Download Pascal*s Triangle and Binomial Theorem

Pascal’s Triangle and
Binomial Theorem
Binomial Expansion
Warm up
• If a card from a standard deck of 52 cards is drawn, what is the
probability that a 4, 10 or Jack is selected?
• If the probability of having a baby boy is .55, what is the probability of
having three baby girls?
• Solutions : (1)
12
52
=
3
13
(2) The probability of having a girl is .45, so the probability
of having three girls in a row is .45 x .45 x .45 = .0911 =
about 9%
Review
• A random sample of 30 people are asked if they prefer the color red
over the color blue. If the probability of each individual is .5, what is
the probability that:
• 12 will prefer red?
• At most 13 will prefer red?
• At least 20 will prefer blue?
Review (continued)
• Solutions:
• (1) REMEMBER – This is a PDF problem because it computes the exact
probability for one outcome or value
• In the calculator: 2nd VARS -> binompdf -> Enter the following: (sample size, probability,
number of successes or failures
• In this example it will look like this: binompdf(30,.5,12) = .08 = 8%
• (2) REMEMBER – This is a CDF problem because we are looking at a range of
values (not looking for one exact amount)
• In the calculator: 2nd VARS -> binomcdf -> Enter the following: (sample size, probability,
number of successes or failure
• In this example it will look like this: binomcdf(30,.5,13) = .2923 = 29%
Review (continued)
• (3) CDF or PDF problem?
• Since CDF always finds probabilities that are ‘less than or equal to’, and we are looking
for values that are 20 or greater, we need to do 1-CDF
• In the calculator: 1 - 2nd VARS -> binomcdf -> Enter the following: (sample size,
probability, number of successes or failures
• In this example, it will look like this: 1-binomcdf(30,.5, 20) = .021 = 2%
• *NOTE – Questions that ask for ‘less than 6’ or ‘more than
6’ (or any other number) DO NOT include the 6 when
finding the probability because we are only want numbers
less than or more than 6
Pascal’s Triangle
• What are some things you notice?
Pascal’s Triangle (continued)
• Start each triangle with a triangle of three 1s
• The outside edges of the triangle (start and end of each row) will all be
ones
• The rest of the numbers are the two numbers above it added together
• https://en.wikipedia.org/wiki/Pascal%27s_triangle#/media/File:PascalTrian
gleAnimated2.gif
How can we use this? – Binomial Expansion
• Recall –
• Find the following values: 21 , 22 , 20
• 21 = 2
• 22 = 4
• 20 = 1
• Try the following: (𝑥 + 3)1 and (𝑥 + 3)2
• What if we wanted to find (𝑥 + 3)3 ?
Binomial Expansion (continued)
• We can use Pascal’s triangle to expand binomials more easily
Let’s try (𝑥 + 6)4
- The row we will be using corresponds to the power of the equation
- In this case we use row 4
Binomial Expansion (continued)
(𝑥 + 6)4 :
1
4
6
4
1
*
*
*
*
*
𝑥4
𝑥3
𝑥2
𝑥1
𝑥0
*
*
*
*
*
60
61
62
63
64
=
=
=
=
=
𝑥4
24𝑥 3
216𝑥 2
864𝑥 1
1296
= 𝒙𝟒 + 𝟐𝟒𝒙𝟑 + 𝟐𝟏𝟔𝒙𝟐 + 𝟖𝟔𝟒𝒙 + 𝟏𝟐𝟗𝟔
**NOTE 1 – If we were expanding (2𝑥 + 6)4 , the second column would have
2x instead of x
**NOTE 2 – Sign DOES matter: If we had (𝑥 − 6)4 , we would use -6 in the last
column instead of 6
**Note 3 – Order DOES matter!
Fill in the blanks
• Use Pascal’s Triangle and Binomial expansion to fill in the missing
values for (𝑥 + 3)3
30
1
𝑥2
3
1
32
𝑥0
Now solve:
= 𝑥 3 + 9𝑥 2 + 27𝑥 + 27
Practice
• Use binomial expansion to solve (𝑥 + 4)4
= 𝑥 4 + 16𝑥 3 + 96𝑥 2 + 256𝑥 + 256
- Find the 4th term of (3 − 2𝑥)5
1st term
1
x
35 x
−2𝑥 0
2nd term
5
x
34 x
−2𝑥 1
3rd term
10 x
33 x
−2𝑥 2
4th term
10 x
32 x
−2𝑥 3
5th term
5
x
31 x
−2𝑥 4
6th term
1
x
30 x
−2𝑥 5
4th Term = (10)(32 ) −2𝑥 3 = 10 9 −8𝑥 3 = −𝟕𝟐𝟎𝒙𝟑
Practice (continued)
- Continuing our previous problem of (3 − 2𝑥)5 , what is our 3rd coefficient?
1st term
1 x
35 x −2𝑥 0
2nd term
5 x
34 x −2𝑥 1
3rd term
10 x 33 x −2𝑥 2
4th term
10 x 32 x −2𝑥 3
5th term
5 x
31 x −2𝑥 4
6th term
1 x
30 x −2𝑥 5
= (10)(33 )(−2𝑥 2 )
= (10)(27)(4𝑥 2 ) = 1,080𝑥 2
Our coefficient is 1,080
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