4.5: At Least One Rule and Tree Diagrams
- Page ID
- 49080
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- Use tree diagrams as visual tools to map all possible outcomes step-by-step.
- Organize and compute complex probabilities effectively with these methods.
- Solve multi-step probability problems and reduce the risk of missing outcomes.
Assume there is a 25% chance that people will favor a new education policy proposed by a state government. If 5 people are randomly selected, what is the chance that at least one of the 5 will favor the policy? To directly compute this probability by looking at all possible variations of one or more factors favoring the policy would be cumbersome. It turns out that a quick way to calculate this type of probability is to compute the chance that none of them favors the policy and then subtract it from 1. This method is a general rule that can be applied to a wider range of applications. The rule is defined below.
At Least One Rule for Probabilities
The probability of "at least one" refers to the likelihood that an event occurs at least once in multiple trials. It means that the event happens one or more times, as opposed to not happening at all.
P(at least one time) = 1 – P(none)
Where,
- P(none) is the probability that the event does not happen in any trial.
- By subtracting this from 1, we find the probability that the event occurs at least once.
In the example described at the beginning of this section, selecting a small sample from a large population will make the events independent. Thus, to compute the probability, the complement rule must be used to calculate the probability that one person does not favor the policy. This is P(not in favor) = 1 – 0.25 = 0.75. The chance that all five are not in favor is computed using the probability of the complement and the multiplication rule. P(All three are not in favor) = 0.75 \(\cdot\) 0.75 \(\cdot\) 0.75 \(\cdot\) 0.75 \(\cdot\) 0.75 = 0.237. Finally, to compute the probability that at least one favors the probability, use the formula. P(at least one favors the policy) = 1 – 0.237 = 0.763. Thus, the answer is 0.763.
The chance that a person is a "night owl" is around 12%. Night owls are those people who believe they are more productive in the late hours of the night. If a sample of 3 people from the population is selected, compute the following probabilities.
- All three are night owls.
- None of the three is a night owl.
- At least one of the three is a night owl.
Solutions
- These events are independent. We can use the multiplication rule to compute the answer. P(all three are night owls) = 0.12 \(\cdot\) 0.12 \(\cdot\) 0.12 = 0.0017
- The complement must be computed first. It is P(not a night owl) = 1 – 0.12 = 0.88. The answer is P(all three are not night owls) = 0.88 \(\cdot\) 0.88 \(\cdot\) 0.88 = 0.681
- We use at least one rule and the result of part 2 to get the answer. P(At least one is a night owl) = 1 – 0.681 = 0.319.
For a standard deck of cards, assume five cards are selected without replacement. Compute the probability of getting at least one spade.
Solution
Since the cards are selected without replacement, the events are dependent. That means for each selection, a card must be removed, and this must be reflected in the probability. To use at least one rule, compute the probability that five cards are not spades. There are 13 spades and 39 cards that are not spades. The answer will be rounded to a decimal with three place places.
P(none is a spade) = \(\dfrac{39}{52}\) \(\cdot\) \(\dfrac{38}{51}\) \(\cdot\) \(\dfrac{37}{50}\) \(\cdot\) \(\dfrac{36}{49}\)\(\cdot\) \(\dfrac{35}{48}\)= 0.222
The computation of the final answer is provided below.
P(at least one is a spade) = 1 – 0.222 = 0.778
Some probability problems may be extremely complex and have multiple outcomes. A tree diagram can be constructed for these problems. The tree diagram is used to help map out all the possibilities for the sequence of events in the problem. The events are represented by dots called nodes, and the outcomes of each event are represented by the branches or lines of the diagram. The first step is to construct the tree diagram. The second step is to assign probabilities to each branch in the tree diagram. The third step requires multiplying all probabilities along each pathway of the tree diagram that leads to the desired result. The final step is to add all the probabilities of the different pathways that resulted in the desired outcome.
A coin is flipped two times. Compute the chance of getting exactly two tails.
Solution
- First, construct the tree diagram.
Figure \(\PageIndex{1}\) Tree Diagram
- Second, assign probabilities to each branch. In this case, each branch will have a probability of 1/2 or 0.5 because there are only two possibilities per toss.
Figure \(\PageIndex{2}\) Tree Diagram with Probabilities
- Third, follow each branch that leads to the desired outcome of two tails and multiply probabilities along the pathway.
Figure \(\PageIndex{3}\) Tree Diagram with Desired Pathway
P( two tails) = 0.5 \(\cdot\) 0.5 = 0.25
Exercises
A literature professor randomly selects 2 books from a collection of 10 books:
- 4 are poetry books.
- 6 are novels.
The books are selected without replacement.
What is the probability that at least one of the selected books is a poetry book?
- Answer
-
Use the "at least one" formula to compute the probability. Compute the probability that none of the books is a poetry book.
P(none is a poetry book) = \(\dfrac{6}{10} \cdot \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}\)
P(at least one of the selected books is a poetry book ) = 1 – P(none is a poetry book) = \(1 - \dfrac{1}{3} = \dfrac{2}{3}\)
A standard deck of 52 playing cards is used. Two cards are drawn without replacement.
What is the probability that at least one of the two cards drawn is a heart?
- Answer
-
Use the "at least one" formula to compute the probability. Compute the probability that none of the books is a heart.
P(none is a heart) = \(\dfrac{39}{52} \cdot \dfrac{38}{51} = \dfrac{19}{34}\)
P(at least one of the cards is a heart) = 1 – P(none is a heart) = \(1 - \dfrac{19}{34} = \dfrac{15}{34}\)
A student takes 3 multiple-choice quiz questions. Each question has 4 answer choices, and the student randomly guesses on each question. Each question is independent. What is the probability that the student gets at least one question correct?
- Answer
-
First, calculate the probability of guessing correctly. It is p(correct) = \(\dfrac{1}{4}\)
Second, calculate the probability of guessing incorrectly. It is p (incorrect) = \(1 - \dfrac{1}{4}=\dfrac{3}{4}\)
Third, calculate the probability that all three are guessed incorrectly. It is \(P(\text{all three are incorrect}) = \left( \dfrac{3}{4} \right)^3 = \dfrac{27}{64} \)
Finally, calculate the probability that at least one problem is guessed correctly.
P(at least one guessed correctly) = 1 – P(none is correct) = \(1 - \dfrac{27}{64} = \dfrac{37}{64}\)
A sociologist is studying college students and whether they participate in community service and student clubs. At a local college, 60% of students participate in community service. Of the students who participate in community service, 70% also belong to a student club. Of the students who do not participate in community service, 40% belong to a student club. Create a tree diagram and use it to answer the following questions:
- What is the probability that a student participates in community service and belongs to a student club?
- What is the probability that a student does not participate in community service and does not belong to a student club?
- What is the probability that a student belongs to a student club?
- Answer
-
Figure \(\PageIndex{4}\): Tree Diagram of Student Involvement and Club Participation: This tree diagram models student involvement in community service and student clubs. It begins by splitting students into those who participate in community service (60%) and those who do not (40%). Each group then branches into students who belong to a club and those who do not, with corresponding probabilities shown at each stage. The diagram illustrates how to multiply along branches to find joint probabilities and combine outcomes to determine the overall probability that a student belongs to a club.
The answers are provided below using the tree diagram and multiplying adjacent probabilities.
- P(community service and student club) = \(0.6 \cdot 0.4 = 0.42\)
- P(community service and not student club) = \(0.6 \cdot 0.3 = 0.18\)
- P(Student Club) = \(0.6 \cdot 0.4 +0.4 \cdot 0.4= 0.42+0.16=0.58\)
Author
"4.5: At Least One Rule and Tree Diagrams" by Alfie Swan is licensed under CC BY 4.0