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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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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 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
Three Die, Two Sixes 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
Three Die, Two Sixes 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
Three Die, Two or Three Sixes At least two 6’s with three die Don’t care about third die! 3 * (1/6)(1/6)(6/6) ? No! Double counts trip 6’s
Three Die, Two or Three Sixes 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
What About n Die, x Sixes? 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))
What About n Die, >= x Sixes? 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)) }
Binomial Coefficient Given population n Given x samples (nx ) = BiCoef(n,x) = n! / x!(n-x)! http://mathworld.wolfram.com/BinomialCoefficient.html
Chose 2 from 4 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 5 Chose 3 from { A, B, C, D, E } ABC,ABD,ABE,ACD,ACE,ADE,BCD,BCE,BDE,CDE BiCoef(5,3) = 10 5! 3!(5-3)!
Chose 3 from 6 Chose 3 from { A, B, C, D, E, F } BiCoef(6,3) = 20 6! 3!(6-3)!
Chose 4 from 8 Chose 4 from { A, B, C, D, E, F, G, H } BiCoef(8,4) = 70 8! 4!(8-4)!
Example 35 users Each user active 10% of the time Find probability that 10 users are active BiCo(35,10) (.1)10(.9)25
Example 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 +…