Lesson #8

Introduction toProbability

Flip a single coin:

S = { T , H }

Or, if we let X = # heads,

S = { 0 , 1 }

P = P(Event) = P(A) = mn

0 m n

0 p 1

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 }

n! = n(n - 1)(n - 2) … 1

Combinations - order irrelevant

Choose r objects from n, without replacement

nCr or n

r

Permutations - order is important

nPr

AB BAAC CAAD DABC CBBD DBCD DC

4P2 = 12

4C2 = = 64

2

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)

!n!

n - r

For any combination (subset) of r objects,

there are r! arrangements or permutations.

nnCr = =

r

nPr =

r!n!

(n - r)! r!

52 52! = =

47! 5!5

(52)(51)(50)(49)(48)(47!)

(5)(4)(3)(2)(1)(47!)

311,875, 200=

120 2,598 = ,960

51 =

4

249,900

P(Ace of hearts) = 2,5249

98,900,960

= .0962