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Lesson #8 Introduction to Probability. Flip a single coin:. S = { T , H }. Or, if we let X = # heads, . S = { 0 , 1 }. P = P(Event) = P(A) =. 0 m n. 0 p 1. . REMEMBER!. 0 p 1. S = { TT , TH , HT , HH }. S = { 0 , 1 , 2 }.
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Lesson #8 Introduction to Probability
Flip a single coin: S = { T , H } Or, if we let X = # heads, S = { 0 , 1 }
P = P(Event) = P(A) = 0 m n 0 p 1
REMEMBER! 0 p 1
S= { TT , TH , HT , HH } S= { 0 , 1 , 2 }
S= { TT , TH , HT , HH } S= { 0 , 1 , 2 }
S= { TT , TH , HT , HH } S= { 0 , 1 , 2 }
S= { TT , TH , HT , HH } S= { 0 , 1 , 2 }
S= { TT , TH , HT , HH } S= { 0 , 1 , 2 }
S= { TT , TH , HT , HH } S= { 0 , 1 , 2 }
S= { TT , TH , HT , HH } S= { 0 , 1 , 2 }
nCr or n! = n(n - 1)(n - 2) … 1 Choose r objects from n, without replacement Combinations - order irrelevant Permutations - order is important nPr
4C2= = 6 AB BA AC CA AD DA BC CB BD DB CD DC 4P2= 12
In general, nPr = n(n-1)(n-2) … (n-r+1) (n-r)(n-r-1) … (1) (n-r)(n-r-1) … (1)
In general, nPr = n(n-1)(n-2) … (n-r-1) (n-r)(n-r-1) … (1) (n-r)(n-r-1) … (1)
For any combination (subset) ofrobjects, there arer!arrangements or permutations.
(47!) (5)(4)(3)(2)(1) (47!) 249,900 = .0962 P(Ace of hearts) =