This section covers:
 Average, Mean, Median, Mode, Range and Mean Absolute Deviation
 Box and Whisker Plot
 Stem and Leaf Plot
 Frequency Tables and Graphs
 Pie Chart
 Probability
 More Practice
Note that we’ll cover Scatter Plots, Correlation, and Regression here.
Somewhere in the prealgebra stages, you’ll get a short, but fun introduction to Statistics, where you’ll cover such topics as finding the average and median of numbers, the median of numbers, organizing data (numbers) with different types of graphs, and performing analyses on data. Typically, this type of mathematics falls under the realm of Statistics, or the study of the collection, organization, and presentation of data.
You may also be introduced to Probability, which is the study of how likely events are to occur, or happen.
Let’s first talk about some types of measurement you’ll see in Statistics.
Average, Mean, Median, Mode, Range and Mean Absolute Deviation (MAD)
Let’s say you go around and ask your friends to keep track of how much time your friends spend each week doing their homework. Fortunately, you have a lot of friends online that you can ask quite easily. 20 of your friends reply with the following numbers:
\(10, 2, 15, 28, 1, 32, 12, 14, 8, 17, 22, 6, 42, 3, 14, 7, 12, 23, 20, 8\) hours of studying each week
(I see that you have fairly studious friends!)
Mean (Average)
Just by glancing at the numbers, can you guess the average? Probably not. But it’s easy to get the average: you just add all the numbers up and divide by the total number of responses:
Average hours per week of your 20 friends =
\(\frac{{\text{1}0+\text{2}+\text{15}+\text{28}+\text{1}+\text{32}+\text{12}+\text{14}+\text{8}+\text{17}+\text{22}+\text{6}+\text{42}+\text{3}+\text{14}+\text{7}+\text{12}+\text{23}+\text{2}0+8}}{{20}}\,\,=\frac{{296}}{{20}}\,\,=\,\,14.8\)
Another word for the average is the mean.
Now let’s do some other fun stuff with this data. What if someone asked you for a number that is exactly in the middle of the data; in other words, the number that has just as many answers above it as below it. Let’s sort the data (put it in order) to get this number:
\(1\,\,\,\,\,2\,\,\,\,\,3\,\,\,\,\,6\,\,\,\,\,7\,\,\,\,\,8\,\,\,\,\,8\,\,\,\,\,\,10\,\,\,\,\,\,12\,\,\,\,\,12\,\,\,\,\,14\,\,\,\,\,14\,\,\,\,\,15\,\,\,\,\,17\,\,\,\,\,20\,\,\,\,\,22\,\,\,\,\,23\,\,\,\,\,28\,\,\,\,\,32\,\,\,\,\,42\)
Median
To get the middle number or median, cross out numbers from both ends until you arrive at the middle number or numbers. Since we have an even number, we’ll get two “middles”, so to get the median, we’ll have to take the mean or average of those numbers:
\(\require{cancel} \displaystyle \begin{array}{l}\xcancel{{1\,\,\,\,\,2\,\,\,\,\,3\,\,\,\,\,6\,\,\,\,\,7\,\,\,\,\,8\,\,\,\,\,\,8\,\,\,\,\,\,10\,\,\,\,\,\,12}}\,\,\,\,\,\,\,\,\,\,\,\left[\!\left[ {12\,\,\,\,\,\,14} \right]\!\right]\,\,\,\,\,\,\,\xcancel{{14\,\,\,\,\,15\,\,\,\,\,17\,\,\,\,\,20\,\,\,\,\,22\,\,\,\,\,23\,\,\,\,\,28\,\,\,\,\,32\,\,\,\,\,42\,}}\\\\\,\,\,\text{Start crossing out from this end }\triangleright \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\triangleleft \,\,\,\text{Cross out the same number from this side}\end{array}\)
We end up with two numbers in the middle: 12 and 14. If we just had one number, that number would be the median, but since we have two numbers, we take the average of 12 and 14, which is \(\displaystyle \frac{{12+14}}{2}=13\); the median is 13. Remember that the word “median” sounds like the word “middle”.
Mode
The mode is the number or numbers that occur most often; you can have more than one mode. In the case of our data, the modes are 8, 12, and 14, all of which occur twice in the data. Remember that the word “mode” sounds like the word “most” (often).
Range
The range, which is the difference between the largest number and smallest number, is 41 (\(421\)).
Mean Absolute Deviation (MAD)
Let’s find one more statistic that may be helpful to know. The mean absolute deviation (MAD) of a set of numbers tells us on average, how far our numbers are from the middle (mean) of the numbers. This measures the variability (dispersion) of the numbers. The mean absolute deviation is also called the average absolute deviation.
In our situation, this might give us an indication of how similar the study habits are with our friends; in other words, the higher the mean absolute deviation, the greater the spread or variability of the students’ study habits. Note that such a statistic is probably most useful when comparing different data sets.
To get this, we get the mean of the data (which we already have: 14.8). Then we take each value and find the distance (absolute value of the difference) from this mean, and then take the average of them. For example, the first value (from the ordered values) is \(\left {114.8} \right=13.8\).
\(\displaystyle \begin{array}{c}\frac{{13.8+1\text{2}\text{.8}+\text{11}\text{.8}+\text{8}\text{.8}+\text{7}\text{.8}+\text{6}\text{.8}+\text{6}\text{.8}+\text{4}\text{.8}+\text{2}\text{.8}+\text{2}\text{.8}+\text{.8}+\text{.8}+\text{.2}+\text{2}\text{.2}+\text{5}\text{.2}+\text{7}\text{.2}+\text{8}\text{.2}+\text{13}\text{.2}+17.2+27.2}}{{20}}\,\,\\=\frac{{162.2}}{{20}}\,\,=\,\,8.11\end{array}\)
Then the mean absolute deviation is 8.11 hours. What this means is among the friends, there is a lot of variability (dispersion) among homework time each week.
Later, we’ll see how to use a similar dispersion measure, the standard deviation (which is more complicated to measure but more often used), and also how to get a lot of these measurements in the graphing calculator! These can be found in the Scatter Plots, Correlation, and Regression section.
Box and Whisker Plot
There are a couple different ways of viewing data graphically that you’ll learn in your math classes. First, a box and whisker plot is a visual picture of the data that shows where the middle of the data is (the median), and how far away from the middle the other points lie.
To draw what we call a box and whisker plot, or box plot, we also need to get the lower and upper quartiles, which are like the medians of the numbers to the left and right of the median, including the median. (In our case, we include the 12 since we didn’t have a true middle number, or median). To get the lower quartile, let’s start crossing out again:
\(\displaystyle \xcancel{{1\,\,\,\,\,2\,\,\,\,\,3\,\,\,\,\,6\,\,\,\,\,}}\,\,\,\,\,\,\,\,\left[\!\left[ {7\,\,\,\,\,\,8} \right]\!\right]\,\,\,\,\,\,\,\xcancel{{8\,\,\,\,\,10\,\,\,\,\,12\,\,\,\,\,12}}\)
We are left with 7 and 8 in the middle, and the mean is 7.5. So the lower quartile is 7.5.
We do the same to get the upper quartile, but use the numbers to the right of the median.
\(\displaystyle \xcancel{{14\,\,\,\,\,14\,\,\,\,\,15\,\,\,\,\,17\,}}\,\,\,\,\,\,\,\,\left[\!\left[ {20\,\,\,\,\,\,22} \right]\!\right]\,\,\,\,\,\,\,\xcancel{{23\,\,\,\,\,28\,\,\,\,\,32\,\,\,\,\,42}}\)
We are left with 20 and 22 in the middle, and the mean is 21. So the upper quartile is 21.
Note that the word “quartile” is related to the word “quarter”; we have divided up the data into four quarters.
Then we plot the lowest number, lower quartile, median, upper quartile, and highest number like this; see how it looks like a box with whiskers on each side?
Can you see that \(\displaystyle \frac{1}{4}\) of your 20 friends (5 friends) study less than 7.5 hours per week, \(\displaystyle \frac{1}{2}\) (10 friends) less than 13 hours, and \(\displaystyle \frac{3}{4}\) (15 friends) less than 21 hours a week? You can also see that \(\displaystyle \frac{1}{2}\) of your friends (10 friends) study between 7.5 hours (the lower quartile) and 21 hours (the upper quartile) per week. You can also see that the highest point (42 hours) is somewhat of an outlier; this means that this point may not fit in with the rest of the data.
(Note that we could have used our graphing calculator to derive some of these values, such as the median and quartiles. See Basic Stats on Data from Calculator in the Scatter Plots, Correlation, and Regression section to see how to do this).
Stem and Leaf Plot
We could also draw a stem and leaf graph with the data, which resembles plant stems and leaves:
For this plot, you put the first digit (the tens) of all the numbers on the lefthand side, and then put the ones on the righthand side, in order. Here you can see the same thing – since we know (from earlier) that the median is 7.5, we can see that there is a larger difference between the median and the largest number (42) than the median and the smallest number (1).
We can also see how the smaller data is more clumped together; we’ll see this next in Frequency Tables and Graphs.
Frequency Tables and Graphs
We could also draw what we call a frequency table that shows us how many of your friends studied less than or equal to 10 hours per week, 11 to 20 hours per week, 21 to 30 hours per week, 31 to 40 hours per week and 41 to 50 hours per week. These are called “buckets” or “classes” and each class has 10 hours in it (0 to 10 hours, and so on). Notice that our buckets of data are a little different than the stem and leaf table above:
Frequency Table 

Number of hours studying 
Number of your friends in that “bucket” 
0 – 10  8 
11 – 20  7 
21 – 30  3 
31 – 40  1 
41 – 50  1 
Total  20 
Then we could draw a histogram from this data, as shown below. A histogram, or a frequency graph is a graph showing the distribution of the data – where it lies.
Or even better, we can draw a relative frequency histogram, where we divide the number of friends in each “bucket” or “class” by the total number of friends, as shown below. The reason this is a better graph is that if you add up all the amounts for each bucket, we’ll get a grand total of 1 (all the decimals on the left add up to 1), so we can compare different sets of data together.
Relative Frequency Table 

Number of hours studying 
\(\displaystyle \frac{{\mathbf{Number}\text{ }\mathbf{of}\text{ }\mathbf{your}\text{ }\mathbf{friends}\text{ }\mathbf{in}\text{ }\mathbf{that}\text{ }\mathbf{bucket}\text{ }}}{{\mathbf{total}\text{ }\mathbf{number}\text{ }\mathbf{of}\text{ }\mathbf{friends}}}\) 
0 – 10  .40 (or 40%) 
11 – 20  .35 (or 35 %) 
21 – 30  .15 (or 15%) 
31 – 40  .05 (or 5%) 
41 – 50  .05 (or 5%) 
Total  1 (or 100%) 
This data is skewed to the right, or positively skewed (the righthand side has a longer “tail”). When data is skewed to the left (or negatively skewed), the left side has a longer “tail”. When data isn’t skewed left or right, we call the data “symmetric”.
When the data is skewed right or left, the median is a better measure of the central tendency (average) for the data, since the mean could be misleading. The mean tends to go out in the “tail”. Here are some examples of other data that shows this (means and medians may not be accurate – just giving an idea):
Histogram  Observation 
When the data is symmetric, the mean and median are typically close together.  
When the data is skewed left (or negatively skewed), it is “pulled” to the left, and this drags the mean too low.  
When the data is skewed right (or positively skewed), it is “pulled” to the right, and this drags the mean too high. 
Pie Chart
One more type of graph that’s fun to draw is a circle graph, or pie chart. We could divide up the “buckets” from our data above (numbers of hours our friends are studying per week), and compute how big to make the pieces of the pie with the use of a little bit of Geometry.
Since we know the percentages of friends who fall into each category from the relative frequency chart above, we can get the angle measurements of each piece of the pie (from the center of the pie) by using proportions and the fact that there are 360 degrees in a circle. By using proportions (for example, \(\displaystyle \frac{{40}}{{100}}=\frac{?}{{360}}\)), we find that we can just multiply the relative frequency by 360 degrees to get each angle measurement. Here’s the table again, with the degrees for each bucket:
Relative Frequency Table with Angle Measurements 

Number of hours studying 
\(\displaystyle \frac{{\text{Number of your friends in that bucket }}}{{\text{total number of friends}}}\) 
Angle measurement (degrees) 
0 – 10  .40 (or 40%)  .4 x 360 = 144 
11 – 20  .35 (or 35 %)  .35 x 360 = 126 
21 – 30  .15 (or 15%)  .15 x 360 = 54 
31 – 40  .05 (or 5%)  .05 x 360 = 18 
41 – 50  .05 (or 5%)  .05 x 360 = 18 
Total  1 (or 100%)  360 
So (using our handydandy protractor that we learned how to use in Elementary School Geometry, and will see later in our Geometry section) – here is our pie chart:
Don’t worry if you don’t totally get all this now! Later on in more advanced Algebra we’ll learn even more ways to display and interpret data (like when we compare two sets of data), and also how to make a lot of these displays on your graphing calculator.
Probability
Before Algebra, you may also have studied a topic called Probability, which is related to Statistics.
Basically, a probability is a number between 0 and 1 that tells us how likely something is about to occur. Have you ever heard the expression “The probability of my passing this course is about 0“? That’s not a good sign for passing that course. Note that a lot of times, probability is given in a percent (0 to 100%) instead of a decimal.
Also important is the concept that the probability that something happens and it doesn’t happen (the complement) adds up to 1. Therefore, the probability of something happening is (1 – the probability of the exact opposite happening).
Probability can get complicated in advanced courses (we’ll visit some of this later), but we’ll just talk about a few “counting” techniques and how to compute some basic probabilities.
A probability is defined as a fraction with the number of times something occurs over the number of possible ways something can occur (possible ways are called outcomes, the set of all possible outcomes is the sample space, and this is typically in the denominator in a probability).
For example, the probability of getting a head if you flip a coin is \(\displaystyle \frac{1}{2}\), since only one thing happens (either a head or a tail), but 2 things could have happened (the head or the tail). It’s a little confusing, but you’ll get it after a while. This is an experiment, since it involves chance.
When events are independent (not related), we can actually multiply to get the probability of both happening! We do have to be careful though, since if the events are dependent on one another (like choosing again, without replacing), the formulas become more complicated; you you may want to check out probability on StatisticsHowTo.com to know more about this. I also address this below.
Something that has no chance of happening has a probability of 0 (like the probability of getting a 7 when you roll a die), and something that will always occur has a probability of 1 (like the probability of getting a number in between and including 1 and 6 when you roll a die).
Experimental probabilities are those you get by actually doing an experiment (like flipping the coin above). You’d have to this for many, many times (try it!) to get close to the theoretical probability, which is the probability we get through mathematics.
An example of trying an experimental probability is to flip a coin 40 times and record whether you get a head or a tail. At each coin toss, add up the number of heads so far, and divide by number of flips so far, to get the experimental probability each time. Notice how it gets closer to the theoretical probability .5 (more reliably – less variability) as you get closer to 40 coin tosses.
I just did this experiment with flipping a penny and checking the experimental probability that I get heads at each coin toss (total number of heads so far, divided by total number of flips so far). Notice how, even though the experimental probability doesn’t end up at exactly .5, the trend is that it gets closer to .5 (with less variance or deviation) the more times I flip the coin:
Coin Flip 
Heads or Tails  # of Heads/
# of Coin Tosses So Far 
Experimental Probability of Heads So Far  Coin Flip  Heads or Tails  # of Heads/
# of Coin Tosses So Far 
Experimental Probability of Heads So Far 
1  Tails  0 over 1  0  21  Tails  10 over 21  .476 
2  Heads  1 over 2  .500  22  Heads  11 over 22  .500 
3  Heads  2 over 3  .667  23  Tails  11 over 23  .478 
4  Heads  3 over 4  .750  24  Tails  11 over 24  .458 
5  Tails  3 over 5  .600  25  Heads  12 over 25  .480 
6  Tails  3 over 6  .500  26  Heads  13 over 26  .500 
7  Heads  4 over 7  .571  27  Tails  13 over 27  .481 
8  Heads  5 over 8  .625  28  Heads  14 over 28  .500 
9  Tails  5 over 9  .556  29  Tails  14 over 29  .483 
10  Heads  6 over 10  .600  30  Heads  15 over 30  .500 
11  Tails  6 over 11  .545  31  Heads  16 over 31  .516 
12  Tails  6 over 12  .500  32  Heads  17 over 32  .531 
13  Heads  7 over 13  .538  33  Heads  18 over 33  .545 
14  Heads  8 over 14  .571  34  Tails  18 over 34  .529 
15  Tails  8 over 15  .533  35  Tails  18 over 35  .514 
16  Heads  9 over 16  .563  36  Heads  19 over 36  .528 
17  Tails  9 over 17  .529  37  Tails  19 over 37  .513 
18  Tails  9 over 18  .500  38  Heads  20 over 38  .526 
19  Tails  9 over 19  .474  39  Tails  20 over 39  .513 
20  Heads  10 over 20  .500  40  Heads  21 over 40  .525 
If we did this experiment say for 2000 times, our experimental probability each time would be reliably very, very close to the theoretical probability of .5. (You might try this for a science experiment!)
Counting Principles
Probabilities usually involve some sort of counting to put on the top or bottom of the fraction.
Fundamental Counting Principle
Here’s an example of the Fundamental Counting Principle, which says that you have a certain number of ways to do something and another number of ways to do something else, you can just multiply those numbers to get the numbers of ways to do both.
Let’s say we have 3 shirts, 2 skirts, and 2 pairs of shoes that we’ve taken on a vacation. We want to know the probability of picking our sleeveless blue shirt, with our purple skirt, with our platform sandal shoes for that day.
Do you see that the total number of things that you can get, or outcomes, is 3 times 2 times 2, which would be 12? Think about it – for the first shirt, you could wear one of two skirts, and one of two pairs of shoes, for the second shirt, the same thing, and so on. You could draw a “tree” diagram like this:
To get that combination (order doesn’t matter: sleeveless blue shirt, purple skirt, and sandals), there would be 1 way out of 12 possible ways, so the probability would be \(\displaystyle \frac{1}{12}\)!
Combinations and Permutations
Let’s now talk about more advanced probability techniques, including combinations and permutations. Combinations and permutations exist since most of the time probability concerns picking a subset (smaller set) of things from a larger set of things, and how we pick the sets is important. As we saw above, this ratio of the desired subset to the number of all possible subsets is between 0 and 1, and this is the probability.
When the order of the subset matters, we have a permutation. An example of this is wanting the number of ways of picking a president, vicepresident, and secretary from a group of people.
When order doesn’t matter, we have a combination. (I remember this since we don’t care about “order” when we have an “o” in the word: combination.) An example of this is wanting the number of ways that any three people can be chosen from a group of people. We’ll actually use combinations again in the Binomial Expansion section.
The math is a little difficult for calculating permutations and combinations. Remember that \(n!=n\times \left( {n1} \right)\times \left( {x2} \right)\times …..\left( 1 \right)\). For example, \(4!=4\times 3\times 2\times 1=24\).
Counting Technique  Formula and Example  Calculator 
PERMUTATION order matters
\(n\) items, ordered \(k\) ways
different orders of \(k\) count as separate cases 
\(\displaystyle P\left( {n,k} \right)={}_{n}{{P}_{k}}=\frac{{n!}}{{\left( {nk} \right)!}}\)
Number of ways to choose a president, vicepresident, and secretary from a 10person club. Order matters since a person being president is different than that person being vicepresident, for example.
\(\require{cancel} \begin{align}P\left( {10,3} \right)={}_{{10}}{{P}_{3}}&=\frac{{10!}}{{\left( {103} \right)!}}\\&=\frac{{10\times 9\times 8\times \cancel{{7!}}}}{{\cancel{{7!}}}}\\&=720\end{align}\) 
For \({}_{{10}}{{P}_{3}}\), press 10, then MATH, then Right Arrow three times to PROB, then 2, for nPr, then 3:
(To repeat with different numbers, you can use 2^{nd} ENTER, then Left Arrow to edit.) 
COMBINATION order doesn’t matter
\(n\) items, chosen \(k\) ways
different orders of \(k\) don’t count as separate cases 
\(\displaystyle C\left( {n,k} \right)={}_{n}{{C}_{k}}=\left( {\begin{array}{*{20}{c}} n \\ k \end{array}} \right)=\frac{{n!}}{{\left( {nk} \right)!k!}}\)
Number of ways to choose any three people from a 10person club. Order doesn’t matter since a person doesn’t have a specific role; there will be fewer ways to do this.
\(\begin{align}C\left( {10,3} \right)={}_{{10}}{{C}_{3}}&=\frac{{10!}}{{\left( {103} \right)!3!}}\\&=\frac{{10\times 9\times 8\times \cancel{{7!}}}}{{\cancel{{7!}}\cdot 3\times 2\times 1}}\\&=120\end{align}\) 
Press 10, then MATH, then Right Arrow three times to PROB, then 3, for nCr, then 3:
(To repeat with different numbers, you can use 2^{nd} ENTER, then Left Arrow to edit.)

Here are some examples of permutation and combination problems:
Permutation/Combination Examples  Explanation 
In how many ways could you make a pizza using 4 toppings, if 15 toppings are available?  Since there is no specific order of the toppings on a pizza, use the combination \({}_{{15}}{{C}_{4}}=1365\). (Note that this may also be written as \(\displaystyle \left( {\begin{array}{*{20}{c}} {15} \\ 4 \end{array}} \right)=1365\)). 
If 200 students each have a numbered ticket to win a lottery, how many ways could cash prizes of $500, $200, and $100 each be awarded?  There are three types of cash prizes; therefore, the order of choosing the prizes matters. Use the permutation \({}_{{200}}{{P}_{3}}=7,880,400\). That’s a lot! 
11 girls try out for the 11 positions on their school’s soccer team. How many ways could the positions be filled, if any girl can play any position?  This one’s a little tricky, since we are choosing the same number of girls that we have in the group. Order does matter, since any girl can play any position. Let’s try it: \({}_{{11}}{{P}_{{11}}}=39,916,800\). This turns out to be the same as \(11!\), which is using the Fundamental Counting Principle above, since we start out with 11 positions, fill one and have 10 left, fill one and have 9 left, and so on (multiply choices).
Note that if the girls could only play one position in this problem, we’d use \({}_{{11}}{{C}_{{11}}}=1\), which makes sense since order wouldn’t matter and there’d only be one way to assign 11 girls to the team. 
How many different threeletter “words” can be made out of the “word” TEXAS?  On this one, we’ll assume that any combination three letters will make a real word, so we’re taking any group of three. There are 5 letters in TEXAS. Order does matter when we pick out words, so there are \({}_{5}{{P}_{3}}=60\) words. 
How many different sevenletter words can be made from the word ARRANGE?  Order matters, but we also have some duplicate letters. There are 7 letters in ARRANGE, but there are 2 “A“’s and 2 “R“’s. Just picking 7 letters out, we have \(7!\) or \(\displaystyle {}_{7}{{P}_{7}}\) (order matters).
But with the duplicate “A“’s and “R“’s, we need eliminate some of these cases. We can do this by dividing by the ways that these can be arranged, which is \(\left( {2!} \right)\left( {2!} \right)\). The answer is \(\displaystyle \frac{{{}_{7}{{P}_{7}}}}{{\left( {2!} \right)\left( {2!} \right)}}=1260\) Later on, there is an example with MISSISSIPPI; more complicated! 
In a normal deck of cards, how many different 5card hands are possible?  Since order doesn’t matter in drawing a hand of cards, we have \({}_{{52}}{{C}_{5}}=2,598,960\). That’s a lot of hands! 
Supposed 5 players from a track team of 40 players are chosen for random drug testing.
(a) What is the number of combinations of players chosen? (b) There are two captains on the team. How many combinations in (a) include both these two people? 
(a) Order doesn’t matter, so use the combination \({}_{{40}}{{C}_{5}}=658,008\).
(b) This is tricky. We’ll multiply probabilities here, using the Fundamental Counting Principle. The number of ways to pick the two captains from the two captains on the team is \({}_{2}{{C}_{2}}=1\), and the number of ways to pick the remaining 38 players is \({}_{{38}}{{C}_{3}}=8,436\). So, the total number of combinations with the two captains is \({}_{2}{{C}_{2}}\times {}_{{38}}{{C}_{3}}=8,436\). Notice how, even though we’re multiplying, the numbers before the C add to 40, and the number after add to 5.

Probability Problems
Here are examples of probability problems that contain the counting techniques we’ve looked at.
Important tip for probability problems: Generally, you add probabilities when the events happen to be alternatives, like an “either/or” situation (and they are mutually exclusive). You multiply probabilities when you want two or more things to happen, either at the same time, or one after another (and they are independent). See below for more on this.
Probability Problem 
Explanation 

What is probably of getting either a 2 or a 6 when you roll a die?  Since there are 2 ways to get either a 2 or a 6, and 6 possible ways to get numbers, the probability is \(\displaystyle \frac{2}{6}\) or \(\displaystyle \frac{1}{3}\).  
(a) What is the probability of getting heads on a coin 3 times in a row?
(b) What is the probability of getting exactly 2 heads and 1 tail? 
(a) Since the probability of getting a heads the first time is \(\displaystyle \frac{1}{2}\), and this is unrelated to the next 2 tosses, the probability is \(\displaystyle \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}\). This is similar to the counting we did when we counted the shirts with skirts with shoes.
(b) The best way to see this is to draw tree diagrams: We can see that there are \(\displaystyle \frac{3}{8}\) ways to get 2 heads and 1 tail; this is the probability. 

(a) What is the probability of landing on pink with this spinner?
(b) What is the probability of spinning a red and then a blue? 
(a) Since there are 2 areas where the color is pink, and there are 10 areas total, the probability is \(\displaystyle \frac{2}{{10}}\) or \(\displaystyle \frac{1}{5}\).
(b) The probability of spinning a red and then a blue (independent events) is \(\displaystyle \frac{2}{{10}}\times \frac{4}{{10}}=\frac{8}{{100}}=\frac{2}{{25}}\). 

What is the probability of drawing either an ace or a red card from a deck of cards?  OK, this one’s a little trickier. A deck has 52 cards; 4 of which are aces, but 26 of which are red. But the total number of cards that you are looking for is 26 (red) plus 4 (aces) minus the 2 cards that are both an ace and red (since we don’t want to doublecount them). This amounts to 28 cards.
The probability is \(\displaystyle \frac{{28}}{{52}}\) or \(\displaystyle \frac{7}{{13}}\). 

A bag has 10 red marbles and 8 blue marbles.
(a) (with replacement) What is the probability of drawing a red marble, putting it back in the bag, and then drawing another red marble? (b) (without replacement) What is the probability of drawing a red, leaving it out of the bag, and then drawing another red marble? 
(a) The probability is \(\displaystyle \frac{{10}}{{18}}\times \frac{{10}}{{18}}\approx .309\), since both times, there are 10 red marbles, and 18 total marbles.
(b) The probability of drawing a red marble is \(\displaystyle \frac{{10}}{{18}}\) and if the red marble is not put back in, the sample space changes from 18 to 17, and number of marbles from 10 to 9. The probability is then \(\displaystyle \frac{{10}}{{18}}\times \frac{9}{{17}}\approx .294\). 

A college tutoring lab provides tutors for three math subjects: Algebra, Trig and Calculus. The following contingency table shows the number of students in each subject with the type of help they request:
(a) Find the probability that a Trig student wants homework help? (b) Find the probability that a student wants test review help or takes calculus. 
(a) There are 50 trig students and 35 want homework help. This probability is \(\displaystyle \frac{{35}}{{50}}=.7\).
(b) There are a total 75 students who want test review help, and a total of 25 students that take calculus. But some of these overlap; specifically, those students who both want test review help and take calculus (10 students). And note that we have 190 students total. So, we have:
\(\displaystyle \frac{{75+2510}}{{190}}\approx .474\) 

The 5 letters in the word TEXAS are randomly arranged.
(a) Find the probability that the 3^{rd} letter is a “T” and the last is a vowel? (b) Find the probability that the 4^{th} and 5^{th} letters are consonants. (c) Find the probability that the arrangement spells “SAXET”. 
The total number of ways that the letters can be arranged (order matters) is \({}_{5}{{P}_{5}}=5!=5\times 4\times 3\times 2\times 1=120\). Note that you could also use a tree diagram to figure this out.
(a) To find the probability that the 3^{rd} letter is a “T” and last is a vowel, we can just pretend the 1^{st} letter is a “T” and next is a vowel; this is easier. Use the Fundamental Counting Principle, knowing there are 3 consonants and 2 vowels (and 3 letters remain): \(\displaystyle \frac{{\overset{\text{T}}{\mathop{1}}\times \overset{{\text{vowel}}}{\mathop{2}}\times \overset{{\text{remaining letters}}}{\mathop{{3\times 2\times 1}}}}}{{5!}}=\frac{{12}}{{120}}=.1\) (b) Assume that the first 2 letters are consonants instead of the 4^{th} and 5^{th}; this is easier: \(\displaystyle \frac{{\overset{{\text{consonant}}}{\mathop{3}}\times \overset{{\text{consonant}}}{\mathop{2}}\times \overset{{\text{remaining letters}}}{\mathop{{3\times 2\times 1}}}}}{{5!}}=\frac{{36}}{{120}}=.3\) (c) There is only one way to get “SAXET”; we have \(\displaystyle \frac{1}{{5!}}=\frac{1}{{120}}\approx .008\). 

Kennedy wants to play tennis tomorrow, but there’s a 30% chance of rain, in which case she can’t play outdoors.
If it doesn’t rain, the probability of her playing is .85, but even if it does rain, she’ll still play tennis (indoors) 25% of the time.
What is the probability that Kennedy will be playing tennis tomorrow? 
Let’s do a probability tree for this problem, starting with the probability of it raining:
The probability that she plays if it doesn’t rain is .595 (multiply each branch), and the probability that she plays if it does rain is .075. Add these to get .67. (The reason we added these probabilities instead of multiplying is because the two events (rain or not rain, play or not play) are alternatives; they aren’t both happening). 

At a middle school, there are 300 students who are either in band, orchestra, or choir.
Of the 90 people in band, 40 are female. Of the 100 males, 35 are in orchestra.
If there are 30 students in choir who are female, what is the probability that a randomly selected student will be enrolled in choir? 
Let’s make a contingency table for this situation; the numbers in red are the ones that we had to come up with. Note that we didn’t have to fill the whole table.
The probability that a random student is enrolled in choir is \(\displaystyle \frac{{45}}{{300}}=.15\).

Here are more advanced probability problems.
Important tip for probability problems: Generally, you add probabilities when the events happen to be alternatives, like an “either/or” situation (and they are mutually exclusive). You multiply probabilities when you want two or more things to happen, either at the same time, or one after another (and they are independent). See below for more on this.
Probability Problem  Explanation 
What is the probability of being dealt a royal ace, if 5 cards of a normal card deck are drawn?
(A royal ace is 10, J, Q, K, and A, all in one suit.) 
The number of 5cards hands in a normal deck (order doesn’t matter) is \({}_{{52}}{{C}_{5}}=2,598,960\). The number of ways a royal ace can be drawn is 4, since there are 4 suits and only one way for each suit to get a royal ace. The probability is \(\displaystyle \frac{4}{{{}_{{52}}{{C}_{5}}}}=\frac{4}{{2,598,960}}=\frac{{649,740}}{{2,598,960}}\approx .0000015\). Pretty small! 
There are 10 pink marbles and 5 blue marbles in a bag. If 4 marbles are drawn at random, find:
(a) Exactly 3 are pink (b) All 4 are pink (c) At least 3 are pink (d) At least 1 is pink 
Let’s first find the total number of ways that 4 marbles are chosen from 15 marbles (order doesn’t matter): \({}_{{15}}{{C}_{4}}=1365\).
(a) The number of possible ways that exactly 3 pink marbles could be drawn is \({}_{{10}}{{C}_{3}}\times {}_{5}{{C}_{1}}=120\times 5=600\) (take the number of ways to pick 3 marbles from 10 pink marbles; multiply this by the number of ways to pick the remaining marble from the blue). The probability is \(\displaystyle \frac{{{}_{{10}}{{C}_{3}}\times {}_{5}{{C}_{1}}}}{{{}_{{15}}{{C}_{4}}}}=\frac{{120\times 5}}{{1365}}=\frac{{600}}{{1365}}\approx .440\). (Note the top indices add up to the bottom: \(10+5=15;\,\,3+1=4\).) (b) Since all 4 are pink, none would be blue: \(\displaystyle \frac{{{}_{{10}}{{C}_{4}}}}{{{}_{{15}}{{C}_{4}}}}=\frac{{210}}{{1365}}\approx .154\). (c) To get at least 3 pink marbles, we could get 3 or 4 pink marbles: \(\displaystyle \frac{{{}_{{10}}{{C}_{3}}\times {}_{5}{{C}_{1}}+{}_{{10}}{{C}_{4}}\times {}_{5}{{C}_{0}}}}{{{}_{{15}}{{C}_{4}}}}=\frac{{810}}{{1365}}\approx .593\). (d) To get the probability that at least 1 is pink, we can get the probability that all are blue, and then subtract this from 1 to get its complement (since anything else would have at least one pink marble): \(\displaystyle 1\frac{{{}_{{10}}{{C}_{0}}\times {}_{5}{{C}_{4}}}}{{{}_{{15}}{{C}_{4}}}}=1\frac{{1\times 5}}{{1365}}\approx .996\). This makes sense since so many are pink in the bag. 
A committee of 6 is selected from a group of 6 boys and 5 girls.
(a) What is the probability that the committee is all boys? (b) What is the probability that exactly 4 out of the 6 are boys? 
We know the total number of ways 6 kids is selected from the group of the boys and girls (11) is \({}_{{11}}{{C}_{6}}=462\); order doesn’t matter, since there are no positions.
(a) There is only way to get all boys, since we are only 6 boys total. The probability then is \(\displaystyle \frac{1}{{{}_{{11}}{{C}_{6}}}}=\frac{1}{{462}}\approx .002\) (low). (b) The number of possible ways 4 out of 6 boys could be chosen is \({}_{6}{{C}_{4}}=15\), and the number of ways to pick 2 girls out of the remaining 5 kids is \({}_{5}{{C}_{2}}=10\). The probability then that exactly 4 out of the 6 are girls is \(\displaystyle \frac{{{}_{6}{{C}_{4}}\times {}_{5}{{C}_{2}}}}{{{}_{{11}}{{C}_{6}}}}=\frac{{150}}{{462}}\approx .325\). 
The letters in the word MISSISSIPPI are written on pieces of paper and put in a bag.
5 cards are chosen at random.
Find the probability that one of each letter is chosen. 
We know the total number of ways 5 letters is selected from the word MISSISSIPPI (11 letters) is \({}_{{11}}{{C}_{5}}=462\); order doesn’t matter.
Let’s first get the number of ways each letter could be chosen individually. For example, the number of ways an “M” is chosen is 1, since there’s only 1 “M”, the number of ways that an “I” is chosen is 4 since there are 4 “I“’s, and so on.
To get the probability that one of each letter is chosen, multiply these together, and then put this over the total number of ways to choose 5 letters: \(\displaystyle \frac{{\overset{\text{M}}{\mathop{1}}\,\,\times \overset{\text{I}}{\mathop{4}}\,\,\,\times \overset{\text{S}}{\mathop{4}}\,\,\,\times \overset{\text{P}}{\mathop{2}}\,}}{{{}_{{11}}{{C}_{5}}}}=\frac{{32}}{{462}}\approx .069\). 
You are dealt 13 cards from a standard deck.
(a) Set up (don’t solve) the probability that you receive all hearts. (b) Set up the probability that you receive no hearts. (c) Set up the probability that you receive at least one heart. 
The total number of ways to choose 13 cards from a deck of 52 cards (order doesn’t matter) is \({}_{{52}}{{C}_{{13}}}\).
(a) There is only one way to get all hearts (\({}_{{13}}{{C}_{{13}}}=1\)) since there are 13 hearts. This probability is \(\displaystyle \frac{1}{{{}_{{52}}{{C}_{{13}}}}}\) . (b) The number of ways to pick no hearts is \(\displaystyle \frac{{{}_{{13}}{{C}_{0}}\times {}_{{39}}{{C}_{{13}}}}}{{{}_{{52}}{{C}_{{13}}}}}\), or \(\displaystyle \frac{{{}_{{39}}{{C}_{{13}}}}}{{{}_{{52}}{{C}_{{13}}}}}\), since there are 39 cards that aren’t hearts. (c) The probability that you receive at least one heart is the complement (opposite) that you receive no hearts (\(\displaystyle 1\frac{{{}_{{39}}{{C}_{{13}}}}}{{{}_{{52}}{{C}_{{13}}}}}\)). 
There are 10 girls and 5 equal scholarships.
(a) Find the probability that Brynn and Quinn (2 of the 10) are chosen. (b) Find the probability that Brynn is chosen first and Quinn second. 
The total number of ways the two girls could get the scholarships is \({}_{{10}}{{C}_{5}}=252\).
(a) The number of ways that Brynn and Quinn could be chosen is \({}_{2}{{C}_{2}}\times {}_{8}{{C}_{3}}=1\times 56=56\), since we’d have to include the number of ways the other 3 scholarships are chosen, and multiply. The probability is \(\displaystyle \frac{{{}_{2}{{C}_{2}}\times {}_{8}{{C}_{3}}}}{{{}_{{10}}{{C}_{5}}}}=\frac{{1\times 56}}{{252}}\approx .222\). (b) Since there are only 2 ways that the girls can be chosen (BrynnQuinn, or QuinnBrynn), we take the probability in (a) and divide by 2: \(\approx .111\). 
9 girls try out for twelve distinct positions on a sports team.
(a) How many ways could the 9 positions be filled if any girl can play any position? (b) If the positions are chosen at random, what is the probability that Erica is chosen as shortstop? 
(a) This one’s a little tricky, since we are choosing the same number of girls that we have in the group. Order does matter, since any girl can play any position. Let’s try it: \({}_{9}{{P}_{9}}=362,880\). This turns out to be the same as \(9!\), which is using the Fundamental Counting Principle above.
(b) If Erica is chosen as shortstop, then there are 8 other positions to be filled (order matters) for the other 8 girls; this is \({}_{8}{{P}_{8}}=40320\). Thus, the probability that Erica is chosen for this position is \(\displaystyle \frac{{{}_{8}{{P}_{8}}}}{{{}_{9}{{P}_{9}}}}=\frac{{40,320}}{{362,880}}\approx .111\). Aha! This is the same as \(\displaystyle \frac{1}{9}\), which makes sense!

A vending machine has 10 different drinks. One day, 8 employees each purchased a drink from the machine.
What is the probability that at least 2 of the employees purchased the same drink? 
There are probably many different ways to solve this, but the easiest would be to find the probability that the employees all purchased different drinks, and then subtract that from 1:
\(\displaystyle 1\frac{{10}}{{10}}\times \frac{9}{{10}}\times \frac{8}{{10}}\times \frac{7}{{10}}\times \frac{6}{{10}}\times \frac{5}{{10}}\times \frac{4}{{10}}\times \frac{3}{{10}}=1.018144\approx .98\)
The probability is pretty high, which makes sense. 
Using Multiplication and Addition to Find Probabilities in Compound Situations
As we said earlier, when events are independent (not related), we can actually multiply to get the probability of both happening! We do have to be careful though, since if the events are dependent on one another (like choosing again, without replacing), the formulas become more complicated.
Here are some formulas with examples:
Probability Formula
\(\bigcap :\,\,\text{Intersection (and)}\,\,\,\,\,\,\,\,\bigcup \text{: Union (or)}\) 
Example  
Complimentary events: \(P\left( A \right)+P\left( {{A}’} \right)=1\) 
Complementary events happen when there are only two outcomes of the event. Since one or the other must happen, their probabilities must add up to 1. For example, on a coin toss, \(\displaystyle P\left( {\text{getting heads}} \right)+P\left( {\text{getting tails}} \right)=1\), since you are guaranteed either heads or tails.  
Independent Events: \(\begin{array}{l}P\left( {A\bigcap B} \right)=P\left( A \right)\cdot P\left( B \right)\\P\left( {A\text{ and }B} \right)=P\left( A \right)\cdot P\left( B \right)\end{array}\) 
Independent events happen when one of the events doesn’t have anything to do with the other; they are totally independent. For example,
\(\displaystyle \begin{align}&P\left( {\text{getting heads on coin toss AND getting a 6 on die toss}} \right)\\\,\,&=P\left( {\text{getting heads on coin toss }} \right)\cdot P\left( {\text{getting a 6 on coin toss }} \right)\\&=\frac{1}{2}\cdot \frac{1}{6}=\frac{1}{{12}}\end{align}\) Note that since the probabilities are fractions, the probability goes down for multiple independent events, which makes sense. 

Addition Rule: \(\begin{array}{l}P\left( {A\bigcup B} \right)=P\left( A \right)+P\left( B \right)P\left( {A\bigcap B} \right)\\P\left( {A\text{ or }B} \right)=P\left( A \right)+P\left( B \right)P\left( {A\text{ and }B} \right)\end{array}\) 
The probabilities of one event happening on its own OR another event happening on its own are added, but you have to subtract out the probability of them BOTH happening. For example, if you take out a card from a deck, what is the probability that it is an ace or a club?
\(\begin{align}&P\left( {\text{ace or a club}} \right)\\&=P\left( {\text{ace }} \right)+P\left( {\text{club}} \right)P\left( {\text{ace and a club}} \right)\\&=\frac{4}{{52}}+\frac{{13}}{{52}}\frac{1}{{52}}=\frac{4}{{13}}\end{align}\) If the events are mutually exclusive (they both can’t happen), then the formula becomes \(P\left( {A\bigcup B} \right)+P\left( A \right)+P\left( B \right)\). For example, if you take out a card from a deck, what is the probability that it is an ace of spades or an ace of clubs? \(\begin{align}&P\left( {\text{ace of spades or ace of clubs}} \right)\\&=P\left( {\text{ace of spades}} \right)+P\left( {\text{ace of clubs}} \right)P\left( {\text{ace of spades and ace of clubs}} \right)\\&=\frac{1}{{52}}+\frac{1}{{52}}0=\frac{1}{{26}}\end{align}\) 

Conditional Probability: \(\begin{align}P\left( {AB} \right)&=\frac{{P\left( {A\bigcap B} \right)}}{{P\left( B \right)}}\\P\left( {AB} \right)&=\frac{{P\left( {A\text{ and }B} \right)}}{{P\left( B \right)}}\end{align}\)
Note that the independent formula above is a result of this formula, where the two events are independent, so conditional probabilities are 0. 
Conditional probability is the probability of an event occurring, but it has a relationship with one or more other previous events. We did a problem like this above, and used a contingency table (you might also see tree or venn diagrams). The table shows the number of students in three different subjects with the type of math lab help they request:
Let’s get the probability that a Trig student wants homework help. The answer may seem obvious ( \(\frac{{35}}{{50}}\)), but let’s use the formula: \(\begin{align}P\left( {\text{Homework Help}\text{Trig Student}} \right)&=\frac{{P\left( {\text{Homework Help}\bigcap \text{Trig Student}} \right)}}{{P\left( {\text{Trig Student}} \right)}}\\&=\frac{{\frac{{35}}{{190}}}}{{\frac{{50}}{{190}}}}=\frac{{35}}{{50}}=.7\end{align}\) 

Bayes Formula: \(\displaystyle P\left( {AB} \right)=\frac{{P\left( {BA} \right)\cdot P\left( A \right)}}{{P\left( B \right)}}\) 
Bayes is still a conditional probability, but in a different form. It is used to find the probability of a certain event, based on prior knowledge that may be related to the event (probabilities that we already know). As an example, the Bayes Formula can be very useful to determine the accuracy of medical tests.

Again, don’t worry if you don’t get all this now; it’s just important to get the main concepts.
Learn these rules, and practice, practice, practice!
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