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# Probability - PowerPoint PPT Presentation

Probability. Using dice. One Die. Chance of rolling a 6 with single die p = 1/6 Chance of not rolling a 6 p’ = 1-p. Two Die. Chance of rolling double 6’s with two die p = (1/6)(1/6) = 1/36 Chance of not rolling a double 6’s p’ = 1-p. Three Die, Three Sixes.

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## PowerPoint Slideshow about 'Probability' - katina

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### Probability

Using dice

Chance of rolling a 6 with single die

p = 1/6

Chance of not rolling a 6

p’ = 1-p

Chance of rolling double 6’s with two die

p = (1/6)(1/6) = 1/36

Chance of not rolling a double 6’s

p’ = 1-p

Chance of rolling all 6’s with three die

p = (1/6)(1/6)(1/6) = 1/216

Chance of not rolling all 6’s

p’ = 1-p

n Die, n Sixes

Chance of rolling all 6’s with n die

p = (1/6)n

Chance of not rolling all 6’s

p’ = 1-p

Chance of rolling exactly two 6’s with three die

p =(1/6)(1/6)(5/6)? No! dice aren’t ordered

This is probability of rolling double 6’s with two die followed by not a six on the third die.

We don’t care whether first, second, or third die is not a 6

Chance of rolling exactly two 6’s with three die

We don’t care whether first, second, or third die is not a 6 so:

p =(5/6)(1/6)(1/6)+(1/6)(5/6)(1/6)+(1/6)(1/6)(5/6)

Or

p = 3 (1/6)(1/6)(5/6)

= 15/216

At least two 6’s with three die

3 * (1/6)(1/6)(6/6) ? No! Double counts trip 6’s

At least two 6’s with three die

(1/6)(1/6)(1)+2(5/6)(1/6)(1/6) = 1/36+10/216

= 16/216

= 2/27

Or easier: (Two 6’s or Three 6’s)

3(5/6)(1/6)(1/6)+(1/6)(1/6)(1/6) = 15/216 + 1/216

= 16/216

= 2/27

Let p = 1/6

Let q = 1- p

Let B be the number of ways we can choose exactly x sixes out of a population of n die

= B (px q(n-x))

Let p = 1/6

Let q = 1- p

Let Bi be the number of ways we can choose exactly i sixes out of a population of n die

Sum for all i in x..n

{ (Bi (pi q(n-i)) }

Given population n

Given x samples

(nx ) = BiCoef(n,x) = n! / x!(n-x)!

http://mathworld.wolfram.com/BinomialCoefficient.html

Chose 2 from { A, B, C, D }

{A, B}, {A, C }, { A, D }, {B, C }, {B, D }, {C, D}

BiCoef(4,2) = 6

4!

2!(4-2)!

Chose 3 from { A, B, C, D, E }

BiCoef(5,3) = 10

5!

3!(5-3)!

Chose 3 from { A, B, C, D, E, F }

BiCoef(6,3) = 20

6!

3!(6-3)!

Chose 4 from { A, B, C, D, E, F, G, H }

BiCoef(8,4) = 70

8!

4!(8-4)!

35 users

Each user active 10% of the time

Find probability that 10 users are active

BiCo(35,10) (.1)10(.9)25

35 users

Each user active 10% of the time

Find probability at least 10 are active

BiCo(35,10) (.1)10(.9)25 +BiCo(35,11) (.1)11(.9)24 +…

Or 1 - probability less than 10 are active

Probability at least 10 are active

+ probability less than 10 are active

= 100%

BiCo(35,9) (.1)9(.9)26 +BiCo(35,8) (.1)8(.9)27 +…