M43 Picturing Probabilities - JaPipeDiagrams

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Picturing probabilities

The idea for this problem set comes fromRepresenting Probabil-ities with Pipe Diagrams By Clifford Konold (May 1996). The

Mathematics Teacher, vol. 89:5, 378-382. NCTM. Alexandria,Virginia.

1. A farmer is irrigating her fields and releases 1000 cubic This problem does not involveprobability, but it provides us witha useful visual metaphor andtechnique that will help us answerprobability questions.

meters of water into an irrigation channel. Upon enteringthe field, the water flow divides with 70% flowing into achannel to the left and 30% going into a smaller channelto the right. Extend the diagram below to show how muchwater will be in each channel if the two smaller channelssplit once more, again with 70% flowing to the left and30% to the right. Label each channel with the amount of

water flowing through them and draw their width to matchtheir capacity (note that the channels already drawn havea width proportional to their amounts).

2. A class is playing an elimination game. Each student picks

a number from 1 to 6 and the teacher rolls a die. The stu-dents who picked the number that is rolled have to sitdown. Everyone left standing picks a number again andplays another round. This situation can be representedwith a channel diagram as well. In this case, however, thechannels do not show the route that water has taken.Instead, they show the probability that a student has fol-

Problems with a Point: December 23, 2002 c EDC 2002

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Picturing probabilities: Problem 2

lowed a particular path. The following diagram showsan initial channel with the student standing. The widthis one, because the probability of a student beginning thegame is 100%. After the teacher rolls the die, the channel

splits to show the probability that the student will have tosit down, 1

6, or remain standing, 5

6.

(a) Extend the diagram to show the probabilities after a Note that any vertical linethrough your channel diagramshould intersect channels whose

total width is 1 (all possible pathsup to that point have beencounted).

second roll of the die and then after a third roll. Label

each channel with the probability of reaching that sit-uation (that is, with the fraction of water that wouldreach that channel) and make the width of that chan-nel proportional to the probability. Be sure to show allstudents at each stage regardless of whether they aresitting or standing.

(b) What is the probability that a student will be sittingby the end of two rolls?

(c) Extending channel diagrams out several levels can bedifficult. However, imagining what the extended dia-gram would look like and what calculations you would

have to make without actually drawing the diagrambeyond a certain point can be helpful. Picture a di-agram showing the probabilities for the possible out-comes showing the first six rolls. What is the prob-ability that a student is still standing after six rolls?What is the probability that they are sitting by the

Problems with a Point: December 23, 2002 c EDC 2002

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Picturing probabilities: Problem 3

end of the sixth roll?

3. Make a channel diagram for the possible outcomes of threeconsecutive flips of an unbiased coin (heads and tails areequally likely). Each flip of the coin is an event and should

correspond to a branching of the channels. What are theprobabilities of getting heads all three times? Two times?Just once? No times?

4. Recent research has demonstrated that teenagers are par-ticularly susceptible to nicotine addiction (see Tobacco Ad-diction Found to Be Nearly Immediate at www.umassmed.edu/pap/news/2002/08 29 02.cfm). Studies have docu-mented changes in brain chemistry after a teenager smokesonly a single cigarette and have shown that heavy smokingis a response to, and not a cause of, addiction. Assume

that the probability of a teenager becoming addicted tosmoking is 8% each time he or she smokes a cigarette.

(a) Make a channel diagram that shows the likilihood ofaddiction for the first four times a teenager smokes.

(b) How many times would a teen need to smoke before Once they become a smoker, theaverage teen takes 18 years tosuccessfully quit.

there was an 80% chance that he or she had becomeaddicted to nicotine? (Hint: The answer is not 10.Can you use your diagram to explain why not?)

Problems with a Point: December 23, 2002 c EDC 2002

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Picturing probabilities: Solutions 1

Solutions

1. The channels will receive 700(0.7) = 490 cubic meters, 700(0.3) =210 cubic meters, 300(0.7) = 210 cubic meters, and 300(0.3) =90 cubic meters respectively. Since the original channel is10 boxes high for 1000 cubic meters, each box represents100 cubic meters and the smaller channels are drawn withwidths reflecting that scale.

2.(a) The top channel will split and the student will get to

Problems with a Point: December 23, 2002 c EDC 2002

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Picturing probabilities: Solutions 2

remain standing 56

of the time. So, 56

of the water in thechannel that is 5

6wide, or 25

36of the water, will flow into

the channel. 2536

is the probability that the student willstill be standing after two rolls of the die. 1

6 times 5

6,

or 536 , of the time the student will have to sit after the

second roll. The lower channel does not branch further,because the student who is already sitting does not getto play.

(b) We can find the probability that a student will be sittingafter two rolls in two ways. We can add the widths (prob-abilities) of the channels representing a sitting student,or we can subtract the probability that a student is stillstanding from 1:

16

+ 536

= 1 2536

= 1136

.(c) Each roll of the die shrinks the width of the channel that

represents the probability that a student is still standingby 5

6. So, the probability of surviving 6 rolls is 5

6 times

itself 6 times, or 56

6, which is approximately 33.5% of the

time. The probability that a student is sitting by the endof the sixth roll is the complement of this result or 66.5%.

3. Since the questions only ask about the cumulative results ofthe three flips, there are two different diagrams that servethe purpose. The first shows each branching and the eightdistinct and equally likely sequences of heads and tails thatcan arise. Each time a channel branches, we multiply by theprobability of getting the next result (0.5). Therefore, theprobability of getting any particular sequence is the same,(12

)(12

)(12

) = 18

.

Problems with a Point: December 23, 2002 c EDC 2002

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Picturing probabilities: Solutions 3

The second version keeps track of the number of heads ortails that have appeared up through the most recent flip. Itkeeps the number of branches to a minimum by merging theflow of channels that represent the same circumstance.

4. This problem is similar to problem 2. Both problems onlyinvolve two outcomes, standing and sitting or not addictedand addicted, and one of the outcomes is permanent. Thepermanent outcomes, sitting or being addicted, are unaf-

Problems with a Point: December 23, 2002 c EDC 2002

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Picturing probabilities: Solutions 4

fected by further events and accumulate greater probabili-ties over time.

(a) Again, there are two possible channel diagrams. Onehas a structure like the solution presented for problem

2a. The other takes advantage of the limited numberof outcomes and is more like the second diagram in thesolution to problem 3. Below is an example of the secondtype. The precise proportions have not been maintainedbecause of the numbers involved (which have also beenrounded to two places). After each cigarette, 92% of theunaddicted channel remains unaddicted. The remaining8% of that channel becomes addicted and accumulatesin the addicted channel. The values for the unaddictedchannel are the powers of 0.92. The bottom channel isone minus these values.

(b) We see from our diagram that the probability of beingaddicted after smoking ncigarettes is 1 0.92n. Solving0.8 = 1 0.92n yields 0.2 = 0.92n. We can solve for nbymaking a table of powers of 0.92 and seeing when we getbelow 0.2 or by using logarithms: n= log 0.2

log 0.92 19.3. So

there will be an 80% chance of addiction after the twen-tieth cigarette. It was not correct to just divide 8% into80% to get 10 cigarettes, because subsequent cigarettesled to addiction in 8% of the already diminished non-addicted channel. In a population of beginning smokers,some will already have become addicted and so you arefinding 8% of a smaller group.

Problems with a Point: December 23, 2002 c EDC 2002