the binomial distribution · 2016-03-03 · vce maths methods - unit 4 - the binomial distribution...

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


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


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.


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.


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


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


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.


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%


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)


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


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).


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


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.



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.



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.

the binomial distribution · 2016-03-03 · vce maths methods - unit 4 - the binomial distribution...

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