Steph makes \(90\%\), percent of the free-throws she attempts. She is going to shoot \(3\) free-throws. Assume that the results of free-throws are independent from each other.
Find the probability that she makes her first \(2\) free-throws and misses her third free-throw. Round your answer to the nearest hundredth if necessary. P(make, make miss)= ? Find the probability that she makes any \(2\) out of \(3\) free throws. Source: khanacademy.org |
We have learned that combinations can be used to determine the number of ways objects can be selected from a group of objects without regard to order. But there are other types of problems where combinations can be used. In this lesson, we will use combinations to help us expand a binomial.
At this point, you might be asking yourself, “What is a binomial?” or “What does it mean to expand a binomial?”
A binomial is an expression with two terms joined by addition or subtraction. For example, \(x+2\) or \(3x-7\) or \(x^2+3x\).
To expand a binomial means to multiply the expression by itself a set number of times, which is indicated by an exponent.
For example, \(\left(x+y\right)^5\) means \(\left(x+y\right)\left(x+y\right)\left(x+y\right)\left(x+y\right)\left(x+y\right)\).
We could multiply this the long way because, of course, there is no distribution rule for exponents when the terms are added or subtracted. But would we really want to do all this work? Probably not, especially as the exponent increases.
Look at the pattern that develops when expanding \(\left(x+y\right)^n\) where \(n=2,3,4,5\)
\(\left(x+y\right)^2=\left(x+y\right)\left(x+y\right)=x^2+2xy+y^2\)
\(\left(x+y\right)^3=\left(x+y\right)^2\left(x+y\right)=\left(x^2+2xy+y^2\right)\left(x+y\right)=x^3+3x^2y+3xy^2+y^3\)
\(\left(x+y\right)^4=\left(x+y\right)^3\left(x+y\right)=\left(x^3+3x^2y+3xy^2+y^3\right)\left(x+y\right)=x^4+4x^3y+6x^2y^2+4xy^3+y^4\)
\(\left(x+y\right)^5=\left(x+y\right)^4\left(x+y\right)=\left(x^4+4x^3y+6x^2y^2+4xy^3+y^4\right)\left(x+y\right)=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\)
The Binomial Theorem (or Binomial Expansion) gives us a method to expand \(\left(x+y\right)^n\), which results in a sum of terms of the form \(ax^by^c\), where \(a\) is called the coefficient of expansion (and we can use combinations for this), \(b\) and \(c\) are integer exponents \(\ge0\). We just need to figure out a way to find the coefficient of expansion, and the powers of each term in the binomial to set this problem up, in addition to the number of terms in the expansion.
You may have noticed a few patterns that occur in the above expansion:
In addition,
Example 1: Expand the binomial \(\left(x+y\right)^5\) (the expansion will have \(6\) terms)
\(\left(x+y\right)^5\ =_5C_0x^5y^0+_5C_1 x^4y^1+_5C_2 x^3y^2+_5C_3x^2y^3+_5C_4x^1y^4+_5C_5x^0y^5\)
\(\left(x+y\right)^5=1x^5y^0+5x^4y^1+10x^3y^2+10x^2y^3+5x^1y^4+1x^0y^5\)
\(\left(x+y\right)^5\ =x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\)
It is actually much easier to understand the process if you hear it explained, so let’s look at a video example.
Example 2: Expand the binomial \(\left(2x-3\right)^4\)
At this point, you might be asking yourself, “What is a binomial?” or “What does it mean to expand a binomial?”
A binomial is an expression with two terms joined by addition or subtraction. For example, \(x+2\) or \(3x-7\) or \(x^2+3x\).
To expand a binomial means to multiply the expression by itself a set number of times, which is indicated by an exponent.
For example, \(\left(x+y\right)^5\) means \(\left(x+y\right)\left(x+y\right)\left(x+y\right)\left(x+y\right)\left(x+y\right)\).
We could multiply this the long way because, of course, there is no distribution rule for exponents when the terms are added or subtracted. But would we really want to do all this work? Probably not, especially as the exponent increases.
Look at the pattern that develops when expanding \(\left(x+y\right)^n\) where \(n=2,3,4,5\)
\(\left(x+y\right)^2=\left(x+y\right)\left(x+y\right)=x^2+2xy+y^2\)
\(\left(x+y\right)^3=\left(x+y\right)^2\left(x+y\right)=\left(x^2+2xy+y^2\right)\left(x+y\right)=x^3+3x^2y+3xy^2+y^3\)
\(\left(x+y\right)^4=\left(x+y\right)^3\left(x+y\right)=\left(x^3+3x^2y+3xy^2+y^3\right)\left(x+y\right)=x^4+4x^3y+6x^2y^2+4xy^3+y^4\)
\(\left(x+y\right)^5=\left(x+y\right)^4\left(x+y\right)=\left(x^4+4x^3y+6x^2y^2+4xy^3+y^4\right)\left(x+y\right)=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\)
The Binomial Theorem (or Binomial Expansion) gives us a method to expand \(\left(x+y\right)^n\), which results in a sum of terms of the form \(ax^by^c\), where \(a\) is called the coefficient of expansion (and we can use combinations for this), \(b\) and \(c\) are integer exponents \(\ge0\). We just need to figure out a way to find the coefficient of expansion, and the powers of each term in the binomial to set this problem up, in addition to the number of terms in the expansion.
You may have noticed a few patterns that occur in the above expansion:
- The number of terms in the expansion of the binomial is equal to \(n + 1\) (one more than the power the binomial is raised to).
- The sum of the exponents in each term will equal \(n\) (the power the binomial is raised to).
- The power of the first term in the binomial will decrease by \(1\) from \(n\) to \(0\).
- The power of the second term in the binomial will increase by \(1\) from \(0\) to \(n\).
In addition,
- The coefficients are found using combinations: \(_nC_0,\ _nC_1,\ _nC_2,\ ...,\ _nC_n\) (we will look at another method to find the coefficients as well).
- If the terms in the binomial are subtracted, the signs in the expansion will alternate between positive and negative.
Example 1: Expand the binomial \(\left(x+y\right)^5\) (the expansion will have \(6\) terms)
\(\left(x+y\right)^5\ =_5C_0x^5y^0+_5C_1 x^4y^1+_5C_2 x^3y^2+_5C_3x^2y^3+_5C_4x^1y^4+_5C_5x^0y^5\)
\(\left(x+y\right)^5=1x^5y^0+5x^4y^1+10x^3y^2+10x^2y^3+5x^1y^4+1x^0y^5\)
\(\left(x+y\right)^5\ =x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\)
It is actually much easier to understand the process if you hear it explained, so let’s look at a video example.
Example 2: Expand the binomial \(\left(2x-3\right)^4\)
There is another way to determine the coefficients of expansion, which is using Pascal’s Triangle. You will need to be able to draw the triangle, so pay attention to the pattern below.
Once you write the first several rows of Pascal’s Triangle, you select the row that contains the same number of terms as the expansions. Instead of using combinations, you can use the numbers in the rows as the coefficients.
Example 3: Expand the binomial \(\left(3x+1\right)^5\) using Pascal’s Triangle.
EXTENSION: Expand the binomial \(\left(B+G\right)^4\) as a model representing the number of boys (\(B\)) and the number of girls (\(G\)) in a family with \(4\) children. Then determine the number of ways that the family could have \(3\) boys and one girl. Can you find the probability that a family with \(4\) children will have \(3\) boys and one girl?
The last type of problem we will investigate is how to find a specific term of the expansion without actually carrying out the entire process.
Example 4: Find the \(8th\) term in the expansion of \(\left(3x-5y\right)^{12}\).
- The coefficient of expansion is \(_{12}C_7\) because when we expand the entire binomial, we begin the \(r\) value at \(0\). So the \(r\) value in the coefficient is one less than the term’s position in the list.
- The power for the first term, \(3x\), is \(n – r\). In this case, \(12 – 7 = 5\).
- Since the exponents must add to \(n\), the power of the second term, \(-5y\), is \(r\). In this case \(7\).
\(\begin{align}&_{12}C_7\left(3x\right)^{12-7}\left(-5y\right)^7\ & \ \ \ &\text{1) Find the coefficient of expansion and powers for the terms of the binomial.}\\
&792\left(3x\right)^5\left(-5y\right)^7\ & \ \ \ &\text{2) Calculate the coefficient.}\\
&792\cdot3^5x^5\cdot\left(-5\right)^7y^7\ & \ \ \ &\text{3) Distribute the exponents.}\\
&792\cdot243\cdot\left(-78,125\right)x^5y^7\ & \ \ \ &\text{4) Find the value of the powers.}\\
&-15,035,625,000x^5y^7\ & \ \ \ &\text{5) Multiply the numbers together.}\end{align}\)
If you like using formulas you can use this: \(_nC_r\left(1^{st}\ term\right)^{n-r}\left(2^{nd\ }term\right)^r\), where \(r\) is the position in the list minus one.
Example 5: Find the \(4th\) term in the expansion of \(\left(3a-5\right)^7\).