The binomial distribution
Throwing five dice
Number of
sixes
Tally:
Relative
frequency
0
1
2
3
4
5
Throwing five dice
Number of
sixes
0
1
2
3
4
5
Probability
What is the probability distribution?
P  A  B  Tree
P  A  Pdiagram,
A and B
are independent
B provided two
dice
5
P(not six) 
6
P(not six) 
5
6
P(six) 
P(six) 
5 1
P(not six, six)  
6 6
1
6
P(not six) 
5 5 5
P(no sixes)     
6 6 6
5
6
1
6
1
P(six) 
6
2

5 1

P
(1
six)

2



6 6


1 5
P(six, not six)  
6 6
1 1 1
P(two sixes)     
6 6 6
2
2
2
5 1 1
5
P(any outcome)     2       1
6 6 6
6
The probability of all the
equivalent paths is the same.
We multiply by the number
of possible paths.
Tree diagram, three dice
NS
5 5 5 5
P(no sixes)      
6 6 6 6
3
NS
2
5
P(not six) 
6
NS
5 1
P(one six)  3    
6 6
S
3
2
3
S2
5 1
5
5 1
1
P(any outcome)     3      3         1
6  6 NS  6 
6
6 6
2
5 1
NS
P(two sixes)  3    
1
S
6 6
P(six) 
6
NS
S
S
1 1 1 1
P(3 sixes)      
6 6 6 6
3
Two dice:
Three dice:
Number of sixes
0
Probability
5
 
6
Number of sixes
0
Probability
“Pascal’s
triangle”
5
 
6
2
1
2
5 1
2 
6 6
1
 
6
1
3
2
2
2
5 1
3   
6 6
5 1
3   
6 6
3
2
1
 
6
3
Four dice:
Number of
sixes
0
1
2
3
4
Probability
Five dice:
Number of
sixes
Probability
0
1
2
3
4
5
Probability distribution for the number of sixes in 30 throws.
Why “binomial”?
5
Let p  = probability of 1, 2, 3, 4, or 5
6
1
Let q  = probability of 6
6
2
2
p

q

p

2
pq

q


2
3
2
2
3
p

q

p

3
p
q

3
pq

q


3
4
3
2 2
3
4
p

q

p

4
p
q

6
p
q

4
pq

q


4
p  q
5
 p  5p q  10 p q  10 p q  5pq  q
5
P(no sixes, five
of anything else)
4
3 2
2 3
4
P(five sixes)
5
P  X  r   Cr p q
n
r
nr
Probability of two sixes from 5 dice:
2
1  5
P  X  2   C2    
6 6
3
5
• X = the result of the experiment (number of
successes)
• r = a number in the range 0 to n
n
• C r "n choose r" is the number of ways of spelling a
word ppqqq with r letter "p"s and n-r letter "q"s.
• n = number of trials
• p = probability of each “success”
• q = 1-p = probability of each “failure”
n
C r calculates the numbers in Pascal's triangle
1
C 0  1,
1
C 0  1,
2
C 0  1,
3
4
C 0  1,
4
C1  2,
2
C1  3,
3
C1  4,
All in this row will be 5C something
C1  1
1
4
C2  1
2
C2  3,
2
C2  6,
4
C3  1
3
C3  4,
4
C4  1
Success or failure?
• With 5 dice if just two are a “six”, the other 3
are “not six”:
P  2 sixes   P  3 not six 
The formula gives the same probability because
Pascal's triangle is symmetrical, 5C2  5C 3 ,
2
3
3
1 5 5  5 1
C2      C 3    
6 6
6 6
5
2
A range of outcomes
• With 20 coins, if we get 1 heads, there must
be 19 tails:
P 1 head  P 19 tails 
• A cumulative probability is for a range of
outcomes:
P  X  2   P  X  0   P  X  1  P  X  2 
P  >2 heads   1  P   2 heads 