Slide1 : Binomial Distributions have to follow these 4 conditions:
Each observation has only 2 possibilities: success or failure.
There is a fixed number n of observations.
Each observation is independent of the others.
The probability of success, p, is the same for each observation. Chapter 8.1 - Binomial Distributions
Binomial Expansion : Binomial Expansion
(x + y)0 = 1
(x + y)1 = 1x + 1y
(x + y)2 = 1x2 + 2xy + 1y2
(x + y)3 = 1x3 + 3 x2y + 3xy2 + 1y3
(x + y)4 = 1x4 + 4 x3y + 6x2y2 +4 xy3 + 1y4
(x + y)5 = ?
Do you see Pascal’s Triangle?
Can you fill in the next line?
Page 2 : Page 2 (x + y)0 = 1
(x + y)1 = 1x + 1y
(x + y)2 = 1x2 + 2xy + 1y2
(x + y)3 = 1x3 + 3 x2y + 3xy2 + 1y3
(x + y)4 = 1x4 + 4 x3y + 6x2y2 +4 xy3 + 1y4
(x + y)5 = 1x5 + 5 x4y + 10x3y2 +10 x2y3 + 5xy4 + 1y5
Pascal’s Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
Slide4 : The coefficients of each term are combinations: the top number is the exponent of the binomial (n) and the bottom number is the term, minus 1.
The first term in the parentheses starts with the highest value of the exponent, and counts down, and the second term starts with 0 and counts up.
0C0
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
4C0 4C1 4C2 4C3 4C4
Slide5 : = 1x5 + 5 x4y +10x3y2 +10 x2y3 + 5xy4 + 1y5 1 +5 +10 +10 +5 +1 x5 x4 x3 x2 x1 x0 y0 y y2 y3 y4 y5 Expanding (x + y)5:
Slide6 : If you want to know the 4th term of the expansion of (x + y)5,
You know this will have 6 terms – one more than the exponent
The coefficient will be 5C3 = 10 (first number is exponent, 2nd is 1 less than term number
The power of x will be the exponent minus the 2nd number in the combination, ie 5-3 = 2
The power of y will be the 2nd number in the combination, ie 3 in this case
Therefore, we get as an answer: 10x2y3
Slide7 : Now, how does this apply to probability?
Well, if you let x represent the probability of success, and y the probability of failure, then: P(X = k) = pk (1-p)n-k Let’s try an example:
If your probability of shooting a free throw and making it is .7 (70%), then the probability of missing is .3 If you shoot 5 free throws, what’s the probability you make 2 of them? P(X = 2) = (.7)2 (.3)3 = 10 (.49) (.027) = .1323