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Learning Objectives
By the end of this section, you will be able to:
- Use Pascal’s Triangle to expand a binomial
- Evaluate a binomial coefficient
- Use the Binomial Theorem to expand a binomial
Before you get started, take this readiness quiz.
- Simplify: (frac{7 cdot 6 cdot 5 cdot 4}{4 cdot 3 cdot 2 cdot 1}).
If you missed this problem, review Example 1.25. - Expand: ((3 x+5)^{2}).
If you missed this problem, review Example 5.32. - Expand: ((x-y)^{2}).
If you missed this problem, review Example 5.32.
Use Pascal's Triangle to Expand a Binomial
In our previous work, we have squared binomials either by using FOIL or by using the Binomial Squares Pattern. We can also say that we expanded ((a+b)^{2}).
((a+b)^{2}=a^{2}+2 a b+b^{2})
To expand ((a+b)^{3}), we recognize that this is ((a+b)^{2}(a+b)) and multiply.
((a+b)^{3})
((a+b)^{2}(a+b))
(left(a^{2}+2 a b+b^{2}
ight)(a+b))
(a^{3}+2 a^{2} b+a b^{2}+a^{2} b+2 a b^{2}+b^{3})
(a^{3}+3 a^{2} b+3 a b^{2}+b^{3})
((a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3})
To find a method that is less tedious that will work for higher expansions like ((a+b)^{7}), we again look for patterns in some expansions.
| Number of Terms | First Term | Last Term | |
|---|---|---|---|
| ((a+b)^{1}=a+b) | (2) | (a^{1}) | (b^{1}) |
| ((a+b)^{2}=a^{2}+2 a b+b^{2}) | (3) | (a^{2}) | (b^{2}) |
| ((a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}) | (4) | (a^{3}) | (b^{3}) |
| ((a+b)^{4}=a^{4}+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+b^{4}) | (5) | (a^{4}) | (b^{4}) |
| ((a+b)^{5}=a^{5}+5 a^{4} b+10 a^{3} b^{2}+10 a^{2} b^{3}+5 a b^{4}+b^{5}) | (6) | (a^{5}) | (b^{5}) |
| ((a+b)^{n}) | (n) | (a^{n}) | (b^{n}) |
Notice the first and last terms show only one variable. Recall that (a^{0}=1), so we could rewrite the first and last terms to include both variables. For example, we could expand ((a+b)^{3}) to show each term with both variables.
Generally, we don’t show the zero exponents, just as we usually write (x) rather than (1x).
Note
Patterns in the Expansion of ((a+b)^{n})
- The number of terms is (n+1).
- The first term is (a^{n}) and the last term is (b^{n}).
- The exponents on (a) decrease by one on each term going left to right.
- The exponents on (b) increase by one on each term going left to right.
- The sum of the exponents on any term is (n).
Let’s look at an example to highlight the last three patterns.
From the patterns we identified, we see the variables in the expansion of ((a+b)^{n}), would be
((a+b)^{n}=a^{n}+\_\_\_a^{n-1}b^{1}+\_\_\_a^{n-2}b^{2}+ldots+\_\_\_a^{1}b^{n-1}+b^{n}).
To find the coefficients of the terms, we write our expansions again focusing on the coefficients. We rewrite the coefficients to the right forming an array of coefficients.
The array to the right is called Pascal’s Triangle. Notice each number in the array is the sum of the two closest numbers in the row above. We can find the next row by starting and ending with one and then adding two adjacent numbers.
This triangle gives the coefficients of the terms when we expand binomials.
Definition (PageIndex{1})
Pascal's Triangle
In the next example, we will use this triangle and the patterns we recognized to expand the binomial.
Example (PageIndex{1})
Use Pascal's Triangle to expand ((x+y)^{6}).
Solution:
We know the variables for this expansion will follow the pattern we identified. The nonzero exponents of (x) will start at six and decrease to one. The nonzero exponents of (y) will start at one and increase to six. The sum of the exponents in each term will be six. In our pattern, (a=x) and (b=y).
(egin{array}{l}{(a+b)^{n}=a^{n}+\_\_\_a^{n-1} b^{1}+\_\_\_a^{n-2} b^{2}+ldots+\_\_\_a^{1}b^{n-1}+b^{n}} {(x+y)^{6}=x^{6}+\_\_\_x^{5} y^{1}+\_\_\_x^{4} y^{2}+\_\_\_x^{3} y^{3}+\_\_\_x^{2} y^{4}+\_\_\_x^{1} y^{5}+y^{6}}end{array})
Exercise (PageIndex{1})
Use Pascal’s Triangle to expand ((x+y)^{5}).
- Answer
(egin{array}{l}{x^{5}+5 x^{4} y+10 x^{3} y^{2}+10 x^{2} y^{3}} {+5 x y^{4}+y^{5}}end{array})
Exercise (PageIndex{2})
Use Pascal’s Triangle to expand ((p+q)^{7}).
- Answer
(egin{array}{c}{p^{7}+7 p^{6} q+21 p^{5} q^{2}+35 p^{4} q^{3}} {+35 p^{3} q^{4}+21 p^{2} q^{5}+7 p q^{6}+q^{7}}end{array})
In the next example we want to expand a binomial with one variable and one constant. We need to identify the (a) and (b) to carefully apply the pattern.
Example (PageIndex{2})
Use Pascal’s Triangle to expand ((x+3)^{5}).
Solution:
We identify the (a) and (b) of the pattern.
In our pattern, (a=x) and (b=3).
We know the variables for this expansion will follow the pattern we identified. The sum of the exponents in each term will be five.
((a+b)^{n}=a^{n}+\_\_\_a^{n-1}b^{1}+\_\_\_a^{n-2}b^{2}+ldots+\_\_\_a^{1}b^{n-1}+b^{n} )
((x+3)^{5}=x^{5}+\_\_\_x^{4}cdot3^{1}+\_\_\_x^{3}cdot3^{2}+\_\_\_x^{2}cdot3^{3}+\_\_\_x^{1}cdot3^{4}+3^{5})
Exercise (PageIndex{3})
Use Pascal's Triangle to expand ((x+2)^{4}).
- Answer
(x^{4}+8 x^{3}+24 x^{2}+32 x+16)
Exercise (PageIndex{4})
Use Pascal's Triangle to expand ((x+1)^{6}).
- Answer
(egin{array}{l}{x^{6}+6 x^{5}+15 x^{4}+20 x^{3}+15 x^{2}} {+6 x+1}end{array})
In the next example, the binomial is a difference and the first term has a constant times the variable. Once we identify the (a) and (b) of the pattern, we must once again carefully apply the pattern.
Example (PageIndex{3})
Use Pascal's Triangle to expand ((3x-2)^{4}).
Solution:
We identify the (a) and (b) of the pattern.
In our pattern, (a=3x) and (b=-2).
((a+b)^{n}=a^{n}+\_\_\_a^{n-1}b^{1}+\_\_\_a^{n-2}b^{2}+ldots+\_\_\_a^{1}b^{n-1}+b^{n} )
((3 x-2)^{4}=81 x^{4}+4left(27 x^{3} ight)(-2)+6left(9 x^{2} ight)(4)+4(3 x)(-8)+1 cdot 16)
Exercise (PageIndex{5})
Use Pascal's Triangle to expand ((2x-3)^{4}).
- Answer
(16 x^{4}-96 x^{3}+216 x^{2}-216 x+81)
Exercise (PageIndex{6})
Use Pascal's Triangle to expand ((2x-1)^{6}).
- Answer
(egin{array}{l}{64 x^{6}-192 x^{5}+240 x^{4}-160 x^{3}} {+60 x^{2}-12 x+1}end{array})
Evaluate a Binomial Coefficient
While Pascal’s Triangle is one method to expand a binomial, we will also look at another method. Before we get to that, we need to introduce some more factorial notation. This notation is not only used to expand binomials, but also in the study and use of probability.
To find the coefficients of the terms of expanded binomials, we will need to be able to evaluate the notation (left( egin{array}{l}{n} {r}end{array} ight)) which is called a binomial coefficient. We read (left( egin{array}{l}{n} {r}end{array} ight)) as “(n) choose (r)” or “(n) taken (r) at a time”.
Definition (PageIndex{1})
A binomial coefficient (left( egin{array}{l}{n} {r}end{array} ight)), where (r) and (b) are integers with (0 leq r leq n), is defined as
(left( egin{array}{l}{n} {r}end{array} ight)=frac{n !}{r !(n-r) !})
We read (left( egin{array}{l}{n} {r}end{array} ight)) as "(n) choose (r)" or "(n) taken (r) at a time."
Example (PageIndex{4})
Evaluate:
- (left( egin{array}{l}{5} {1}end{array} ight))
- (left( egin{array}{l}{7} {7}end{array} ight))
- (left( egin{array}{l}{4} {0}end{array} ight))
- (left( egin{array}{l}{8} {5}end{array} ight))
Solution:
a. We will use the definition of a binomial coefficient,
(left( egin{array}{l}{n} {r}end{array} ight)=frac{n !}{r !(n-r) !})
(left( egin{array}{l}{5} {1}end{array} ight))
Use the definition, (left( stackrel{5}{1} ight)=frac{n !}{r !(n-r) !}), where (n=5, r=1).
(frac{5 !}{1 !(5-1) !})
Simplify.
(frac{5 !}{1 !(4) !})
Rewrite (5!) as (5cdot 4!)
(frac{5 cdot 4 !}{1 ! cdot 4 !})
Simplify, by removing common factors.
(frac{5cdot cancel{4 !}}{1 ! cdot cancel{4 !}})
Simplify.
(5)
(left( egin{array}{l}{5} {1}end{array} ight)=5)
b. (left( egin{array}{l}{7} {7}end{array} ight))
Use the definition, (left( stackrel{5}{1} ight)=frac{n !}{r !(n-r) !}), where (n=7, r=7).
(frac{7 !}{7 !(7-7) !})
Simplify.
(frac{7 !}{7 !(0) !})
Simplify. Remember (0!=1).
(1)
(left( egin{array}{l}{7} {7}end{array} ight)=1)
c. (left( egin{array}{l}{4} {0}end{array} ight))
Use the definition, (left( stackrel{5}{1} ight)=frac{n !}{r !(n-r) !}), where (n=4, r=0).
(frac{4 !}{0 !(4-0) !})
Simplify.
(frac{4 !}{0 !(4) !})
Simplify.
(1)
(left( egin{array}{l}{4} {0}end{array} ight)=1)
d. (left( egin{array}{l}{8} {5}end{array} ight))
Use the definition, (left( stackrel{5}{1} ight)=frac{n !}{r !(n-r) !}), where (n=8, r=5).
(frac{8 !}{5 !(8-5) !})
Simplify.
(frac{8 !}{5 !(3) !})
Rewrite (8!) as (8cdot 7cdot 6cdot 5!) and remove common factors.
(frac{8cdot7cdotcancel{6}cdotcancel{5!}}{cancel{5!}cdotcancel{3}cdotcancel{2}cdot1})
Simplify.
(56)
(left( egin{array}{l}{8} {5}end{array} ight)=56)
Exercise (PageIndex{7})
Evaluate each binomial coefficient:
- (left( egin{array}{l}{6} {1}end{array} ight))
- (left( egin{array}{l}{8} {8}end{array} ight))
- (left( egin{array}{l}{5} {0}end{array} ight))
- (left( egin{array}{l}{7} {3}end{array} ight))
- Answer
- (6)
- (1)
- (1)
- (35)
Exercise (PageIndex{8})
Evaluate each binomial coefficient:
- (left( egin{array}{l}{2} {1}end{array} ight))
- (left( egin{array}{l}{11} {11}end{array} ight))
- (left( egin{array}{l}{9} {0}end{array} ight))
- (left( egin{array}{l}{6} {5}end{array} ight))
- Answer
- (2)
- (1)
- (1)
- (6)
In the previous example, ((a)), ((b)), ((c)) demonstrate some special properties of binomial coefficients.
Definition (PageIndex{2})
Properties of Binomial Coefficients
(left( egin{array}{l}{n} {1}end{array} ight)=n quad left( egin{array}{l}{n} {n}end{array} ight)=1 quad left( egin{array}{l}{n} {0}end{array} ight)=1)
Use the Binomial Theorem to Expand a Binomial
We are now ready to use the alternate method of expanding binomials. The Binomial Theorem uses the same pattern for the variables, but uses the binomial coefficient for the coefficient of each term.
Definition (PageIndex{3})
Binomial Theorem
For any real numbers (a) and (b), and positive integer (n),
((a+b)^{n}=left( egin{array}{c}{n} {0}end{array} ight) a^{n}+left( egin{array}{c}{n} {1}end{array} ight) a^{n-1} b^{1}+left( egin{array}{c}{n} {2}end{array} ight) a^{n-2} b^{2}+ldots+left( egin{array}{c}{n} {r}end{array} ight) a^{n-r} b^{r}+ldots+left( egin{array}{c}{n} {n}end{array} ight) b^{n})
Example (PageIndex{5})
Use the Binomial Theorem to expand ((p+q)^{4}).
Solution:
We identify the (a) and (b) of the pattern.
In our pattern, (a=p) and (b=q).
We use the Binomial Theorem.
((a+b)^{n}=left( egin{array}{c}{n} {0}end{array} ight) a^{n}+left( egin{array}{c}{n} {1}end{array} ight) a^{n-1} b^{1}+left( egin{array}{c}{n} {2}end{array} ight) a^{n-2} b^{2}+ldots+left( egin{array}{c}{n} {r}end{array} ight) a^{n-r} b^{r}+ldots+left( egin{array}{c}{n} {n}end{array} ight) b^{n})
Substitute in the values (a=p, b=q) and (n=4).
((p+q)^{4}=left( egin{array}{c}{4} {0}end{array} ight) p^{4}+left( egin{array}{c}{4} {1}end{array} ight) p^{4-1} q^{1}+left( egin{array}{c}{4} {2}end{array} ight) p^{4-2} q^{2}+left( egin{array}{c}{4} {3}end{array} ight) p^{4-3} q^{3}+left( egin{array}{c}{4} {4}end{array} ight) q^{4})
Simplify the exponents.
((p+q)^{4}=left( egin{array}{l}{4} {0}end{array} ight) p^{4}+left( egin{array}{c}{4} {1}end{array} ight) p^{3} q+left( egin{array}{c}{4} {2}end{array} ight) p^{2} q^{2}+left( egin{array}{c}{4} {3}end{array} ight) p q^{3}+left( egin{array}{c}{4} {4}end{array} ight) q^{4})
Evaluate the coefficients, remember, (left( egin{array}{l}{n} {1}end{array} ight)=n, left( egin{array}{l}{n} {n}end{array} ight)=1, left( egin{array}{l}{n} {0}end{array} ight)=1)
((p+q)^{4}=1 p^{4}+4 p^{3} q^{1}+frac{4 !}{2 !(2) !} p^{2} q^{2}+frac{4 !}{3 !(4-3) !} p^{1} q^{3}+1 q^{4})
((p+q)^{4}=p^{4}+4 p^{3} q+6 p^{2} q^{2}+4 p q^{3}+q^{4})
Exercise (PageIndex{9})
Use the Binomial Theorem to expand ((x+y)^{5}).
- Answer
(egin{array}{l}{x^{5}+5 x^{4} y+10 x^{3} y^{2}+10 x^{2} y^{3}} {+5 x y^{4}+y^{5}}end{array})
Exercise (PageIndex{10})
Use the Binomial Theorem to expand ((m+n)^{6}).
- Answer
(egin{array}{l}{m^{6}+6 m^{5} n+15 m^{4} n^{2}+20 m^{3} n^{3}} {+15 m^{2} n^{4}+6 m n^{5}+n^{6}}end{array})
Notice that when we expanded ((p+q)^{4}) in the last example, using the Binomial Theorem, we got the same coefficients we would get from using Pascal's Triangle.
The next example, the binomial is a difference. When the binomial is a difference, we must be careful in identifying the values we will use in the pattern.
Example (PageIndex{6})
Use the Binomial Theorem to expand ((x-2)^{5}).
Solution:
We identify the (a) and (b) of the pattern.
In our pattern, (a=x) and (b=-2).
We use the Binomial Theorem.
((a+b)^{n}=left( egin{array}{c}{n} {0}end{array} ight) a^{n}+left( egin{array}{c}{n} {1}end{array} ight) a^{n-1} b^{1}+left( egin{array}{c}{n} {2}end{array} ight) a^{n-2} b^{2}+ldots+left( egin{array}{c}{n} {r}end{array} ight) a^{n-r} b^{r}+ldots+left( egin{array}{c}{n} {n}end{array} ight) b^{n})
Substitute in the values (a=x, b=-2), and (n=5).
((x-2)^{5}=left( egin{array}{l}{5} {0}end{array} ight) x^{5}+left( egin{array}{c}{5} {1}end{array} ight) x^{5-1}(-2)^{1}+left( egin{array}{c}{5} {2}end{array} ight) x^{5-2}(-2)^{2}+left( egin{array}{c}{5} {3}end{array} ight) x^{5-3}(-2)^{3}+left( egin{array}{c}{5} {4}end{array} ight) x^{5-4}(-2)^{4}+left( egin{array}{c}{5} {5}end{array} ight)(-2)^{5})
Simplify the coefficients. Remember, (left( egin{array}{l}{n} {1}end{array} ight)=n, left( egin{array}{l}{n} {n}end{array} ight)=1, left( egin{array}{l}{n} {0}end{array} ight)=1).
((x-2)^{5}=left( egin{array}{l}{5} {0}end{array} ight) x^{5}+left( egin{array}{c}{5} {1}end{array} ight) x^{4}(-2)+left( egin{array}{c}{5} {2}end{array} ight) x^{3}(-2)^{2}+left( egin{array}{c}{5} {3}end{array} ight) x^{2}(-2)^{3}+left( egin{array}{c}{5} {4}end{array} ight) x(-2)^{4}+left( egin{array}{c}{5} {5}end{array} ight)(-2)^{5})
((x-2)^{5}=1 x^{5}+5(-2) x^{4}+frac{5 !}{2 ! cdot 3 !}(-2)^{2} x^{3}+frac{5 !}{3 ! 2 !}(-2)^{3} x^{2}+frac{5 !}{4 !1 !}(-2)^{4} x+1(-2)^{5})
((x-2)^{5}=x^{5}+5(-2) x^{4}+10 cdot 4 cdot x^{3}+10(-8) x^{2}+5 cdot 16 cdot x+1(-32))
Exercise (PageIndex{11})
Use the Binomial Theorem to expand ((x-3)^{5}).
- Answer
(egin{array}{l}{x^{5}-15 x^{4}+90 x^{3}-270 x^{2}} {+405 x-243}end{array})
Exercise (PageIndex{12})
Use the Binomial Theorem to expand ((y-1)^{6}).
- Answer
(egin{array}{l}{y^{6}-6 y^{5}+15 y^{4}-20 y^{3}+15 y^{2}} {-6 y+1}end{array})
Things can get messy when both terms have a coefficient and a variable.
Example (PageIndex{7})
Use the Binomial Theorem to expand ((2x-3y)^{4}).
Solution:
We identify the (a) and (b) of the pattern.
In our pattern, (a=2x) and (b=-3y).
We use the Binomial Theorem.
((a+b)^{n}=left( egin{array}{c}{n} {0}end{array} ight) a^{n}+left( egin{array}{c}{n} {1}end{array} ight) a^{n-1} b^{1}+left( egin{array}{c}{n} {2}end{array} ight) a^{n-2} b^{2}+ldots+left( egin{array}{c}{n} {r}end{array} ight) a^{n-r} b^{r}+ldots+left( egin{array}{c}{n} {n}end{array} ight) b^{n})
Substitute in the values (a=2x, b=-3y) and (n=4).
Simplify the exponents.
Evaluate the coefficients. Remember, (left( egin{array}{l}{n} {1}end{array} ight)=n, left( egin{array}{l}{n} {n}end{array} ight)=1, left( egin{array}{l}{n} {0}end{array} ight)=1)
Exercise (PageIndex{13})
Use the Binomial Theorem to expand ((3x-2y)^{5}).
- Answer
(egin{array}{l}{243 x^{5}-810 x^{4} y+1080 x^{3} y^{2}} {-720 x^{2} y^{3}+240 x y^{4}-32 y^{5}}end{array})
Exercise (PageIndex{14})
Use the Binomial Theorem to expand ((4x-3y)^{4}).
- Answer
(egin{array}{l}{256 x^{4}-768 x^{3} y+864 x^{2} y^{2}} {-432 x y^{3}+81 y^{4}}end{array})
The real beauty of the Binomial Theorem is that it gives a formula for any particular term of the expansion without having to compute the whole sum. Let’s look for a pattern in the Binomial Theorem.
Notice, that in each case the exponent on the (b) is one less than the number of the term. The ((r+1)^{st}) term is the term where the exponent of (b) is (r). So we can use the format of the ((r+1)^{st}) term to find the value of a specific term.
Note
Find a Specific Term in a Binomial Expansion
The ((r+1)^{s t}) term in the expansion of ((a+b)^{n}) is
(left( egin{array}{l}{n} {r}end{array} ight) a^{n-r} b^{r})
Example (PageIndex{8})
Find the fourth term of ((x+y)^{7}).
Solution:
| In our pattern, (n=7, a=x) and (b=y). |  |
We are looking for the fourth term. Since (r+1=4), then (r=3). | |
| Write the formula |  |
| Substitute in the values, (n=7, r=3, a=x), and (b=y). |  |
 |  |
| Simplify. |  |
| Simplify. |  |
Exercise (PageIndex{15})
Find the third term of ((x+y)^{6}).
- Answer
(15x^{4}y^{2})
Exercise (PageIndex{16})
Find the fifth term of ((a+b)^{8}).
- Answer
(8ab^{7})
Exercise (PageIndex{17})
Find the coefficient of the (x^{5}) term of ((x+4)^{8}).
- Answer
(7,168)
Exercise (PageIndex{18})
Find the coefficient of the (x^{4}) term of ((x+2)^{7}).
- Answer
(280)
Access these online resources for additional instruction and practice with sequences.
- Binomial Expansion Using Pascal’s Triangle
- Binomial Coefficients
Key Concepts
- Patterns in the expansion of ((a+b)^{n}(
- The number of terms is (n+1).
- The first term is (a^{n}) and the last term is (b^{n}).
- The exponents on (a) decrease by one on each term going left to right.
- The exponents on (b) increase by one on each term going left to right.
- The sum of the exponents on any term is (n).
- Pascal’s Triangle
- Binomial Coefficient (left( egin{array}{l}{mathbf{n}} {mathbf{r}}end{array} ight)) : A binomial coefficient (left( egin{array}{l}{mathbf{n}} {mathbf{r}}end{array} ight)), where (r) and (n) are integers with (0≤r≤n), is defined as
(left( egin{array}{l}{n} {r}end{array} ight)=frac{n !}{r !(n-r) !})
We read (left( egin{array}{l}{n} {r}end{array} ight)) as “(n) choose (r)” or “(n) taken (r) at a time”.
- Properties of Binomial Coefficients
(left( egin{array}{l}{n} {1}end{array} ight)=n quad left( egin{array}{l}{n} {n}end{array} ight)=1 quad left( egin{array}{l}{n} {0}end{array} ight)=1)
- Binomial Theorem:
For any real numbers (a), (b), and positive integer (n),
((a+b)^{n}=left( egin{array}{c}{n} {0}end{array} ight) a^{n}+left( egin{array}{c}{n} {1}end{array} ight) a^{n-1} b^{1}+left( egin{array}{c}{n} {2}end{array} ight) a^{n-2} b^{2}+ldots+left( egin{array}{c}{n} {r}end{array} ight) a^{n-r} b^{r}+ldots+left( egin{array}{c}{n} {n}end{array} ight) b^{n})
14.5: Binomial Theorem
Audio/Videos for Math 102
Flash audio/video lectures for Math 102. (Not for MDEV 102.)
Notes: When running a video, you can select the topic you want to watch by clicking on a topic in the table of contents. Also, each has audio so use headphones or use a computer with speakers. Most of the videos are between 3 & 7 minutes long. If a video is longer than 7 minutes, the time is listed to the side.
Course Introduction
Chapter One and Two Sets
| 1.1 Polya's Four-Step Approach | 2.3 Intersection of Sets |
| 1.1 Handshake Problem (8.5 minutes) | 2.3 Union of Sets |
| 2.3 Complement of a Set | |
| 1.2 Estimation, Rounding, and Notation | 2.3 Set Difference/Relative Complement |
| 2.3 Cardinality of the Union of Sets | |
| 2.1 What is a set? | 2.3 DeMorgan's Laws |
| 2.1 Describing a Set | 2.3 Symbolic Examples |
| 2.1 Basic Notation and Terms | 2.3 Venn Diagrams and Distributive Properties |
| 2 .2 Equality, Subset, and Proper Subset | 2.4 Survey Problem 1 (Basic) |
| 2.2 Number of Subsets of a Set | 2.4 Survey Problem 2 (Or) |
| 2.2 Equivalent Sets | 2.4 Survey Problem 3 (Inconsistent) |
| 2.2 Number of One-to-One Correspondences | |
| 2.2 Pascal's Triangle (Number of Subsets) |
Chapter Three Logic
| 1.3 Inductive Reasoning | 3.3 Truth Tables - Conditional and Biconditional |
| 1.3 Deductive Reasoning | 3.3 Ways of Stating a Conditional |
| 3.3 Define Converse, Inverse, & Contrapositive | |
| 3.1 Statements | 3.3 Equivalence - Con., Inv. & Contra. (Method 1) (Method 2) |
| 3.1 Connectives | 3.3 More Converse, Inverse & Contrapositive Examples |
| 3.1 More Samples with Connectives | 3.3 Biconditional and Equivalence (Method 1) (Method 2) |
| 3.1 Quantifiers | 3.3 Implication Property (Method 1) (Method 2) |
| 3.1 Negating Quantified Statements | 3.3 Negation of Sufficient and Necessary |
| 3.1 Consistency Between Quantified Statements | |
| 3.4 Law of Detachment & Fallacy of Converse (Md1) (Md2) | |
| 3.2 Truth Tables - Negation, Conjunction, Disjunction | 3.4 Law of Contraposition & Fallacy of Inverse (Md1) (Md2) |
| 3.2 Constructing Truth Tables (Method 1) (Method 2) | 3.4 Disjunctive Syllogism (Method 1) (Method 2) |
| 3.2 Number of Cases for a Truth Table | 3.4 Hypothetical Syllogism (Method 1) (Method 2) |
| 3.2 Tautology & Contradiction (Method 1) (Method 2) | 3.4 Truth Table for Complex Arguments (Method 1) (Method 2) |
| 3.2 Equivalence - Distributive (Method 1) (Method 2) | 3.4 T-Proof for an Argument |
| 3.2 DeMorgan's Laws (Method 1) (Method2) | 3.4 Another T-Proof for an Argument |
| 3.5 Syllogisms with Universal & Existential Quantifiers | |
| Summary - "Real" Arguments | 3.5 More Syllogisms with Universal & Existential Quantifiers |
Chapter Thirteen Counting
| 13.1 Introduction to Counting Methods | 13.3 Permutations |
| 13.3 Permutation Formulas | |
| 13.2 Fundamental Counting Principle | 13.3 Permutations with Repetitions |
| 13.2 Fundamental Counting Principle (More Examples) | 13.3 Combinations |
| 13.3 Permutations and Combinations | |
| 13.4 Poker | |
| 13.4 Powerball |
Chapter Fourteen Probability
Chapter Fifteen Descriptive Statistics
| 15.1 Surveys and Data Collection | 15.3 Measures of Dispersion |
| 15.1 Basic Definitions for Statistics | 15.3 More Measures of Dispersion |
| 15.1 Frequency and Relative Frequency Tables (7:20) | 15.3 Coefficient of Variation |
| 15.1 Bar Graphs and Histograms | |
| 15.1 Interpreting Histograms | 15.4 Introduction to the Normal Curve (7:11) |
| 15.1 Stem-and-Leaf Displays | 15.4 Introduction to z-score |
| 15.4 More z-score Examples | |
| 15.2 What is "average"? | 15.4 More z-score Applications |
| 15.2 Advantages & Disadvantages of Types of Averages | |
| 15.2 Compute Mean, Median, Mode, and Midrange | 15.5 Line of Best Fit - Motivation - Mile Run |
| 15.2 Inferences Knowing Average and More Info (7:18) | 15.5 Compute Line of Best Fit |
| 15.2 Five-Number-Summary and Box-and Whisker | 15.5 Compute Linear Correlation Coefficient |
| 15.5 Properties of the Linear Correlation Coefficient | |
| 15.5 Summary Example - First Class Stamps |
Microsoft Excel 2003 for Statistics
Microsoft Excel 2007 for Statistics
Copyright 200 6– 200 7 Timothy Peil
Unless otherwise stated, the above videos were prepared by Timothy Peil . They may be used without permission from the author for home and/or educational (non-profit) purposes only. Any other use must be approved by the author.
5 Binomial Test
The binomial test is used to look for evidence that the proportion of a Binomial distributed random variable may differ from a hypothesized (or previously observed) value.
5.1.1 Test Statistic
The test statistic for a binomial test is the observed frequency of experimental subjects that exhibit the trait of interest.
5.1.2 Definitions
Let (X) be a random variable following a binomial distribution with parameters (n) and (pi) . Let (x) be the observed frequency of experimental subjects exhibiting the trait of interest.
5.1.3 Hypotheses
The hypotheses for the Binomial test may take the following forms:
[egin [egin [egin The decision to reject the null hypothesis is made when the observed value of (x) lies in the critical region that suggests the probability of that observation is low. We define the critical region as the upper bound we are willing to accept for (alpha) , the Type I Error. In a two-sided test, the upper bound is shared equally in both tails. Due to the discrete nature of the distribution, the total probability in the tails may not equal (alpha) . The figures below depict examples of rejection regions for selected values of the Binomial distribution parameters. The decision rule is: Reject (H_0) if (x < Binomial(alpha/2, n, pi_0)) or (x > Binomial(1 - alpha/2, n, pi_0)) Figure 5.1: The examples displayed use (n = 20) . For the top, middle, and bottom examples, (pi) is set at 0.3, 0.5, and 0.75, respectively. Notice that in some cases, the rejection regions for (alpha = 0.10) and (alpha = 0.05) are identical. In the one-sided test, (alpha) is placed in only one tail. The figures below depict examples of rejection regions for selected values of the Binomial distribution parameters. In each case, (alpha) is the area in the tail of the figure. It follows, then, that the decision rule for a lower tailed test is: Reject (H_0) when (x leq Binomial(alpha, n, pi_0)) For an upper tailed test, the decision rule is: Reject (H_0) when (x geq Binomial(1 - alpha, n, pi_0)) Figure 5.2: The examples displayed use (n = 20) . For the top, middle, and bottom examples, (pi) is set at 0.3, 0.5, and 0.75, respectively. The derivations below make use of the following symbols: Two Sided Test [egin Left Sided Test [egin
Right Sided Test [egin Since the Binomial distribution is discrete, the power curve has the interesting characteristic of not being monotonic. It is sometimes described as having a “sawtooth” appearance. This behavior means that a larger sample size is not always preferred. For example, in the following figure, a sample size of 10 has better power than a sample size of 12. Figure 5.3: Power for a Binomial test with (pi_0 = .15) and (pi_a = 0.25) CARLOS A. BRAUMANN is Professor in the Department of Mathematics and member of the Research Centre in Mathematics and Applications, Universidade de Évora, Portugal. He is an elected member of the International Statistical Institute (since 1992), a former President of the European Society for Mathematical and Theoretical Biology ᆠ) and of the Portuguese Statistical Society ᆝ and 2009ᆠ), and a former member of the European Regional Committee of the Bernoulli Society ᆠ). He has dealt with stochastic differential equation (SDE) models and applications (mainly biological). Identity elements are unique. Proof: Suppose (u star a=astar u=a) and (estar a=astar e=a, forall ainRe) . Then (u=ustar e=e) . If (star) is an associative operation, inverse elements for (star) are unique. Suppose (astar a_1=a_1star a=e) and (astar a_2=a_2star a=e) , where (e) is the identity element for the operation (star) . Then (a_1=a_1star e=a_1star(astar a_2)=(a_1star a)star a_2)=estar a_2=a_2) . If (a) has an inverse (a^prime) with respect to the associative operation (star) , and (astar b=astar c) , then (b=c) . Suppose (astar b=astar c) . Then [egin In the field ((Re,+,cdot), a+b=a+cRightarrow b=c) , and if (a
eq 0) and (ab=ac) , then (b=c) . Proof: The Corollary is proved using commutativity and Left Cancellation. By axiom 4 (-0+0=0) . Because 0 is the additive identity, (-0+0=-0) . Therefore (-0=-0+0=0) . (forall ainRe, a>0) if and only if (ainRe^+) . Proof: Suppose (0<a) . Then (a-0inRe^+) , but (a-0=a) , so (ainRe^+) . Conversely, suppose (ainRe^+) . Then (a-0=ainRe^+) , so (0<a) . [egin [egin David Moore pioneered the “data analysis” (conceptual) approach for The Basic Practice Of Statistics 8th Edition Moore Pdf. Moore’s The Basic Practice Of Statistics Pdf became the market-leading bestseller by focusing on how statistics are gathered, analyzed, and applied to real problems and situations—and by confronting student anxieties about the course’s relevance and difficulties head on. This new eighth edition of The Basic Practice Of Statistics Pdf offers the same crystal clear writing and use of real data as past editions, while incorporating new and updated examples and exercises, and presenting new features to support the learning and teaching goals for today’s generation of students. The The Basic Practice Of Statistics 8th Edition Moore Pdf is written specifically to address the College Board AP® Statistics Course Description. Now the overwhelming bestseller for the course returns in a spectacular new edition. For this edition, Josh Tabor joins the author team of veteran AP® teachers who fully understand how to engage and teach high school students. With new problem-solving and test preparation features and a dramatically enhanced suite of media tools, the fifth edition provides everything teachers and students need to succeed in the course and on the AP® Statistics exam. Your students are allowed unlimited access to WebAssign courses that use this edition of the textbook at no additional cost. Additional instructional and learning resources are available with the textbook, and might include testbanks, slide presentations, online simulations, videos, and documents. Access is contingent on use of this textbook in the instructor's classroom. Moore's data analysis approach in The Basic Practice of Statistics 7th edition moves students away from formulas and number-crunching, focusing instead on how working statisticians in a variety of fields collect and analyze data, and use the results to tackle real-world problems. The WebAssign component for this text engages students with immediate feedback, an interactive eBook with online resources, and a question bank of end-of-section exercises. The St. Petersburg game was invented by the mathematician Nicolaus Bernoulli in the (18) th Century. He discussed it with his cousin Daniel Bernoulli, who published a famous solution in the St. Petersburg Academy Proceedings. (Hence the name of the game.) His solution: replace monetary value with real value, utility in other words. Recall that the more money you have, the less value additional money brings. So even though the payoffs in the St. Petersburg game double from ($2) to ($4) to ($8) etc., the real value doesn’t double. It grows more slowly. Figure 14.3: Bernoulli’s logarithmic utility function What is the real value of money then? How much utility does a gain of ($x) bring? Bernoulli proposed that utility increases “logarithmically” with money. In terms of dollars, (u($x)=log(x)) . Look at Figure 14.3 to see what this looks like. Notice how, for example, ($16) doesn’t have twice the utility of ($8) . In fact you have to go all the way up to ($64) to get twice the utility of ($8) . That’s because (64 = 8^2) , and logarithms have the property that (log(a^b) = b imes log(a)) . The difference this makes to the St. Petersburg game is displayed in Figure 14.4. The rectangles don’t all have the same area of (1) anymore, they get smaller instead. In fact, Bernoulli showed that the total area of all the rectangles is only about (1.39) . In other words, the expected utility of the St. Petersburg game is about (1.39) , the equivalent of a ($4) gain (because (log(4) approx 1.39) ). Figure 14.4: Dollars vs. utils in the St. Petersburg game So if Bernoulli is right about the utility of money, the fair price for the St. Petersburg game is only about ($4) . And in fact that’s about what most people say they would pay. Class meets: Section 3 MWF 1:10 - 2:00 in 1068 East Hall Section 7 MWF 2:10 - 3:00 in 1084 East Hall. Office Hours: Tu 3:30-5:00 pm, W 10:30 - 12:00 am, F 10:00 - 11:00 am in 4844 East Hall. Prerequisites: This course makes serious use of the material from the Calculus sequence (Math 215, 255 or 285). You should have solid knowledge of differentiation and integration, in particular of the geometric ideas behind these concepts (tangent lines, local extrema, areas, volumes). This course requires a working knowledge of multivariate differentiation and integration (including change of variables). If you do not have a strong background in Calculus, please do not take this course we do not want you to fail because of your inadequate background. Previous exposure to basic combinatorics (combinations and permutations) is generally expected. If you do not have this background and wish to take this course, please carefully study Sections 1.1-1.4 of the textbook and the self-test problems at the end of Chapter 1. Course Description: Basic concepts of probability are introduced, and applications to other sciences are noted. The emphasis is on concepts, calculations, derivations and problem-solving. Topics include methods of both discrete and continuous probability, conditional probability, independence, random variables, joint distributions, expectations, variances, covariances, and limit laws. Text: Sheldon Ross, A first course in probability. Pearson Prentice Hall, 8th ed. (2010). ISBN: 9780136033134. Missing/late work: There will be no make-up for the quizzes or exams for any reason. A missed exam counts as zero points, with the following exception. If you miss a midterm exam due to a documented medical or family emergency, the exam's weight will be added to the final exam. Late homework will not be accepted. In case of a medical or road emergency that prevents you from attending the class, you can e-mail me your scanned or typed homework as a single pdf file, along with short explanation of what happened, on the day when the homework is collected by 5 pm. Course Schedule: It is a useful practice to read ahead the sections to be covered. To see what material is coming next, have a look at my previous Math/Stats 425 course. After each class, the schedule will be updated with the description of the material we covered in class, examples and self-test problems, and new homework. It is a useful practice to work on the examples and self-test problems. Their solutions are included in the back of the textbook do not turn them in as a part of homework. Wednesday, January 4 Multiplication Principle (a.k.a. the basic principle of counting) (1.1). Permutations (1.3). Combinations (1.4 begin). Lecture notes All examples in Sections 1.2 Examples 3a, 3b, 3c 4a, 4b, 4c. Self-test problems (p.20-21): 1, 2, 3, 4, 6. Homework 1, due January 11 (p. 16-17): 3, 5, 7(a,b), 15, 17, 21. Friday, January 6 Combinations contd. (1.4). Binomial theorem. Multinomial coefficients (1.5). Lecture notes Examples 4d, 4e 5a, 5b, 5c. Self-test problems (p. 20-21):5, 8, 9, 10, 11, 12. Homework 1, due January 11 (p. 16-17): 9, 13, 28. Monday, January 9 Sample space and events. Operatoins on events (2.1-2.2). Lecture notes All Examples in 2.1-2.2. Self-test problems (p. 56-57): 1. Homework 2, due January 18 (p. 50-54): 1, 3, 6. Wednesday, January 11 Quiz 1 Axioms of probability (2.3). Properties of probability. Inclusion-Exclusion Principle (2.4 up to Example 4a). Lecture notes All examples in those sections. Self-test problems (p. 56-57): 2, 4, 14, 15. Homework 2, due January 18 (p. 50-54): 9, 11, 12 (hint: use Venn diagram for this last problem). (Do not do Problem 14 at this time it will be assigned later in Homework 3.) --> Friday, January 13 Inclusion-Exclusion Principle continued. Sample spaces having equally likely ourcomes (2.5 begins). Lecture notes Examples 5a-d, recall the Birthday Problem (Example 5i). Self-test problems (p. 56-57): 5, 6, 7, 8. Homework 2, due January 18 (p. 50-54): 8, 17, 21. Monday, January 16: no class (Martin Luther King Jr. Day) Wednesday, January 18 Quiz 2 Generalized Inclusion-Exclusion Principle (Proposition 4.4). The Matching Problem (Example 5m). More problems on equally likely outcomes. Lecture notes Examples 5e, l, n. Self-test problems (p.56-57): 10, 12, 13, 17, 18, 20. Homework 3, due January 25 (p. 50-54): 14, 23, 25, 28, 32. Friday, January 20 Conditional Probabilities. The Law of Total Probability (3.1-3.2). Lecture notes Examples 2b, e, f, the first example on p.21 of the lecture notes. Self-test problems (p.114-116): 2, 3, 5, 9a. Homework 3, due January 25 (p.102-110): 16, 19b, 21. Monday, January 23 Bayes Formula (3.3). Lecture notes Examples 3a, c, d (covered in class), e, f, k (covered in class), n. Homework 4, due February 1 (p.102-110): 15, 18, 19a, 26, 35, 90 (pass to the complement, use the inclusion-exclusion principle). Wednesday, January 25 Quiz 3 Independent events (3.4). Simple random walk. Lecture notes Examples 4b, c, g, h, i. Self-test problems (p.114-116): 4, 9b, 10, 11, 12, 15, 16 (should refer to problem 3.66b), 21, 23. Homework 4, due February 1 (p.102-110): 50, 57(a,b), 64, 66(a) (in some editions there is a typo -- this problem should refer to Figure 3.4). Friday, January 27 Random variables (4.1). Discrete random variables (4.2). Probability mass function (pmf) and cumulative distribution function (cdf). Lecture notes Examples 4a, b, c, d. No additional homework. Monday, January 30 Expected value (4.3). Examples: lottery group testing. Lecture notes Examples 3a, b, c, d. Self-test problems (p.183-185): 1, 2, 3, 4 (if you dare), 6. Homework 5, due February 8 (p.172-179): 1, 13, 17, 19. Note: the textbook often calls CDF "the distribution function". Wednesday, February 1 Quiz 4 Expected value of a function of a random variable (4.4). Variance (4.5). Lecture notes Examples 4a, 5a. Self-test problems (p.183-185): 5. Homework 5, due February 8 (p.172-179): 21, 25, 30, 35, 37. Friday, February 3 Example: the matching problem (expectation and variance). Bernoulli and binomial random variables (4.6). Lecture notes Examples 6a, b, c, e, f. Self-test problems (p.183-185): 9, 10, 11a, 12, 13. Homework 5, due February 8 (p.172-179): 42, 43, 48. Monday, February 6 Poisson distribution (4.7). Example: People vs. Collins trial. Lecture notes Examples 7a, b, c. Self-test problems (p.183-185): 14, 15. More examples are added for Feb.3 class above Homework 6, due February 15 (p.172-179): 45, 52, 53, 54, 59. Wednesday, February 8 Quiz 5 Binomial and Poisson distributions (continued). Example: jury duty letters. Lecture notes Self-test problems (p.384-387): 3, 4. Homework 6, due February 15 (p.172-179): 64(a,b) (p.373-379): 7 (express the number of chosen objects as a sum of 10 indicators), 8. Friday, February 10 Computing expectations by conditioning (from 7.5). Geometric distribution (4.8.1). Lecture notes Examples: from Section 7.5: 5c, h from section 4.8: 8a, b. Self-test problems (p.384-387): 17 (p.183-185): 22. Homework 6, due February 15 (p.373-379): 48(a,b), 53, 58 (condition on the outcome of the first flip). Monday, February 13 Continuous distributions (5.1). Uniform distribution (5.3). Lecture notes Examples: 1a, b, c (all very useful), 3c. Self-test problems (p.229-213): 1, 2, 7. Homework 7, due February 22 (p.224-227): 1, 4, 11, 13. Wednesday, February 15 Quiz 6 Transformations of random variables. Lecture notes Example from Section 5.2: 1d. Homework 7, due February 22: additional problems. Friday, February 17 Expectation of continuous random variables (5.2). Expectation and variance of the uniform distribution (Ex.3a). Lecture notes Self-test problems (p.229-231): 3, 4, 5, 6. Homework 7, due February 22 (p.224-227): 7, 14. Monday, February 20 Standard normal distribution (5.4). Lecture notes Wednesday, February 22 Midterm Exam. In class. Covers Chapters 1 - 4. Sample problems: Practice Exam (due to Prof. Montgomery): problems 1, 5. Solutions. More exam problems (due to Prof. Montgomery): try problems 1(a), 2, 3, 5, 6, 9, 10, 12. Practice Exam (due to Prof. Derksen): problems 1, 2, 4, 5. Practice Exam (due to Prof. Montgomery): problems 3(except b), 4(except d,f). Solutions. Practice Exam (due to Prof. Khoury): problems 1, 5, 8, 9, 11. Practice Exam (due to Prof. Khoury): problems 3, 4a, 5, 9, 10, 11. Practice Exam (due to Prof. Khoury): problems 2, 3, 7. Practice Exam (due to Prof. Khoury): problems 2, 8(a,b). Practice Exam (due to Prof. Khoury): problems 1, 6, 9. Practice Exam (due to Prof. Khoury): problems 1, 3, 8. Homework 1: all problems, Homework 2: 9, 11, 8, 21 Homework 3: 14, 28, 32, 16, 19, 21 Homework 4: 15, 35, 57, 66a Homework 5: 17, 19, 43, 37 Homework 6: all problems. Friday, February 24 Normal distribution with general mean and varaince (5.4). Lecture notes Examples: 4b, c, d, e. Self-test problems (p.229-231): 8, 9, 10, 11. Homework 8, due March 7 (p.224-227): 15, 16, 17, 18, 22. Winter break: February 27 - March 2 Monday, March 5 Normal approximation to the binomial distribution (5.4.1). Lecture notes Examples 4f, g, h, i. Self-test problems (p.229-231): 12, 17. Homework 9, due March 14 (p.224-227): 20, 27, 28. Wednesday, March 7 Quiz 7 Exponential distribution (5.5). Lecture notes Examples 5a, b, c, d. Self-test problems (p.229-231): 18, 19. Homework 9, due March 14 (p.224-227): 31, 32, 34, 39. Friday, March 9 Joint distributions (6.1). Lecture notes Examples 1a, b, c, d, e. Self-test problems (p.293-296): 6.2, 6.3, 6.6(a,c), 6.7(b,c,d,e,f). Homework 9, due March 14 (p.287-291):6, 8, 9(a,b,c,e), 10. Monday, March 12 Independent random variabes (6.2). Lecture notes Examples 2a, d, f. Homework 10, due March 21 (p.287-291): 15, 16, 20, 22, 23, 27. Wednesday, March 14 Quiz 8 Sums of independent random variables (6.3). Lecture notes No additional homework problems today. Friday, March 16 Sums of independent random variables (6.3). Convolutions. Sums of independent normal random variables. Lecture notes Homework 10, due March 21 (p.287-291): 29, 30, 31. Monday, March 19 Sums of independent random variables (Bimonial, Poisson, exponential). Gamma distribution (6.3 continued). Lecture notes Homework 11, due March 28 (p.287-291): 28, 33(a,b). Wednesday, March 21 Quiz 9 Conditional distributions (6.4-6.5). Lecture notes Homework 11, due March 28 (p.287-291): 40, 41, 42 (compute the conditional pdf of Y given X=x). Friday, March 23 Conditional distributions (6.4-6.5 continued). Lecture notes Homework 11, due March 28 (p.287-291): 48, and an additional problem. Monday, March 26 Transformations of joint distributions (6.7). Lecture notes Homework 12, due April 4 (p.287-291): 52, 54, 55, 56(a). Wednesday, March 28 Quiz 10 Properties of expectation (7.1). Expectation of sums of random variables (7.2-7.3). Lecture notes Homework 12, due April 4 (p.373-379): 4, 5, 11, 21. Friday, March 30 Coupon collector's problem (Section 7, Examples 3d, 2i). Expectation of product of independent random variables (Proposition 4.1). Lecture notes Homework 12, due April 4 (p.373-379): 19(a). Monday, April 2 Covariance, correlation (7.3). Lecture notes Homework 13, due April 11 (p.373-379): 50, 38, 40. Additional problem. Wednesday, April 4 Quiz 11 Properties of covariance. Variance of a sum. Lecture notes Homework 13, due April 11 (p.373-379): 6, 31, 36, 37, 42, 45, 63(a). Friday, April 6 Applications to statistics: sample mean and sample variance (Ex. 4a). Lecture notes Homework 13, due April 11 (p.373-379): 65. Monday, April 9 Moment generating functions (7.7). Lecture notes Wednesday, April 11 Quiz 12 Central Limit Theorem (8.3). Lecture notes Homework 14, will not be collected (p.413-414): 5, 7, 8, 14. Friday, April 13 Markov's and Chebyshev's inequalities. Weak law of large numbers. Monday, April 16 Review. Practice Exam 0 Practice Exam 1 Practice Exam 2 Practice Exam 3 Homework 7: 4, 7, 13, 14, additional problems 1, 2. Homework 8: 16, 22. Homework 9: 28, 32, 8, 9, 10. Homework 10: 16, 20, 22, 27, 31. Homework 11: 28, 41, 42, 48. Homework 12: 54, 56(a), 4, 5, 11, 21. Homework 13: 50, 31, 36, 37, 42, 63(a). Homework 14: 5, 7, 8, 14. Course webpage: http://www-personal.umich.edu/ romanv/teaching/2011-12/425/425.html 5.1.4 Decision Rule
5.1.5 Power
About the Author
33.4 Preliminary Results in the Field of Real Numbers
33.4.1 Theorem: Uniqueness of Identities
33.4.2 Theorem 2: Uniqueness of Inverses
33.4.3 Theorem: Left Cancellation Law
33.4.4 Corollary: Right Cancellation
33.4.5 Lemma
33.4.6 Theorem
33.4.7 Theorem
4.3 Expected Values
(Rightarrow) (x = y + 1) and (m = n - 1)
About The Basic Practice Of Statistics 8th Edition Pdf Free
14.2 Bernoulli’s Solution
Introduction to Probability - MATH/STATS 425, Winter 2012
The course covers most of Chapters 1--7 and a part of Chapter 8.
This formula sheet will be available during the exam. In addition, you may bring two index cards (3x5 in) to the exam. You may write whatever you like on one side of each card. Please only write by hand on the cards photocopyied or printed cards are not allowed. Friday, April 20 Final Exam. 1:30-3:30 pm, in 1400 CHEM. Covers the whole course.
Also see the Ctools class page which contains the archive of e-mail messages to the class, forums and chat room.
Watch the video: Probability: Binomial Theorem (May 2022).
