<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are viewing an older version of this Concept. Go to the latest version.

# Tree Diagrams

## Multiply probabilities along the branches and add probabilities in columns

%
Progress
Practice Tree Diagrams
Progress
%
Tree Diagrams and Probability Distributions

In any given day of the week, there is a 62% likelihood that it will be cloudy in Seattle. What is the probability that there will be less than 3 cloudy days in Seattle during your one-week visit?

### Guidance

Sometimes it is useful to create a tree diagram to illustrate the possible outcomes of multiple events and their individual probabilities, calculate the probabilities of the combined events and the sample space. In other cases, we might use combinations or permutations to create a Probability Distribution table. A Probability Distribution is a table which includes all possible outcomes (sample space) and their respective probabilities.

#### Example A

A game of chance involves flipping a coin and selecting a chip from one of two urns. If the coin toss results in heads, then you select from urn A which contains 8 yellow chips and 5 green chips. If the coin toss results in tails, then you select from urn B which contains 6 yellow chips and 6 green chips. Use this information to create a tree diagram illustrating the possible outcomes and their probabilities and then determine the probability of selecting a yellow chip.

Solution: First, we need to make a tree diagram. The first branches of the diagram show the coin toss results and the second sets of branches show the chip selection results. Notice that each set of branches in the tree diagram has probabilities which sum to 1. This happens because one of the outcomes must occur. In other words, you either select a yellow or a green chip (there is no other outcome in the sample space) so the sum of the probabilities will be 1. These types of events are called Complimentary Events .

By multiplying “across” the branches, we can determine the probabilities of the combined events. Now, look at the sum of the probabilities on the far right: $\frac{4}{13}+\frac{5}{26}+\frac{1}{4}+\frac{1}{4}=1$ . The entire sample space is shown here so the sum of the probabilities of all the possible outcomes should be 1. This is an excellent way to check for accuracy in your tree diagram calculations.

Now, to answer the question: What is the probability of selecting a yellow chip? Looking at the diagram, there are two ways to select a yellow chip. One, we could toss the coin and get heads and then select a yellow chip from urn A and this probability is $\frac{4}{13}$ . Two, we could toss the coin and get tails and then select a yellow chip and this probability is $\frac{1}{4}$ . We can add the probabilities of these two “paths” to the same end result and get $\frac{4}{13}+\frac{1}{4}=\frac{29}{52} \thickapprox 0.5577$ .

#### Example B

In a box of 20 candies, there are 8 which contain nuts. If 5 pieces are randomly selected and consumed, create a probability distribution table to show the probability of selecting 0, 1, 2, 3, 4, or 5 candies which contain nuts in the sample.

Solution: First, let’s create a formula for determining the probability of each of the outcomes. We can use combinations to help us do this. First, how many ways are there to select 5 pieces of candy from a box of 20 pieces? This is a combination, so $_{20}C_5$ or $\dbinom{20}{5}$ . This value will be the total number of possible outcomes and thus the denominator of our probability ratio. Now, how many ways are there to select 0 of the 8 candies with nut and 5 of the 12 candies without nuts? Again, we have combinations and their product can be found for the numerator of our probability ratio: $\dbinom{8}{0} \dbinom{12}{5}$ . Now we can put it all together and find the probability of selecting 0 candies with nuts: $\frac{\dbinom{8}{0} \dbinom{12}{5}}{\dbinom{20}{5}}=\frac{33}{646} \thickapprox 0.05108$ .

Similarly, for 1 candy containing nuts: $\frac{\dbinom{8}{1} \dbinom{12}{4}}{\dbinom{20}{5}}=\frac{165}{646} \thickapprox 0.25542$ .

For 2 candies with nuts we get: $\frac{\dbinom{8}{2} \dbinom{12}{3}}{\dbinom{20}{5}}=\frac{385}{969} \thickapprox 0.39732$ , and so on.

The table to the right shows all of the final probabilities for each outcome in the sample space. This is called a Probability Distribution Table.

What happens if we add up all of the probabilities in this table?

$0.05108 + 0.25542 + 0.39732 + 0.23839 + 0.05418 + 0.00361 = 1$

This means that the probability of getting one of these outcomes is 100%. Also, this shows that our probability distribution is correct because we have included all of the possible outcomes and the sum of their probabilities is 1. In other words, this illustrates that there are no other possible outcomes since there is a 100% chance of getting one of these results.

Number of Candies Selected Containing Nuts Probability
0 $\frac{33}{646} \thickapprox 0.05108$
1 $\frac{165}{646} \thickapprox 0.25542$
2 $\frac{385}{969} \thickapprox 0.39732$
3 $\frac{77}{323} \thickapprox 0.23839$
4 $\frac{35}{646} \thickapprox 0.05418$
5 $\frac{7}{1938} \thickapprox 0.00361$

#### Example C

Over time, Ronald has shown that in 2 of 5 attempts he makes a bulls eye with a bow and arrow. Create a probability distribution table which shows the possible outcomes and the associated probabilities when Ronald shoots three arrows.

Solution: First, we will consider each bulls eye a success and each non bulls eye a failure. So, the probability of a success is $\frac{2}{5}$ and the probability of a failure is $\frac{3}{5}$ . The probability of zero successes is $\left(\frac{3}{5}\right)^3=\frac{27}{125}=0.216$ . Similarly, the probability of three successes is $\left(\frac{2}{5}\right)^3=\frac{8}{125}=0.064$ . With three shots, there are two other possibilities to consider. Ronald could also have one success and two failures or two successes and one failure. In these cases, we must consider that any one or two of the shots could be successes so we will multiply by the number of “combinations” that are possible. For example, if Ronald has one success, then there are $\dbinom{3}{1}$ or 3 ways this could occur: SFF, FSF or FFS. So we will multiple the combinations by the probability of one success and two failures: $\tbinom{3}{1} \left(\frac{2}{5}\right)^1 \left(\frac{3}{5}\right)^2=\frac{54}{125}=0.432$ . For two successes and one failure we have $\tbinom{3}{2} \left(\frac{2}{5}\right)^2 \left(\frac{3}{5}\right)^1=\frac{36}{125}=0.288$ . Now, make a probability distribution table:

Number of Bulls Eyes Probability
0 0.216
1 0.432
2 0.288
3 0.064

This is an example of a special type of probability called a Binomial Probability because its rule resembles the Binomial Theorem. In order for a problem to be a binomial probability it must consist of multiple independent trials, called Bernoulli Trials, in which there is either a success or a failure. In other words, $P(\text{success}) + P(\text{failure}) = 1$ and the result of each trial is independent of the result of a previous trial.

If we let $n = \text{number of trials}$ , $p = \text{probability of a success}$ and $r = \text{number of successes}$ , we can use the following formula to determine the probability of any number of successes.

$P( r \ \text{successes}) = \dbinom{n}{r}(p)^r(1-p)^{n-r}.$

Notice that this formula is exactly what we did to find the probability of Ronald shooting two bulls eyes:

For two bulls eyes, $n=3$ , $p=\frac{2}{5}$ , and $r=2$ : $\dbinom{3}{2} \left(\frac{2}{5}\right)^2 \left(1- \frac{2}{5}\right)^{3-2}=\dbinom{3}{2} \left(\frac{2}{5}\right)^2 \left(\frac{3}{5}\right)^1=\frac{36}{125}=0.288$ .

Intro Problem Revisit Less than seven cloudy days implies that there could be 0, 1, or 2 cloudy days during your one-week visit. We can add these individual probabilities to determine the probability. For the Binomial Probability, $n=7$ (since there are 7 days in a week), $p=0.62$ and $r$ takes on the values 0, 1 and 2.

$\dbinom{7}{0}(0.62)^0(0.38)^7+\dbinom{7}{1}(0.62)^1(0.38)^6+ \dbinom{7}{2}(0.62)^2(0.38)^5=0.0.001144 + 0.01306774 + 0.06396902568 = 0.07818076568$

Therefore, there is only about a 7.8% chance that less than 3 days of your trip will be cloudy. Better make sure to pack an umbrella!

### Guided Practice

1. Sarah either walks or rides the bus to school. When she walks she is more likely to be late to school than when she rides the bus. Complete the tree diagram and find the probability that Sarah is late for school.

2. In a case of 15 light bulbs there are 2 defective bulbs. Create a probability distribution which illustrates the possible outcomes and their respective probabilities if we randomly select 3 bulbs from the box. Show that the sum of the probabilities is 1.

3. On any given workday, there is a 15% chance that Professor Calculus will cause an explosion in his laboratory. Use the Binomial Probability formula to determine the probability that Professor Calculus will cause less than three explosions in a five day work week.

1.

To find the probability that Sarah is late we need to add the probabilities of the two different ways she can be late. She can walk and be late or she can ride the bus and be late:

$P(\text{walk and late}) + P(\text{bus and late}) =\left(\frac{2}{3}\right) \left(\frac{1}{10}\right)+ \left(\frac{1}{3}\right) \left(\frac{1}{20}\right)=\frac{1}{12} \thickapprox 0.08333$

2.

Number of defectiteve bulbs Probability
0 $\frac{\dbinom{2}{0} \dbinom{13}{3}}{\dbinom{15}{3}}=0.628571$
1 $\frac{\dbinom{2}{1} \dbinom{13}{2}}{\dbinom{15}{3}}=0.342857$
2 $\frac{\dbinom{2}{2} \dbinom{13}{1}}{\dbinom{15}{3}}=0.028571$

$0.62857 + 0.342857 + 0.028571 = 0.999998$

Since we rounded the individual probabilities, the sum may reflect this inaccuracy. For all intents and purposes, this answer is 1.

Note that it is not possible to select a sample containing 3 defective bulbs because there are only 2 defective bulbs in the box.

3. Less than three explosions implies that Professor Calculus could cause 0, 1 or 2 explosions in the work week. We can add these individual probabilities to determine the probability. For the Binomial Probability, $n=5$ (since there are 5 days in the work week), $p=0.15$ and $r$ takes on the values 0, 1 and 2.

$\dbinom{5}{0}(0.15)^0(0.85)^5+\dbinom{5}{1}(0.15)^1(0.85)^4+ \dbinom{5}{2}(0.15)^2(0.85)^3=0.443705 + 0.391505 + 0.138178 = 0.973388$

### Vocabulary

Probability Distribution
A summary, usually a table, of the possible outcomes of an experiment or series of events and all the corresponding probabilities.
Complimentary Events
A pair of events such that if one event does not occur the other one will. As a result, the sum of their probabilities must be equal to 1.

### Practice

Jamie and Olivia are best friends and neighbors. As such they often eat dinner together at one of their houses. About 30% of the time, they eat at Jamie’s house and her mother makes a non vegetarian meal 65% of the time. The rest of the time, they eat at Olivia’s house and her mother serves a vegetarian meal 55% of the time.

1. Make a tree diagram, including the appropriate probabilities to illustrate this scenario.
2. What is the probability of going to Olivia’s house and eating a meal containing meat?
3. What is the probability that the girls will eat a vegetarian dinner?

Tommy has 20 water balloons in bucket. There is a $\frac{1}{8}$ chance of each balloon exploding on him before he even has a chance to throw it. Use a Binomial Probability Distribution to answer the following questions.

1. What is the probability that Tommy throws eight without any breaking on him?
2. What is the probability that exactly five of the eight balloons he attempts to throw explode on him?
3. What is the probability that more than two of the eight explode on him?

A vaccine has a 92% success rate. The vaccine is given to 50 patients in a medical practice. Use a Binomial Probability Distribution to answer the following questions.

1. What is the probability that it works for all patients?
2. What is the probability that is fails for exactly 9 patients?
3. What is the probability that it fails for more than 1 patient?

Five cards are randomly selected from a deck of cards.

1. Create a probability distribution table for the number of “high” cards (jack, queen, king or ace) in a 5 card hand chosen at random.
2. What is the probability of getting at least one high card?
3. What is the probability of getting at least two?

### Vocabulary Language: English

If events A and B are mutually inclusive, then P(A or B) = P(A) + P(B) – P(A and B)
Multiplication Rule

Multiplication Rule

States that for 2 events (A and B), the probability of A and B is given by: P(A and B) = P(A) x P(B).
tree diagrams

tree diagrams

Tree diagrams are a way to show the outcomes of simple probability events where each outcome is represented as a branch on a tree.