Solving Eq & Ineq        Graphs & Func.        Systems of Eq.        Polynomials        Frac. Express.        Powers & Roots        Complex Numbers        Quadratic Eq.        Quadratic Func.       Coord. Geo.        Exp. & Log. Func.    Probability        Matrices        Trigonometry        Trig. Identities        Equations & Tri. On this page we hope to clear up problems you might have with probability and things related to it, such as factorials and sigma notation.  Sigma notation is especially useful to know because you use it a lot in calculus when you find area under curves.  Click any of the links below or scroll down to better your understanding of probability. Sigma notation Permutations Combinations Probability Quiz on Probability Σ, the Greek letter sigma (if you have an old browser that can't display Sigma, it looks like a mix between an E and a Z), can be used to simplify sequences of numbers.  The sum of terms in the sequence is found using sigma notation (also called summations).  Example: ``` 1. Problem: 4 Σ (2k + 1) k = 1 Solution: This is a sum of (2k + 1) from 1 to 4. Plug all numbers from 1 to 4 into the general term ((2k + 1) in this case) and then add the terms together. (2(1) + 1) = 3 (2(2) + 1) = 5 (2(3) + 1) = 7 (2(4) + 1) = 9 3 + 5 + 7 + 9 = 24``` A permutation of a set of objects is an arrangement of the objects in a certain order.  For example, take the set of four objects {pepperoni, sausage, onions, mushrooms}.  They can be arranged on a pizza many different ways.  Below are a few of the ways. pepperoni, sausage, onions, mushrooms sausage, onions, mushrooms, pepperoni onions, mushrooms, pepperoni, sausage mushrooms, pepperoni, sausage, onions pepperoni, sausage, mushrooms, onions There are some more, but we won't list them.  To find the number of different arrangements of the set we select a first choice; there are 4 possible choices.  Now we take a second choice; there are 3 choices.  Now pick a third choice; there are 2 choices.  Finally, there is 1 choice for the last selection.  Thus, there are 4 * 3 * 2 * 1 or 24 different ordered arrangements of the toppings.   This product can also be written as 4! (read: 4-factorial). The total number of permutations of a set of n objects is given by n!.  Example: ``` 1. Problem: 5! Solution: 5 * 4 * 3 * 2 * 1 120``` When you have a set of objects and only want to arrange part of them, you have a permutation of n objects r at a time.  For example, if you have 6 toppings for a pizza, and a customer calls and tells you to put any 3 toppings on the pizza, you might want to know how many different pizzas you can make.  You can select the first topping in 6 ways, the second in 5, and the third in 4.  As we learned above, this can be written as 6 * 5 * 4.  There is a theorem that tells us about a formula for the situation above.  It says the number of permutations of a set of n objects taken r at a time is given by the following formula: nPr = (n!)/(n - r)!.  Example: ``` 2. Problem: If a school has lockers with 50 numbers on each combination lock, how many possible combinations using three numbers are there. Solution: Recognize that n, or the number of objects is 50 and that r, or the number of objects taken at one time is 3. Plug those numbers in the permutation formula. 50! 50P3 = -------- (50 - 3)! Use a calculator to find the final answer. 117600``` Things are immensely simplified when you can repeat the objects.  For example, if you are making license plates with only 4 letters on them, and you can repeat the letters, you can take the first letter from 26 options, the same for the second, third, and fourth.   Therefore, there are 264 or 456976 available license plates using 4 letters if you can repeat letters.  There is a special theorem that tells us the number of arrangements of n objects taken r at a time, with repetition is given by nr.  Example: ``` 3. Problem: How many 4 digit license plates can you make using the numbers from 0 to 9 while allowing repetitions. Solution: Realize there are 10 objects taken 4 at a time. Plug that information into the formula for repeated use. 104 10000``` Unordered arrangements of objects are called combinations.  For example, by the definition of combinations, a pizza with the left half pineapple and the right half pepperoni is the same thing as a pizza with the left half pepperoni and the right half pineapple. The number of combinations of a set of n objects taken r at a time is given by nCr = (n!)/(r!(n - r)!).   Example: ``` 1. Problem: For a study, 4 people are chosen at random from a group of 10 people. How many ways can this be done? Solution: Since you're going to have the same group of people no matter the order you choose the people in, you set up the problem as a combination. 10! 10C4 = ----------- 4!(10 - 4)! Use a calculator to find the answer. There are 210 different groups of people you can choose. ``` If an event, E can occur m ways out of n possible outcomes, the probability of that event is given by P(E) = (m/n).  Example: ``` 1. Problem: What is the probability of rolling a 3 on a die (plural, dice). Solution: On a fair die (not the kind you play with in Vegas, where everything is rigged), there are six equally likely outcomes when you roll. Also, there is only one way to get a 3. By the definition of proba- bility, P(3) = (1/6). ``` Probability will always be a fraction, 0, or 1.  If an event cannot happen, the probability is 0.  If an event is certain to happen, the probability is 1. Take the Quiz on probability.  (Very useful to review or to see if you've really got this topic down.)  Do it!

Math for Morons Like Us - Algebra II: Probability
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