“Bi” means “two” (like a bicycle has two wheels) … … so this is about things with two results. 

Tossing a Coin:


We say the probability of the coin landing H is ½And the probability of the coin landing T is ½ 
Throwing a Die:


We say the probability of a four is 1/6 (one of the six faces is a four).And the probability of not four is 5/6 (five of the six faces are not a four) 
Let’s Toss a Coin!
Toss a fair coin three times … what is the chance of getting two Heads?
We will use these terms:
 Outcome: the result of three coin tosses
 Event: “Two Heads” out of three coin tosses
Tosing a coin three times could get any one of these outcomes (H is for heads and T for Tails):
HHH  
HHT  
HTH  
HTT  
THH  
THT  
TTH  
TTT 
Which outcomes do we want?
“Two Heads” could be in any order: “HHT”, “THH” and “HTH” all have two Heads (and one Tail).
So 3 of the outcomes produce “Two Heads”.
What is the probability of each outcome?
Each outcome is equally likely, and there are 8 of them. So each has a probability of 1/8
So the probability of event “Two Heads” is:
Number ofoutcomes we want  Probability ofeach outcome  
3  ×  1/8  = 3/8 
Let’s Calculate Them All:
The calculations are (P means “Probability of”):
 P(Three Heads) = P(HHH) = 1/8
 P(Two Heads) = P(HHT) + P(HTH) + P(THH) = 1/8 + 1/8 + 1/8 = 3/8
 P(One Head) = P(HTT) + P(THT) + P(TTH) = 1/8 + 1/8 + 1/8 = 3/8
 P(Zero Heads) = P(TTT) = 1/8
We can write this in terms of a Random Variable, X, = “The number of Heads from 3 tosses of a coin”:
 P(X = 3) = 1/8
 P(X = 2) = 3/8
 P(X = 1) = 3/8
 P(X = 0) = 1/8
And we can also draw a Bar Graph:
It is symmetrical!
Making a Formula
Now … what are the chances of 5 heads in 9 tosses … to list all outcomes (512) would take a long time!
So let’s make a formula.
In our previous example, how could we get the values 1, 3, 3 and 1 ?
They are actually in the third row of Pascal’s Triangle … ! 
Can we make them using a formula?
Sure we can, and here it is:  


It is often called “n choose k” and you can read more about it at Combinations and Permutations. 
Note: the “!” means “factorial”, for example 4! = 1×2×3×4 = 24
Let’s use it:
Example: 3 tosses getting 2 Heads
We have n=3 and k=2
n!  =  3!  =  3×2×1  = 3 
k!(nk)!  2!(32)!  2×1 × 1 
So there are 3 outcomes for “2 Heads”
(We knew that already, but now we have a formula for it.)
Let’s use it for a harder question:
Example: what are the chances of 5 heads in 9 tosses?
We have n=9 and k=5
n!  =  9!  =  9×8×7×6×5×4×3×2×1  = 126 
k!(nk)!  5!(95)!  5×4×3×2×1 × 4×3×2×1 
And for 9 tosses there are 2^{9} = 512 total outcomes, so we get the probability:
Number ofoutcomes we want  Probability ofeach outcome  
126  ×  1  =  126 
512  512 
P(X=5) =  126  =  63  = 0.24609375 
512  256 
About a 25% chance.
(Easier than listing them all.)
Bias!
So far the chances of success or failure have been equally likely.
But what if the coins are biased (land more on one side than another) or choices are not 50/50.
Example: You sell sandwiches. 70% of people choose chicken, the rest choose pork.
What is the probability of selling 2 chicken sandwiches to the next 3 customers?
This is just like the heads and tails example, but with 70/30 instead of 50/50.
Let’s draw a tree diagram:
The “Two Chicken” cases are highlighted.
Notice that the probabilities for “two chickens” all work out to be 0.147, because we are multiplying two 0.7s and one 0.3 in each case.
Can we get the 0.147 from a formula? What we want is “two 0.7s and one 0.3”
 0.7 is the probability of each choice we want, call it p
 2 is the number of choices we want, call it k
And
 The probability of the opposite choice is: 1p
 The total number of choices is: n
 The number of opposite choices is: nk
So all choices together is:
p^{k}(1p)^{(nk)}
Example: (continued)
 p = 0.7 (chance of chicken)
 n = 3
 k = 2
So we get:
p^{k}(1p)^{(nk)} = 0.7^{2}(10.7)^{(32)} = 0.7^{2}(0.3)^{(1)} = 0.7 × 0.7 × 0.3 = 0.147
which is the probability of each outcome.
And the total number of those outcomes is:
n!  =  3!  =  3×2×1  = 3 
k!(nk)!  2!(32)!  2×1 × 1 
And we get:
Number of outcomes we want  Probability of each outcome  
3  ×  0.147  =  0.441 
So the probability of event “2 people out of 3 choose chicken” = 0.441
OK. That was a lot of work for something we knew already, but now we can answer harder questions.
Example: You say “70% choose chicken, so 7 of the next 10 customers should choose chicken” … what are the chances you are right?
 p = 0.7
 n = 10
 k = 7
So we get:
p^{k}(1p)^{(nk)} = 0.7^{7}(10.7)^{(107)} = 0.7^{7}(0.3)^{(3)} = 0.0022235661
That is the probability of each outcome.
And the total number of those outcomes is:
n!  =  10!  =  10×9×8  = 120 
k!(nk)!  7!(103)!  3×2×1 
And we get:
Number of outcomes we want  Probability of each outcome  
120  ×  0.0022235661  =  0.266827932 
In fact the probability of 7 out of 10 choosing chicken is only about 27%
Moral of the story: even though the longrun average is 70%, don’t expect 7 out of the next 10.
Putting it Together
Now we know how to calculate how many:
n! 
k!(nk)! 
And the probability of each:
p^{k}(1p)^{(nk)}
We can multiply them together:
Probability of k out of n ways:
P(k out of n) =  n!  p^{k}(1p)^{(nk)}  
k!(nk)! 
The General Binomial Probability Formula
Important Notes:
 The trials are independent,
 There are only two possible outcomes at each trial,
 The probability of “success” at each trial is constant.
Throw the Die
A fair die is thrown four times. Calculate the probabilities of getting:
 0 Twos
 1 Two
 2 Twos
 3 Twos
 4 Twos
In this case n=4, p = P(Two) = 1/6
X is the Random Variable ‘Number of Twos from four throws’.
Substitute x = 0 to 4 into the formula:
P(k out of n) =  n!  p^{k}(1p)^{(nk)} 
k!(nk)! 
Like this (to 4 decimal places):
 P(X = 0) = (4!/0!4!) × (1/6)^{0}(5/6)^{4} = 1 × 1 × (5/6)^{4} = 0.4823
 P(X = 1) = (4!/1!3!) × (1/6)^{1}(5/6)^{3} = 4 × (1/6) × (5/6)^{3} = 0.3858
 P(X = 2) = (4!/2!2!) × (1/6)^{2}(5/6)^{2} = 6 × (1/6)^{2} × (5/6)^{2} = 0.1157
 P(X = 3) = (4!/3!1!) × (1/6)^{3}(5/6)^{1} = 4 × (1/6)^{3} × (5/6) = 0.0154
 P(X = 4) = (4!/4!0!) × (1/6)^{4}(5/6)^{0} = 1 × (1/6)^{4} × 1 = 0.0008
Summary: “for the 4 throws, there is a 48% chance of no twos, 39% chance of 1 two, 12% chance of 2 twos, 1.5% chance of 3 twos, and a tiny 0.08% chance of all throws being a two (but it still could happen!)”
This time the Bar Graph is not symmetrical:
It is not symmetrical!
It is skewed because p is not 0.5
Sports Bikes
Your company makes sports bikes. 90% pass final inspection (and 10% fail and need to be fixed).
What is the expected Mean and Variance of the 4 next inspections?
First, let’s calculate all probabilities.
 n = 4,
 p = P(Pass) = 0.9
X is the Random Variable “Number of passes from four inspections”.
Substitute x = 0 to 4 into the formula:
P(k out of n) =  n!  p^{k}(1p)^{(nk)} 
k!(nk)! 
Like this:
 P(X = 0) = (4!/0!4!) × 0.9^{0}0.1^{4} = 1 × 1 × 0.0001 = 0.0001
 P(X = 1) = (4!/1!3!) × 0.9^{1}0.1^{3} = 4 × 0.9 × 0.001 = 0.0036
 P(X = 2) = (4!/2!2!) × 0.9^{2}0.1^{2} = 6 × 0.81 × 0.01 = 0.0486
 P(X = 3) = (4!/3!2!) × 0.9^{3}0.1^{1} = 4 × 0.729 × 0.1 = 0.2916
 P(X = 4) = (4!/4!0!) × 0.9^{4}0.1^{0} = 1 × 0.6561 × 1 = 0.6561
Summary: “for the 4 next bikes, there is a tiny 0.01% chance of no passes, 0.36% chance of 1 pass, 5% chance of 2 passes, 29% chance of 3 passes, and a whopping 66% chance they all pass the inspection.”
Mean, Variance and Standard Deviation
Let’s calculate the Mean, Variance and Standard Deviation for the Sports Bike inspections.
There are (relatively) simple formulas for them. They are a little hard to prove, but they do work!
The mean, or “expected value”, is:
μ = np
For the sports bikes:
μ = 4 × 0.9 = 3.6
So we would expect 3.6 bikes (out of 4) to pass the inspection.
Makes sense really … 0.9 chance for each bike times 4 bikes equals 3.6
The formula for Variance is:
Variance: σ^{2} = np(1p)
And Standard Deviation is the square root of variance:
σ = √(np(1p))
For the sports bikes:
Variance: σ^{2} = 4 × 0.9 × 0.1 = 0.36
Standard Deviation is:
σ = √(0.36) = 0.6
Note: we could also calculate them manually, by making a table like this:
X  P(X)  X × P(X)  X^{2} × P(X) 
0  0.0001  0  0 
1  0.0036  0.0036  0.0036 
2  0.0486  0.0972  0.1944 
3  0.2916  0.8748  2.6244 
4  0.6561  2.6244  10.4976 
SUM:  3.6  13.32 
The mean is the Sum of (X × P(X)):
μ = 3.6
The variance is the Sum of (X^{2} × P(X)) minus Mean^{2}:
Variance: σ^{2} = 13.32 − 3.6^{2} = 0.36
Standard Deviation is:
σ = √(0.36) = 0.6
And we got the same results as before (yay!)
Summary
The General Binomial Probability Formula
P(k out of n) =  n!  p^{k}(1p)^{(nk)} 
k!(nk)! 
Mean value of X: μ = np
Variance of X: σ^{2} = np(1p)
Standard Deviation of X: σ = √(np(1p))