The Additive Rule and the Binomial Distribution – An Example

We have a two color bowl, with equal probabilities for each color: blue at 50% and green at 50%.

Our experiment consists of five draws with replacement from the bowl – we track the number of blue draws among the five.

How likely is a blue count of zero?

There is only one way to get zero blues, and that is to get five greens. Under the mutual independence of the five draws induced by sampling with replacement, we can use the multiplication rule and write

Pr{Blue Count = 0} = Pr{GGGGG} = Pr{Draw Green on 1^st Draw}* Pr{Draw Green on 2^nd Draw}* Pr{Draw Green on 3^rd Draw}* Pr{Draw Green on 4^th Draw}* Pr{Draw Green on 5^th Draw}.

But the probability of drawing green on any particular draw is .50, so:

Pr{Blue Count = 0} = (.50)* (.50)* (.50)* (.50)* (.50) = (1/2)⁵ = 1/2⁵ = 1/(2*2*2*2*2) = 1/32 » .03125 (3.125%)

How likely is a blue count of one?

There are five ways to get exactly one blue draw in five: BGGGG, GBGGG, GGBGG, GGGBG and GGGGB. We can then write:

Pr{Blue Count = 1} = Pr{BGGGG}+Pr{GBGGG}+Pr{GGBGG}+Pr{GGGBG}+Pr{GGGGB}.

Under the mutual independence of the five draws induced by sampling with replacement, we can use the multiplication rule and write:

Pr{BGGGG} = Pr{Draw Blue on 1^st Draw}* Pr{Draw Green on 2^nd Draw}* Pr{Draw Green on 3^rd Draw}* Pr{Draw Green on 4^th Draw}* Pr{Draw Green on 5^th Draw}.

But the probability of drawing green on any particular draw is .50, and the probability of drawing blue on any particular draw is .50:

Pr{BGGGG} = (.50)* (.50)* (.50)* (.50)* (.50) = (1/2)⁵ = 1/2⁵ = 1/(2*2*2*2*2) = 1/32 » .03125 (3.125%).

Pr{GBGGG} = (.50)*(.50)* (.50)* (.50)* (.50) = (1/2)⁵ = 1/2⁵ = 1/(2*2*2*2*2) = 1/32 » .03125 (3.125%).

Pr{GGBGG} = (.50)* (.50)* (.50)* (.50)* (.50) = (1/2)⁵ = 1/2⁵ = 1/(2*2*2*2*2) = 1/32 » .03125 (3.125%).

Pr{GGGBG}= (.50)* (.50)* (.50)* (.50)* (.50) = (1/2)⁵ = 1/2⁵ = 1/(2*2*2*2*2) = 1/32 » .03125 (3.125%).

Pr{GGGGB} = (.50)* (.50)* (.50)* (.50)* (.50) = (1/2)⁵ = 1/2⁵ = 1/(2*2*2*2*2) = 1/32 » .03125 (3.125%).

So then:

Pr{Blue Count = 1} = Pr{BGGGG} + Pr{GBGGG} +Pr{GGBGG} + Pr{GGGBG}+ Pr{GGGGB}

and then

Pr{Blue Count = 1} = (.50)* (.50)* (.50)* (.50)* (.50) + (.50)*(.50)* (.50)* (.50)* (.50) + (.50)* (.50)* (.50)* (.50)* (.50) + (.50)* (.50)* (.50)* (.50)* (.50) + (.50)* (.50)* (.50)* (.50)* (.50) = 5*(1/2)⁵ = 5*(1/2⁵)= 5/(2*2*2*2*2) = 5/32 » .15625 (15.625%).

The position of the blue and greens among the five draws is irrelevant: if we have b blue draws and g green draws, then the probability is (.50)^b(.50)^g

How likely is a blue count of two?

How can two blue draws show among five draws? Ten ways: BBGGG, GBBGG, GGBBG, GGGBB; BGBGG,GBGBG, GGBGB, BGGBG,GBGGB, BGGGB

The probability for each way is (.50)²(.50)³, so

Pr{Blue Count = 2} = Pr{BBGGG}+ Pr{GBBGG}+ Pr{GGBBG}+ Pr{GGGBB}+ Pr{BGBGG}+ Pr{GBGBG}+ Pr{GGBGB}+ Pr{BGGBG}+ Pr{GBGGB}+ Pr{BGGGB}

Pr{Blue Count = 2} = 10*(.50)²(.50)³ = 10*(1/32) = 10/32 = .3125 (31.25%)

How likely is a blue count of three?

How can three blue draws show among five draws? Ten ways, since this is the same as placing two green draws among the five draws.

GGBBB, BGGBB, BBGGB, BBBGG; GBGBB,BGBGB, BBGBG, GBBGB,BGBBG, GBBBG

The probability for each way is (.50)³(.50)², so

Pr{Blue Count = 3} = Pr{GGBBB}+Pr{BGGBB}+Pr{BBGGB}+Pr{BBBGG}+Pr{GBGBB}+Pr{BGBGB}+Pr{BBGBG}+Pr{GBBGB}+Pr{BGBBG}+Pr{GBBBG}

Pr{Blue Count = 3} = 10*(.50)³(.50)², = 10*(1/32) = 10/32 = .3125 (31.25%)

How likely is a blue count of four?

This is the same as getting exactly one green, and there are five ways to get exactly one green draw in five: GBBBB, BGBBB, BBGBB, BBBGB and BBBBG. We can then write:

Pr{Green Count = 1} = Pr{GBBBB}+Pr{BGBBB}+Pr{BBGBB}+Pr{BBBGB}+Pr{BBBBG}.

Under the mutual independence of the five draws induced by sampling with replacement, we can use the multiplication rule and write:

Pr{GBBBB} = Pr{Draw Green on 1^st Draw}* Pr{Draw Blue on 2^nd Draw}* Pr{Draw Blue on 3^rd Draw}* Pr{Draw Blue on 4^th Draw}* Pr{Draw Blue on 5^th Draw}.

But the probability of drawing green on any particular draw is .50, and the probability of drawing blue on any particular draw is .50:

Pr{GBBBB} = (.50)* (.50)* (.50)* (.50)* (.50) = (1/2)⁵ = 1/2⁵ = 1/(2*2*2*2*2) = 1/32 » .03125 (3.125%).

Pr{BGBBB} = (.50)*(.50)* (.50)* (.50)* (.50) = (1/2)⁵ = 1/2⁵ = 1/(2*2*2*2*2) = 1/32 » .03125 (3.125%).

Pr{BBGBB} = (.50)* (.50)* (.50)* (.50)* (.50) = (1/2)⁵ = 1/2⁵ = 1/(2*2*2*2*2) = 1/32 » .03125 (3.125%).

Pr{BBBGB}= (.50)* (.50)* (.50)* (.50)* (.50) = (1/2)⁵ = 1/2⁵ = 1/(2*2*2*2*2) = 1/32 » .03125 (3.125%).

Pr{BBBBG} = (.50)* (.50)* (.50)* (.50)* (.50) = (1/2)⁵ = 1/2⁵ = 1/(2*2*2*2*2) = 1/32 » .03125 (3.125%).

So then:

Pr{Blue Count = 4} = Pr{Green Count = 1} = Pr{ GBBBB } + Pr{ BGBBB } + Pr{ BBGBB } + Pr{ BBBGB } + Pr{ BBBBG }

and then

Pr{Blue Count = 4} = Pr{Green Count = 1} = 5*(.50)⁴(.50)¹, = 5*(1/32) = 5/32 = .15625 (15.625%)

How likely is a blue count of five?

There is only one way to get five blues, and that is to get zero greens. Under the mutual independence of the five draws induced by sampling with replacement, we can use the multiplication rule and write

Pr{Blue Count = 5} = Pr{BBBBB} = Pr{Draw Blue on 1^st Draw}* Pr{Draw Blue on 2^nd Draw}* Pr{Draw Blue on 3^rd Draw}* Pr{Draw Blue on 4^th Draw}* Pr{Draw Blue on 5^th Draw}.

But the probability of drawing blue on any particular draw is .50, so:

Pr{Blue Count = 5} = (.50)* (.50)* (.50)* (.50)* (.50) = (1/2)⁵ = 1/2⁵ = 1/(2*2*2*2*2) = 1/32 » .03125 (3.125%)

In total:

Pr{Blue Count = 0} = 1/32 » .03125 (3.125%)

Pr{Blue Count = 1} = 5/32 » .15625 (15.625%).

Pr{Blue Count = 2} = 10/32 = .3125 (31.25%)

Pr{Blue Count = 3} = 10/32 = .3125 (31.25%)

Pr{Blue Count = 4} = 5/32 = .15625 (15.625%)

Pr{Blue Count = 5} = 1/32 » .03125 (3.125%)

There is a general formula for probabilities in this sort of experiment:

Pr{Event Count = k of n Trials} = C(n,k)*(Pr{Event})^k* ( 1 – Pr{Event})^{(n – k)}

where C(n,k) = (n*(n-1)*(n-2)*…*1)/[(k*(k-1)*(k-2)*…*1)((n-k)*(n-k-1)*(n-k-2)*…*1)]

There is shorthand for n*(n-1)*(n-2)*…*1: n!, known as n factorial.

Pr{Event Count = k of n Trials} = [n!/((n-k)!k!]*(Pr{Event})^k* ( 1 – Pr{Event})^{(n – k)}

For example:

5! = 5*4*3*2*1 = 20*3*2*1 = 60*2*1 = 120

4! = 4*3*2*1 = 12*2*1 = 24*1 = 24

3! = 3*2*1 = 6*1 = 6

2! = 2*1 = 2

1! = 1

0! =1 (By definition)

Expected Count for Blue

Expected Blue Count for n=5 is

E_Blue,n=5 =

0*Pr{Blue Count = 0} + 1*Pr{Blue Count = 1} + 2*Pr{Blue Count = 2} + 3*Pr{Blue Count = 3} + 4*Pr{Blue Count = 4}

+ 5*Pr{Blue Count = 5} =

0*(1/32) + 1*(5/32) + 2*(10/32) + 3*(10/32) + 4*(5/32) + 5*(1/32) = 80/32 = 2.5

E = nP = 5(.50) = 2.5