4.5: At Least One Rule and Tree Diagrams
- Page ID
- 49080
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Apply the "at least one" rule to calculate the probability that an event occurs one or more times, often by first finding the probability that it never occurs.
- 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
Author
"4.5: At Least One Rule and Tree Diagrams" by Alfie Swan is licensed under CC BY 4.0