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VCE Maths Methods - Unit 4 - The binomial distribution
The binomial distribution
• The coin toss - three coins• The coin toss - four coins• The binomial probability distribution• Rolling dice• Using the TI nSpire• Graph of binomial distribution• Mean & standard deviation
1
VCE Maths Methods - Unit 4 - The binomial distribution
H
T
H
T
H
T
The coin toss - three coins
2
• Three coins are tossed. What is the probability distribution of X (the number of heads?)
H
T
H
T
H
T
H
T
Pr(X=1)=
38
Pr(X=0)=
18
Pr(X=2)=
38
Pr(X=3)=
18
Heads & tails are equally likely. All outcomes are equally likely.
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
VCE Maths Methods - Unit 4 - The binomial distribution
The binomial distribution
3
• The coin toss is an example of a Bernoulli sequence.• Repeated trials where only two distinct outcomes (success or failure)
are possible.• Trials are independent; each trial has the same probability of a successful
outcome.• The binomial distribution describes the chances of geing each possible
value of X; the number of successful outcomes from the number of trials.• The binomial distribution is an example of a discrete probability
distribution.• Coin toss, dice roll or yes / no options will follow a binomial distribution.
VCE Maths Methods - Unit 4 - The binomial distribution
The coin toss - four coins
4
• What if there were four coins?• There are now 16 (24) possible outcomes.• There are now 6 possible outcomes with two heads.
HHTT
Pr(X=2)=
616
= 38
HTHT
HTTH
THHT
THTH
TTHH
There are six different ways that the two heads can be
arranged in the four places.
VCE Maths Methods - Unit 4 - The binomial distribution
The binomial probability distribution
5
• This can be found using the idea of Combinations.• There are six combinations that can be made by choosing two from four
objects.
4C2 =
4!2!2!
=244
=6 Pr(X =2)= 4C2 ×
12
⎛⎝
⎞⎠
2
× 12
⎛⎝
⎞⎠
2
= 616
Chance of success (heads)
Chance of failure (tails)
Number of possible arrangements
Number of failures (tails)Number of successes
(heads)
Pr(X = x )= nC x × p( )x× 1− p( )n−x
Number of trials
VCE Maths Methods - Unit 4 - The binomial distribution
Pascal’s triangle
6
2 trials
3 trials
4 trials
1 trial
(0 trials)
5 trials
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
5 10 10 51 1
X=0
X=1
X=2
X=3
X=4
X=5
VCE Maths Methods - Unit 4 - The binomial distribution
The binomial probability distribution
7
• Four coins: The probability distribution for the number of heads has #ve possible values (0-4).
Pr(X =0)= 4C0 ×
12
⎛⎝
⎞⎠
0
× 12
⎛⎝
⎞⎠
4
= 116
Pr(X =1)= 4C1×
12
⎛⎝
⎞⎠
1
× 12
⎛⎝
⎞⎠
3
= 416
= 14
Pr(X =2)= 4C2 ×
12
⎛⎝
⎞⎠
2
× 12
⎛⎝
⎞⎠
2
= 616
= 38
Pr(X =3)= 4C3 ×
12
⎛⎝
⎞⎠
3
× 12
⎛⎝
⎞⎠
1
= 416
= 14
Pr(X =4)= 4C 4 ×
12
⎛⎝
⎞⎠
4
× 12
⎛⎝
⎞⎠
0
= 116
This distribution is symmetrical when the chances
of success (head) or failure (tail) are equal.
The sum of all probabilities must be equal to one.
VCE Maths Methods - Unit 4 - The binomial distribution
Rolling dice
8
• Four dice are rolled. What is the distribution that describes the number of sixes rolled?
Pr(X =0)= 4C0 ×
16
⎛⎝
⎞⎠
0
× 56
⎛⎝
⎞⎠
4
Pr(X =1)= 4C1×
16
⎛⎝
⎞⎠
1
× 56
⎛⎝
⎞⎠
3
Pr(X =2)= 4C2 ×
16
⎛⎝
⎞⎠
2
× 56
⎛⎝
⎞⎠
2
Pr(X =3)= 4C3 ×
16
⎛⎝
⎞⎠
3
× 56
⎛⎝
⎞⎠
1
Pr(X =4)= 4C 4 ×
16
⎛⎝
⎞⎠
4
× 56
⎛⎝
⎞⎠
0
=1× 625
1296=48.2%
=4× 125
1296=38.6%
=6× 25
1296=11.6%
=4× 5
1296=1.5%
=1× 1
1296=0.08%
VCE Maths Methods - Unit 4 - The binomial distribution
How many trials?
9
• Each roll of the die has 1/6 chance of rolling a 6. How many trials (rolls) are needed to have a 90% probability of rolling a 6 at least once?
• At least one 6 rolled in x trials includes all possible options expect for rolling none.
Pr(X =0)= 1
6⎛⎝
⎞⎠
0
× 56
⎛⎝
⎞⎠
x
= 56
⎛⎝
⎞⎠
x
Pr(X >0)=1− 5
6⎛⎝
⎞⎠
x
0.9=1− 5
6⎛⎝
⎞⎠
x
0.1= 5
6⎛⎝
⎞⎠
x
log(0.1)= x log
56
⎛⎝
⎞⎠
x = log(0.1)log(5 / 6)
≈13
(Or solve on a CAS calculator.)
(Always round up)
VCE Maths Methods - Unit 4 - The binomial distribution
How many trials?
10
• Each roll of the die has 1/6 chance of rolling a 6. How many rolls are needed to have a 90% probability of rolling a 6 at least twice?
Pr(X ≥2)=90%
n ≈22 (Solve on a CAS calculator.)
(Always round up)
Pr(X <2)=10%=Pr(X =0)+Pr(X =1)
10%= nC0 ×
16
⎛⎝
⎞⎠
0 56
⎛⎝
⎞⎠
n
+ nC1×16
⎛⎝
⎞⎠
1 56
⎛⎝
⎞⎠
n−1
10%=1×1× 5
6⎛⎝
⎞⎠
n
+n× 16
⎛⎝
⎞⎠
1 56
⎛⎝
⎞⎠
n−1
0.10= 5
6⎛⎝
⎞⎠
n
+n× 16
⎛⎝
⎞⎠
1 56
⎛⎝
⎞⎠
n−1
VCE Maths Methods - Unit 4 - The binomial distribution
Using the TI-nSpire
11
Pr(X =0)= 4C0 ×
16
⎛⎝
⎞⎠
0
× 56
⎛⎝
⎞⎠
4
=48.2%
• CAS calculators and spreadsheets have a built in function for binomial distributions.
• For example Pr (X=0) from four dice:Menu > 5: Probability > 5: Distributions > D: Binomial Pdf
• The cumulative distribution (Cdf) is used to #nd the sum of all probabilities above or below a value of X eg Pr (X≤3).
VCE Maths Methods - Unit 4 - The binomial distribution
Mean & standard deviation
12
• A probability distribution has a mean (expected value) and standard deviation (average variation from mean). For the binomial distribution:
E(X )=np Var(X )=np(1− p )
The highest variance is when there is 50% chance
of success.
Four coin toss: E(X )=4×0.5=2
Var(X )=4×0.5×0.5=1
SD(X )= 1=1
Rolling four dice: E(X )=4× 1
6= 2
3=0.67
Var(X )=4× 1
6× 5
6= 20
36=0.56
SD(X )= 0.56 =0.75
VCE Maths Methods - Unit 4 - The binomial distribution
Graphs of probability distributions
13
0%
10%
20%
30%
40%
0 1 2 3 4 5 6 7 8 9 10
Pr (X=x)
x
Pr = 0.5
• This distribution is for ten trials.• If the chance of success is 50%, the distribution is symmetrical.
VCE Maths Methods - Unit 4 - The binomial distribution
Graphs of probability distributions
14
0%
10%
20%
30%
40%
0 1 2 3 4 5 6 7 8 9 10
Pr (X=x)
x
Pr = 0.2
• If the chance of success is less than 50%, the distribution is skewed towards the right.
• The expected value of X is less than the middle value.
Values too small to see.
VCE Maths Methods - Unit 4 - The binomial distribution
Graphs of probability distributions
15
0%
10%
20%
30%
40%
0 1 2 3 4 5 6 7 8 9 10
Pr (X=x)
x
Pr = 0.8
• If the chance of success is more than 50%, the distribution is skewed towards the left.
• The expected value of X is greater than the middle value.
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