20pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-43.20pt, -108.00pt), text(size: 9pt)[S])
  circle((0.00pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((0.00pt, -108.00pt), text(size: 9pt)[W])
  circle((43.20pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((43.20pt, -108.00pt), text(size: 9pt)[R])
  circle((86.40pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((86.40pt, -108.00pt), text(size: 9pt)[S])
  circle((129.60pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((129.60pt, -108.00pt), text(size: 9pt)[W])
  circle((172.80pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((172.80pt, -108.00pt), text(size: 9pt)[R])
  content((-172.80pt, -127.80pt), anchor: "north", text(size: 9pt)[SS])
  content((-129.60pt, -127.80pt), anchor: "north", text(size: 9pt)[SW])
  content((-86.40pt, -127.80pt), anchor: "north", text(size: 9pt)[SR])
  content((-43.20pt, -127.80pt), anchor: "north", text(size: 9pt)[WS])
  content((0.00pt, -127.80pt), anchor: "north", text(size: 9pt)[WW])
  content((43.20pt, -127.80pt), anchor: "north", text(size: 9pt)[WR])
  content((86.40pt, -127.80pt), anchor: "north", text(size: 9pt)[RS])
  content((129.60pt, -127.80pt), anchor: "north", text(size: 9pt)[RW])
  content((172.80pt, -127.80pt), anchor: "north", text(size: 9pt)[RR])
})]),
  alt: "A tree diagram with three stages. The diagram shows a node in the first stage branching into three nodes labeled S, W, and R in the second stage with the probabilities, two-sevenths, two-sevenths, and three-sevenths, respectively. The second stage represents the first card. The third stage representing the second card is as follows. Node, S branches into three nodes labeled S, W, and R with the probabilities, one-fourth, three-tenths, and nine-twentieths. Node, W branches into three nodes labeled S, W, and R. The node, R branches into three nodes labeled S, W, and R. The possible outcomes are as follows: S S, S W, S R, W S, W W, W R, R S, R W, and R R.", caption: none,
)


        #strong[Step 6:] Let’s continue filling in the conditional probabilities at the other nodes, always checking to make sure the sum of the probabilities coming out of any node is equal to 1 .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((-0.00pt, 0.00pt), (-129.60pt, -54.00pt), stroke: 0.7pt + black)
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  line((-0.00pt, 0.00pt), (0.00pt, -54.00pt), stroke: 0.7pt + black)
  content((6.48pt, -27.00pt), anchor: "west", text(size: 8pt)[2/7])
  line((-0.00pt, 0.00pt), (129.60pt, -54.00pt), stroke: 0.7pt + black)
  content((71.28pt, -27.00pt), anchor: "west", text(size: 8pt)[3/7])
  line((-129.60pt, -54.00pt), (-172.80pt, -108.00pt), stroke: 0.7pt + black)
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  line((-129.60pt, -54.00pt), (-129.60pt, -108.00pt), stroke: 0.7pt + black)
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  line((-129.60pt, -54.00pt), (-86.40pt, -108.00pt), stroke: 0.7pt + black)
  content((-101.52pt, -81.00pt), anchor: "west", text(size: 8pt)[9/20])
  line((0.00pt, -54.00pt), (-43.20pt, -108.00pt), stroke: 0.7pt + black)
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  line((0.00pt, -54.00pt), (0.00pt, -108.00pt), stroke: 0.7pt + black)
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  line((0.00pt, -54.00pt), (43.20pt, -108.00pt), stroke: 0.7pt + black)
  content((28.08pt, -81.00pt), anchor: "west", text(size: 8pt)[9/20])
  line((129.60pt, -54.00pt), (86.40pt, -108.00pt), stroke: 0.7pt + black)
  content((101.52pt, -81.00pt), anchor: "east", text(size: 8pt)[3/10])
  line((129.60pt, -54.00pt), (129.60pt, -108.00pt), stroke: 0.7pt + black)
  content((136.08pt, -81.00pt), anchor: "west", text(size: 8pt)[3/10])
  line((129.60pt, -54.00pt), (172.80pt, -108.00pt), stroke: 0.7pt + black)
  content((157.68pt, -81.00pt), anchor: "west", text(size: 8pt)[2/5])
  circle((-0.00pt, 0.00pt), radius: 3.00pt, stroke: 0.7pt + black, fill: black)
  circle((-129.60pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-129.60pt, -54.00pt), text(size: 9pt)[S])
  circle((0.00pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((0.00pt, -54.00pt), text(size: 9pt)[W])
  circle((129.60pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((129.60pt, -54.00pt), text(size: 9pt)[R])
  circle((-172.80pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-172.80pt, -108.00pt), text(size: 9pt)[S])
  circle((-129.60pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-129.60pt, -108.00pt), text(size: 9pt)[W])
  circle((-86.40pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-86.40pt, -108.00pt), text(size: 9pt)[R])
  circle((-43.20pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-43.20pt, -108.00pt), text(size: 9pt)[S])
  circle((0.00pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((0.00pt, -108.00pt), text(size: 9pt)[W])
  circle((43.20pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((43.20pt, -108.00pt), text(size: 9pt)[R])
  circle((86.40pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((86.40pt, -108.00pt), text(size: 9pt)[S])
  circle((129.60pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((129.60pt, -108.00pt), text(size: 9pt)[W])
  circle((172.80pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((172.80pt, -108.00pt), text(size: 9pt)[R])
  content((-172.80pt, -127.80pt), anchor: "north", text(size: 9pt)[SS])
  content((-129.60pt, -127.80pt), anchor: "north", text(size: 9pt)[SW])
  content((-86.40pt, -127.80pt), anchor: "north", text(size: 9pt)[SR])
  content((-43.20pt, -127.80pt), anchor: "north", text(size: 9pt)[WS])
  content((0.00pt, -127.80pt), anchor: "north", text(size: 9pt)[WW])
  content((43.20pt, -127.80pt), anchor: "north", text(size: 9pt)[WR])
  content((86.40pt, -127.80pt), anchor: "north", text(size: 9pt)[RS])
  content((129.60pt, -127.80pt), anchor: "north", text(size: 9pt)[RW])
  content((172.80pt, -127.80pt), anchor: "north", text(size: 9pt)[RR])
})]),
  alt: "A tree diagram with three stages. The diagram shows a node in the first stage branching into three nodes labeled S, W, and R in the second stage with the probabilities, two-sevenths, two-sevenths, and three-sevenths, respectively. The second stage represents the first card. The third stage representing the second card is as follows. Node, S branches into three nodes labeled S, W, and R with the probabilities, one-fourth, three-tenths, and nine-twentieths. Node, W branches into three nodes labeled S, W, and R with the probabilities, three-tenths, one-fourth, and nine-twentieths. The node, R branches into three nodes labeled S, W, and R with the probabilities, three-tenths, three-tenths, and two-fifths. The possible outcomes are as follows: S S, S W, S R, W S, W W, W R, R S, R W, and R R.", caption: none,
)


        #strong[Step 7:] We can compute the probability of landing on any final node by multiplying the probabilities along the path we would take to get there. For example, the probability of drawing a suspect first and a weapon second (i.e., ending up on the node labeled “SW”) is #math.equation(block: false, alt: "the fraction 2 over 7 times the fraction 3 over 10 equals the fraction 3 over 35")[$frac(2, 7) × frac(3, 10) = frac(3, 35)$], as illustrated in .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((-0.00pt, 0.00pt), (-129.60pt, -54.00pt), stroke: 0.7pt + black)
  content((-71.28pt, -27.00pt), anchor: "east", text(size: 8pt)[2/7])
  line((-0.00pt, 0.00pt), (0.00pt, -54.00pt), stroke: 0.7pt + black)
  content((6.48pt, -27.00pt), anchor: "west", text(size: 8pt)[2/7])
  line((-0.00pt, 0.00pt), (129.60pt, -54.00pt), stroke: 0.7pt + black)
  content((71.28pt, -27.00pt), anchor: "west", text(size: 8pt)[3/7])
  line((-129.60pt, -54.00pt), (-172.80pt, -108.00pt), stroke: 0.7pt + black)
  content((-157.68pt, -81.00pt), anchor: "east", text(size: 8pt)[1/4])
  line((-129.60pt, -54.00pt), (-129.60pt, -108.00pt), stroke: (paint: rgb("#D62128"), thickness: 1.6pt))
  content((-129.60pt, -81.00pt), anchor: "south", text(size: 8pt)[3/10])
  line((-129.60pt, -54.00pt), (-86.40pt, -108.00pt), stroke: 0.7pt + black)
  content((-101.52pt, -81.00pt), anchor: "west", text(size: 8pt)[9/20])
  line((0.00pt, -54.00pt), (-43.20pt, -108.00pt), stroke: 0.7pt + black)
  content((-28.08pt, -81.00pt), anchor: "east", text(size: 8pt)[3/10])
  line((0.00pt, -54.00pt), (0.00pt, -108.00pt), stroke: 0.7pt + black)
  content((0.00pt, -81.00pt), anchor: "south", text(size: 8pt)[1/4])
  line((0.00pt, -54.00pt), (43.20pt, -108.00pt), stroke: 0.7pt + black)
  content((28.08pt, -81.00pt), anchor: "west", text(size: 8pt)[9/20])
  line((129.60pt, -54.00pt), (86.40pt, -108.00pt), stroke: 0.7pt + black)
  content((101.52pt, -81.00pt), anchor: "east", text(size: 8pt)[3/10])
  line((129.60pt, -54.00pt), (129.60pt, -108.00pt), stroke: 0.7pt + black)
  content((136.08pt, -81.00pt), anchor: "west", text(size: 8pt)[3/10])
  line((129.60pt, -54.00pt), (172.80pt, -108.00pt), stroke: 0.7pt + black)
  content((157.68pt, -81.00pt), anchor: "west", text(size: 8pt)[2/5])
  circle((-0.00pt, 0.00pt), radius: 3.00pt, stroke: 0.7pt + black, fill: black)
  circle((-129.60pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-129.60pt, -54.00pt), text(size: 9pt)[S])
  circle((0.00pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((0.00pt, -54.00pt), text(size: 9pt)[W])
  circle((129.60pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((129.60pt, -54.00pt), text(size: 9pt)[R])
  circle((-172.80pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-172.80pt, -108.00pt), text(size: 9pt)[S])
  circle((-129.60pt, -108.00pt), radius: 9.00pt, stroke: (paint: rgb("#D62128"), thickness: 1.4pt), fill: white)
  content((-129.60pt, -108.00pt), text(size: 9pt)[W])
  circle((-86.40pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-86.40pt, -108.00pt), text(size: 9pt)[R])
  circle((-43.20pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-43.20pt, -108.00pt), text(size: 9pt)[S])
  circle((0.00pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((0.00pt, -108.00pt), text(size: 9pt)[W])
  circle((43.20pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((43.20pt, -108.00pt), text(size: 9pt)[R])
  circle((86.40pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((86.40pt, -108.00pt), text(size: 9pt)[S])
  circle((129.60pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((129.60pt, -108.00pt), text(size: 9pt)[W])
  circle((172.80pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((172.80pt, -108.00pt), text(size: 9pt)[R])
  content((-172.80pt, -127.80pt), anchor: "north", text(size: 9pt)[SS])
  content((-129.60pt, -127.80pt), anchor: "north", text(size: 9pt)[SW])
  content((-86.40pt, -127.80pt), anchor: "north", text(size: 9pt)[SR])
  content((-43.20pt, -127.80pt), anchor: "north", text(size: 9pt)[WS])
  content((0.00pt, -127.80pt), anchor: "north", text(size: 9pt)[WW])
  content((43.20pt, -127.80pt), anchor: "north", text(size: 9pt)[WR])
  content((86.40pt, -127.80pt), anchor: "north", text(size: 9pt)[RS])
  content((129.60pt, -127.80pt), anchor: "north", text(size: 9pt)[RW])
  content((172.80pt, -127.80pt), anchor: "north", text(size: 9pt)[RR])
})]),
  alt: "A tree diagram with three stages. The diagram shows a node in the first stage branching into three nodes labeled S, W, and R in the second stage with the probabilities, two-sevenths, two-sevenths, and three-sevenths, respectively. The second stage represents the first card. The third stage representing the second card is as follows. Node, S branches into three nodes labeled S, W, and R with the probabilities, one-fourth, three-tenths, and nine-twentieths. Node, W branches into three nodes labeled S, W, and R with the probabilities three-tenths, one-fourth, and nine-twentieths. The node, R branches into three nodes labeled S, W, and R with the probabilities three-tenths, three-tenths, and two-fifths. The possible outcomes are as follows: S S, S W, S R, W S, W W, W R, R S, R W, and R R. The path from the start node through the nodes, S and W is highlighted. The probability for the outcome, S W reads, P of S W equals 2 over 7 times 3 over 10 equals 3 over 35.", caption: none,
)


        #strong[Step 8:] Let’s fill in the rest of the probabilities .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((-0.00pt, 0.00pt), (-129.60pt, -54.00pt), stroke: 0.7pt + black)
  content((-71.28pt, -27.00pt), anchor: "east", text(size: 8pt)[2/7])
  line((-0.00pt, 0.00pt), (0.00pt, -54.00pt), stroke: 0.7pt + black)
  content((6.48pt, -27.00pt), anchor: "west", text(size: 8pt)[2/7])
  line((-0.00pt, 0.00pt), (129.60pt, -54.00pt), stroke: 0.7pt + black)
  content((71.28pt, -27.00pt), anchor: "west", text(size: 8pt)[3/7])
  line((-129.60pt, -54.00pt), (-172.80pt, -108.00pt), stroke: 0.7pt + black)
  content((-157.68pt, -81.00pt), anchor: "east", text(size: 8pt)[1/4])
  line((-129.60pt, -54.00pt), (-129.60pt, -108.00pt), stroke: 0.7pt + black)
  content((-129.60pt, -81.00pt), anchor: "south", text(size: 8pt)[3/10])
  line((-129.60pt, -54.00pt), (-86.40pt, -108.00pt), stroke: 0.7pt + black)
  content((-101.52pt, -81.00pt), anchor: "west", text(size: 8pt)[9/20])
  line((0.00pt, -54.00pt), (-43.20pt, -108.00pt), stroke: 0.7pt + black)
  content((-28.08pt, -81.00pt), anchor: "east", text(size: 8pt)[3/10])
  line((0.00pt, -54.00pt), (0.00pt, -108.00pt), stroke: 0.7pt + black)
  content((0.00pt, -81.00pt), anchor: "south", text(size: 8pt)[1/4])
  line((0.00pt, -54.00pt), (43.20pt, -108.00pt), stroke: 0.7pt + black)
  content((28.08pt, -81.00pt), anchor: "west", text(size: 8pt)[9/20])
  line((129.60pt, -54.00pt), (86.40pt, -108.00pt), stroke: 0.7pt + black)
  content((101.52pt, -81.00pt), anchor: "east", text(size: 8pt)[3/10])
  line((129.60pt, -54.00pt), (129.60pt, -108.00pt), stroke: 0.7pt + black)
  content((136.08pt, -81.00pt), anchor: "west", text(size: 8pt)[3/10])
  line((129.60pt, -54.00pt), (172.80pt, -108.00pt), stroke: 0.7pt + black)
  content((157.68pt, -81.00pt), anchor: "west", text(size: 8pt)[2/5])
  circle((-0.00pt, 0.00pt), radius: 3.00pt, stroke: 0.7pt + black, fill: black)
  circle((-129.60pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-129.60pt, -54.00pt), text(size: 9pt)[S])
  circle((0.00pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((0.00pt, -54.00pt), text(size: 9pt)[W])
  circle((129.60pt, -54.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((129.60pt, -54.00pt), text(size: 9pt)[R])
  circle((-172.80pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-172.80pt, -108.00pt), text(size: 9pt)[S])
  circle((-129.60pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-129.60pt, -108.00pt), text(size: 9pt)[W])
  circle((-86.40pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-86.40pt, -108.00pt), text(size: 9pt)[R])
  circle((-43.20pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((-43.20pt, -108.00pt), text(size: 9pt)[S])
  circle((0.00pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((0.00pt, -108.00pt), text(size: 9pt)[W])
  circle((43.20pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((43.20pt, -108.00pt), text(size: 9pt)[R])
  circle((86.40pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((86.40pt, -108.00pt), text(size: 9pt)[S])
  circle((129.60pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((129.60pt, -108.00pt), text(size: 9pt)[W])
  circle((172.80pt, -108.00pt), radius: 9.00pt, stroke: 0.7pt + black, fill: white)
  content((172.80pt, -108.00pt), text(size: 9pt)[R])
  content((-172.80pt, -127.80pt), anchor: "north", text(size: 9pt)[SS])
  content((-129.60pt, -127.80pt), anchor: "north", text(size: 9pt)[SW])
  content((-86.40pt, -127.80pt), anchor: "north", text(size: 9pt)[SR])
  content((-43.20pt, -127.80pt), anchor: "north", text(size: 9pt)[WS])
  content((0.00pt, -127.80pt), anchor: "north", text(size: 9pt)[WW])
  content((43.20pt, -127.80pt), anchor: "north", text(size: 9pt)[WR])
  content((86.40pt, -127.80pt), anchor: "north", text(size: 9pt)[RS])
  content((129.60pt, -127.80pt), anchor: "north", text(size: 9pt)[RW])
  content((172.80pt, -127.80pt), anchor: "north", text(size: 9pt)[RR])
})]),
  alt: "A tree diagram with three stages. The diagram shows a node in the first stage branching into three nodes labeled S, W, and R in the second stage with the probabilities two-sevenths, two-sevenths, and three-sevenths, respectively. The second stage represents the first card. The third stage representing the second card is as follows. Node, S branches into three nodes labeled S, W, and R with the probabilities, one-fourth, three-tenths, and nine-twentieths. Node, W branches into three nodes labeled S, W, and R with the probabilities three-tenths, one-fourth, and nine-twentieths. The node, R branches into three nodes labeled S, W, and R with the probabilities three-tenths, three-tenths, and two-fifths. The possible outcomes are as follows: S S, S W, S R, W S, W W, W R, R S, R W, and R R. The probabilities for the outcomes are as follows. P of S S equals 2 over 7 times 1 over 4 equals 1 over 14. P of S W equals 2 over 7 times 3 over 10 equals 3 over 35. P of S R equals 2 over 7 times 9 over 20 equals 9 over 70. P of W S equals 2 over 7 times 3 over 10 equals 3 over 35. P of W W equals 2 over 7 times 1 over 4 equals 1 over 14. P of W R equals 2 over 7 times 9 over 20 equals 9 over 70. P of R S equals 3 over 7 times 3 over 10 equals 9 over 70. P of R W equals 3 over 7 times 3 over 10 equals 9 over 70. P of R R equals 3 over 7 times 2 over 5 equals 6 over 35.", caption: none,
)


        #strong[Step 9:] A helpful feature of tree diagrams is that the final outcomes are always mutually exclusive, so the Addition Rule can be directly applied. For example, the probability of drawing one suspect and one room (in any order) would be #math.equation(block: false, alt: "P open parenthesis S R close parenthesis plus P open parenthesis R S close parenthesis equals the fraction 9 over 70 plus the fraction 9 over 70 equals the fraction 9 over 35")[$P ( S R ) + P ( R S ) = frac(9, 70) + frac(9, 70) = frac(9, 35)$]. We can find the probabilities of the other outcomes in a similar fashion, as shown in the following table:


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Outcome], [Probability]),
  [2 suspects], [#math.equation(block: false, alt: "the fraction 1 over 14")[$frac(1, 14)$]],
  [2 weapons], [#math.equation(block: false, alt: "the fraction 1 over 14")[$frac(1, 14)$]],
  [2 rooms], [#math.equation(block: false, alt: "the fraction 6 over 35")[$frac(6, 35)$]],
  [1 suspect and 1 weapon], [#math.equation(block: false, alt: "the fraction 3 over 35 plus the fraction 3 over 35 equals the fraction 6 over 35")[$frac(3, 35) + frac(3, 35) = frac(6, 35)$]],
  [1 suspect and 1 room], [#math.equation(block: false, alt: "the fraction 9 over 70 plus the fraction 9 over 70 equals the fraction 9 over 35")[$frac(9, 70) + frac(9, 70) = frac(9, 35)$]],
  [1 weapon and 1 room], [#math.equation(block: false, alt: "the fraction 9 over 70 plus the fraction 9 over 70 equals the fraction 9 over 35")[$frac(9, 70) + frac(9, 70) = frac(9, 35)$]],
))


        Checking once again, the sum of these 6 probabilities is 1, as expected.
]
]


    
    #notebox("Work It Out", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
#emph[The Monty Hall Problem]


      On the original version of the game show #emph[Let’s Make a Deal], originally hosted by Monty Hall and now hosted by Wayne Brady, one contestant was chosen to play a game for the grand prize of the day (often a car). Here’s how it worked: On the stage were three areas concealed by numbered curtains. The car was hidden behind one of the curtains; the other two curtains hid worthless prizes (called “Zonks” on the show). The contestant would guess which curtain concealed the car. To build tension, Monty would then reveal what was behind one of the #emph[other] curtains, which was always one of the Zonks (Since Monty knew where the car was hidden, he always had at least one Zonk curtain that hadn’t been chosen that he could reveal). Monty then turned to the contestant and asked: “Do you want to stick with your original choice, or do you want to switch your choice to the other curtain?” What should the contestant do? Does it matter?


      With a partner or in a small group, simulate this game. You can do that with a small candy (the prize) hidden under one of three cups, or with three playing cards (just decide ahead of time which card represents the “Grand Prize”). One person plays the host, who knows where the prize is hidden. Another person plays the contestant and tries to guess where the prize is hidden. After the guess is made, the host should reveal a losing option that wasn’t chosen by the contestant. The contestant then has the option to stick with the original choice or switch to the other, unrevealed option. Play about 20 rounds, taking turns in each role and making sure that both contestant strategies (stick or switch) are used equally often. After each round, make a note of whether the contestant chose “stick” or “switch” and whether the contestant won or lost. Find the empirical probability of winning under each strategy. Then, see if you can use tree diagrams to verify your findings.
]


  
  
  
  
    
=== Key Terms


    - conditional probability


  
  
    
=== Key Concepts


    - Conditional probabilities are computed under the assumption that the condition has already occurred.
- The Multiplication Rule for Probability is used to find the probability that two events occur in sequence.


  
  
    
=== Formulas


    - If #math.equation(block: false, alt: "E")[$E$] and #math.equation(block: false, alt: "F")[$F$] are events associated with the first and second stages of an experiment, then #math.equation(block: false, alt: "P open parenthesis E and ⁢ F close parenthesis equals P open parenthesis E close parenthesis times P open parenthesis F | E close parenthesis")[$P ( E #h(0.28em) "and" F ) = P ( E ) × P ( F | E )$].