1. Binomial Distribution
Elements of Binomial Distribution (Bernouli Process)
n Identical Trials
2 Outcomes: Success and Failure P(S) = p P(F) = q
Constant Probability of Success
Independent Trials
( ½ + ½)2 = 1/ /4 + 1/4
( ½ + ½ )3 = 1/8 + 3/8 + 3/8 + 1/8

2. Binomial Probability Formula:
Example: S = Bus Major p = n = 4 q = .80
P4(0) = 4C0(.2)0(.8)4 =
P4(1) = 4C1(.2)1(.8)3 =
P4(2) = 4C2(.2)2(.8)2 =
P4(3) = 4C3(.2)3(.8)1 =
P4(4) = 4C4(.2)4(.8)0 =

3. Example: 5 Coin Toss p = ½ n = 5 q = ½
P5(0) = 5C0(1/2)0(1/2)5 =
P5(1) = 5C1(1/2)1(1/2)4 =
P5(2) = 5C2(1/2)2(1/2)3 =
P5(3) = 5C3(1/2)3(1/2)2 =
P5(4) = 5C4(1/2)4(1/2)1 =
P5(5) = 5C5(1/2)5(1/2)0 =

4. Mean for Binomial Distribution - µ = n•p
Variance for Binomial Distribution – σ2 = n•p•q
S – Bus Student n = 20 p = .2 q = .8
P20(5) = 20C5(.2)5(.8)15 =
S – Female Student n = 20 p = .6 q = .4
P20(6) = 20C6(.6)6(.4)14 =
Binomial Probability Table -

5. S – Business Major n = 10 p = .2 q = .8
P10(x = 2) =
P10(x > 2) = 1 – P10(x≤2) =
P10(x ≤ 2) = P(0) + P(1) + P(2)
P10(4 ≤ x ≤ 7) = P(4) + P(5) + P(6) + P(7)

6. Poisson Distribution –
QC Defect Rate
Inventory Withdrawals
Service Center Arrivals