3.5 Tree and venn diagrams

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Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities.

Tree diagrams

A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.

In an urn, there are 11 balls. Three balls are red ( R ) and eight balls are blue ( B ). Draw two balls, one at a time, with replacement . "With replacement" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.

The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R 1, R 2, and R 3 and each blue ball as B 1, B 2, B 3, B 4, B 5, B 6, B 7, and B 8. Then the nine RR outcomes can be written as:

• R 1 R 1
• R 1 R 2
• R 1 R 3
• R 2 R 1
• R 2 R 2
• R 2 R 3
• R 3 R 1
• R 3 R 2
• R 3 R 3

The other outcomes are similar.

There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space .

a. List the 24 BR outcomes: B 1 R 1, B 1 R 2, B 1 R 3, ...

a.

• B 1 R 1
• B 1 R 2
• B 1 R 3
• B 2 R 1
• B 2 R 2
• B 2 R 3
• B 3 R 1
• B 3 R 2
• B 3 R 3
• B 4 R 1
• B 4 R 2
• B 4 R 3
• B 5 R 1
• B 5 R 2
• B 5 R 3
• B 6 R 1
• B 6 R 2
• B 6 R 3
• B 7 R 1
• B 7 R 2
• B 7 R 3
• B 8 R 1
• B 8 R 2
• B 8 R 3

b. Using the tree diagram, calculate P ( RR ).

b. P ( RR ) = $\left(\frac{3}{11}\right)\left(\frac{3}{11}\right)$ = $\frac{9}{121}$

c. Using the tree diagram, calculate P ( RB OR BR ).

c. P ( RB OR BR ) = $\left(\frac{3}{11}\right)\left(\frac{8}{11}\right)$ + $\left(\frac{8}{11}\right)\left(\frac{3}{11}\right)$ = $\frac{48}{121}$

d. Using the tree diagram, calculate P ( R on 1st draw AND B on 2nd draw).

d. P ( R on 1st draw AND B on 2nd draw) = P ( RB ) = $\left(\frac{3}{11}\right)\left(\frac{8}{11}\right)$ = $\frac{24}{121}$

e. Using the tree diagram, calculate P ( R on 2nd draw GIVEN B on 1st draw).

e. P ( R on 2nd draw GIVEN B on 1st draw) = P ( R on 2nd| B on 1st) = $\frac{24}{88}$ = $\frac{3}{11}$

This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB ). Twenty-four of the 88 possible outcomes are BR . $\frac{24}{88}$ = $\frac{3}{11}$ .

f. Using the tree diagram, calculate P ( BB ).

f. P ( BB ) =  $\frac{64}{121}$

g. Using the tree diagram, calculate P ( B on the 2nd draw given R on the first draw).

g. P ( B  on 2nd draw| R  on 1st draw) =  $\frac{8}{11}$

There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB ). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then $\frac{24}{33}$ .

Try it

In a standard deck, there are 52 cards. 12 cards are face cards (event F ) and 40 cards are not face cards (event N ). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P ( FF ).

Total number of outcomes is 144 + 480 + 480 + 1600 = 2,704.

P ( FF ) =

An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. "Without replacement" means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, $\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}$ .

Note

If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement , so that on the second draw there are ten marbles left in the urn.

Calculate the following probabilities using the tree diagram.

a. P ( RR ) = ________

a. P ( RR ) = $\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}$

b. Fill in the blanks:

P ( RB OR BR ) =

b. P ( RB OR BR ) = $\left(\frac{3}{11}\right)\left(\frac{8}{10}\right)$ + $\left(\frac{8}{11}\right)\left(\frac{3}{10}\right)$ = $\frac{48}{110}$

c. P ( R on 2nd| B on 1st) =

c. P ( R on 2nd| B on 1st) = $\frac{3}{10}$

d. Fill in the blanks.

P ( R on 1st AND B on 2nd) = P ( RB ) = (___)(___) = $\frac{24}{100}$

d. P ( R on 1st AND B on 2nd) = P ( RB ) = $\left(\frac{3}{11}\right)\left(\frac{8}{10}\right)$ = $\frac{24}{100}$

e. Find P ( BB ).

e. P ( BB ) = $\left(\frac{8}{11}\right)\left(\frac{7}{10}\right)$

f. Find P ( B on 2nd| R on 1st).

f. Using the tree diagram, P ( B on 2nd| R on 1st) = P ( R | B ) = $\frac{8}{10}$ .

If we are using probabilities, we can label the tree in the following general way.

• P ( R | R ) here means P ( R on 2nd| R on 1st)
• P ( B | R ) here means P ( B on 2nd| R on 1st)
• P ( R | B ) here means P ( R on 2nd| B on 1st)
• P ( B | B ) here means P ( B on 2nd| B on 1st)

Example of discrete variable
Gbenga
hi
Kachalla
what's up here ... am new here
Kachalla
sorry question a bit unclear...do you mean how do you analyze quantitative data? If yes, it depends on the specific question(s) you set in the beginning as well as on the data you collected. So the method of data analysis will be dependent on the data collecter and questions asked.
Bheka
how to solve for degree of freedom
saliou
Quantitative data is the data in numeric form. For eg: Income of persons asked is 10,000. This data is quantitative data on the other hand data collected for either make or female is qualitative data.
Rohan
*male
Rohan
Degree of freedom is the unconditionality. For example if you have total number of observations n, and you have to calculate variance, obviously you will need mean for that. Here mean is a condition, without which you cannot calculate variance. Therefore degree of freedom for variance will be n-1.
Rohan
data that is best presented in categories like haircolor, food taste (good, bad, fair, terrible) constitutes qualitative data
Bheka
vegetation types (grasslands, forests etc) qualitative data
Bheka
I don't understand how you solved it can you teach me
solve what?
Ambo
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lower and upper endpoints
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Sarbaz
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phoenix
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Caleb
Assume you are in a class where quizzes are 20% of your grade, homework is 20%, exam _1 is 15%,exam _2 is 15%, and the final exam is 20%.Suppose you are in the fifth week and you just found out that you scored a 58/63 on the fist exam. You also know that you received 6/9,8/10,9/9 on the first
quizzes as well as a 9/11,10/10,and 4.5/7 on the first three homework assignment. what is your current grade in the course?
Diamatu
Abdul
if putting y=3x examine that correlation coefficient between x and y=3x is 1.
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You have to plot the class midpoint and the frequency
Wydny
ok so you use those two to draw the histogram right.
Amford
yes
Wydny
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Amford
how do you calculate cost effectiveness?
George
Hi everyone, this is a very good statistical group and am glad to be part of it. I'm just not sure how did I end up here cos this discussion just popes on my screen so if I wanna ask something in the future, how will I find you?
Bheka
To make a histogram, follow these steps: On the vertical axis, place frequencies. Label this axis "Frequency". On the horizontal axis, place the lower value of each interval. ... Draw a bar extending from the lower value of each interval to the lower value of the next interval.
Divya
I really appreciate that
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yes of course must have use f test and also use t test individually multple coefficients
rishi
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umar
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akeem
Is R square cannot analysis linear regression of X vs Y relationship?
Mok
To be an economist you have to be professional in maths
umar
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Shehu
what is random sampling what is sample error
@Nistha Kashyap Random sampling is the selection of random items (or random numbers) from the group. A sample error occurs when the selected samples do not truely represent the whole group. The can happen when most or all of the selected samples are taken from only one section of the group;
Ron
Thus the sample is not truely random.
Ron
What is zero sum game?
A game in which there is no profit & no loss to any of the both player.
Milan
Differences between sample mean & population mean
***keydifferences.com/difference-between-sample-mean-and-population-mean.html
Lucien
Not difference in the formula except the notation, sample mean is denoted by x bar and population mean is denoted by mu symbol. There is formula as well as notation between difference variance and standard deviations
Akash
Likely the difference would be in the result, unless the sample is an exact representation of the population (which is unlikely.)
Ron
what is data
Nii
Nii Avin - Data is just a simple way to refer to the numbers in the population, or in the sample used in your calculations.
Ron
what are the types of data
Nii
Data is the very pale android from the Star Trek Enterprise
Andrew
Am Emmanuel from Nigeria
Emmanuel
Am Qudus from Nigeria
Rasak
am Handson from Cameroon
Handson
what is a mode?
Handson
Nii - data is whatever you are sampling. Such as the number of students in each classroom.
Ron
Handson Ndintek - the mode is the number appearing most frequently. Example: 7 9 11 7 4 6 3 7 2. 7 is the mode. In a group such as 7 9 1 4 6 3, there is no mode because no number appears more often than any other.
Ron
hi I want to know how to find class boundary
Baalisi
give me the two types of data
qualitative and quantitative
phoenix
primary and secondary data
Peace
qualitative and quantitative
Prince