If the probability that James is late home from work on any
day is 0.4, what is the probability that he is late home
three times in a five-day working week?
L L L L’ L’ = 0.4 x 0.4 x 0.4 x 0.6 x 0.6 = 0.02304
How many ways could James be late for 3 days out of 5?
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 5 10 10 5 1
5L 4L
3L 2L 1L 0L
P(3 Lates in 5 days) = 10 x 0.02304
= 0.2304
This example is one of a particular type of problem where
there are only two possible outcomes – a BINOMIAL
problem.
If there are a large number of stages (or trials), it can be
difficult to rely on Pascal’s triangle
The number of paths giving r occurrences out of n cases is:
nC
r
=
𝑛
𝑟
𝑛!
=
𝑟! 𝑛 − 𝑟 !
The BINOMIAL PROBABILITY DISTRIBUTION is defined as:
𝒏
𝑷 𝑿=𝒓 =
𝒑𝒓 (𝟏 − 𝒑)𝒏−𝒓 𝒇𝒐𝒓 𝒓 = 𝟎, 𝟏, 𝟐, 𝟑 … 𝒏
𝒓
n stands for the number of trials
p stands for the probability of ‘success’
The particular model can be summarised as:
𝑿 ~ 𝑩(𝒏, 𝒑)
The binomial distribution is a DISCRETE distribution.
If 25 dice are thrown, find the probability that three
sixes are obtained.
n = 25
p=
𝟏
𝑿 ~ 𝑩 𝟐𝟓,
𝟔
1
6
𝑷 𝑿=𝟑 =
𝟐𝟓
𝟑
𝟏
𝟔
𝟑
𝟓
𝟔
𝟐𝟐
= 𝟎. 𝟏𝟗𝟑
You can use the binomial distribution for any situation
where you want to count the number of times a particular
outcome is observed out of a fixed number of cases –
provided certain conditions are satisfied…




There is a fixed number of trials
Each trial has the same two possible outcomes
The outcomes of the trials are independent of one another
The probability of ‘success’ remains constant
You can use standard tables to answer some binomial
problems…
𝑿 ~ 𝑩(𝟓, 𝟎. 𝟑𝟓)
What is (a) P(X ≤ 3)
(a) P(X ≤ 3) = 0.9460
(b) P(X = 3)
(c) P(X ≥ 3)
(b) P(X = 3) = P(X ≤ 3) – P(X ≤ 2)
= 0.9460 – 0.7648
(c) P(X ≥ 3) = 1 – P(X ≤ 2)
= 0.1812
= 1 – 0.7648 = 0.2352
If 𝑋 ~ 𝐵
𝑋~𝐵
3
30,
4
3
30,
4
find (a) P(X ≤ 17)
(b) P(X > 24)
𝑌~𝐵
1
30,
4
(a) P(X ≤ 17)
(a) P(Y ≥ 13)
(b) P(X > 24)
= 1 – P(Y ≤ 12)
= 1 – 0.9784
= 0.0216
(b) P(Y ≤ 5)
= 0.2026
On 40% if the days that Sean travels to work he finds he
has to stop at a particular set of traffic lights.
Find the probability that he has to stop at these lights
no more that five times during a month in which he
works 20 days.
𝑋 ~ 𝐵 20, 0.4
𝑃(𝑋 ≤ 5) = 0.1256
MEAN AND VARIANCE OF THE BINOMIAL DISTRIBUTION
If 𝑿 ~ 𝑩(𝒏, 𝒑)
E(X) = np
Var(X) = np(1 – p)
If X ~ B(10, 0.2) find the mean and variance of X
E(X) = np = 10 x 0.2 = 2
Var(X) = np(1 – p)
= 10 x 0.2 x 0.8 = 1.6
If X ~ B(80, 0.4) find the mean and standard deviation of X
E(X) = np = 80 x 0.4 = 32
Var(X) = np(1 – p)
= 80 x 0.4 x 0.6 = 19.2
Standard Deviation = 𝑉𝑎𝑟(𝑋) = 19.2 = 4.38
X is a binomial distribution with mean 8 and variance 6.4.
Find P(X ≤ 3).
np = 8
np(1 – p) = 6.4
8(1 – p) = 6.4
1 – p = 0.8
𝑋 ~ 𝐵 40, 0.2
P(X ≤ 3) = 0.0285
p = 0.2
n = 40
For the following random variables state whether they can
be modelled by a binomial distribution. If they can, give the
model, if they cannot then explain why.
(a) A dice is thrown repeatedly until a 1 is seen.
X = number of throws.
(b) A dice is thrown 10 times. X = number of 1’s seen
(c) A bag has 25 red and 25 blue balls in it. Five balls are
taken out without replacement.
X = number of red balls taken
(d) X = number of boys in a family of five children
(e) A pair of dice is thrown 25 times.
X = number of times a double is thrown
(f) A pair of dice is thrown 25 times.
X = average score of the sum of the
numbers showing.
(a) A dice is thrown repeatedly until a 1 is seen.
X = number of throws.
NOT binomial
No fixed number of trials




There is a fixed number of trials
Each trial has the same two possible outcomes
The outcomes of the trials are independent of one another
The probability of ‘success’ remains constant
(b) A dice is thrown 10 times. X = number of 1’s seen
Binomial
𝟏
𝑿 ~ 𝑩 𝟏𝟎,
𝟔




There is a fixed number of trials
Each trial has the same two possible outcomes
The outcomes of the trials are independent of one another
The probability of ‘success’ remains constant
(c) A bag has 25 red and 25 blue balls in it. Five balls are
taken out without replacement.
X = number of red balls taken
NOT binomial
P(success) NOT constant
Outcomes of trials NOT independent




There is a fixed number of trials
Each trial has the same two possible outcomes
The outcomes of the trials are independent of one another
The probability of ‘success’ remains constant
(d) X = number of boys in a family of five children
Binomial
𝟏
𝑿 ~ 𝑩 𝟓,
𝟐




There is a fixed number of trials
Each trial has the same two possible outcomes
The outcomes of the trials are independent of one another
The probability of ‘success’ remains constant
(e) A pair of dice is thrown 25 times.
X = number of times a double is thrown
Binomial
𝟏
𝑿 ~ 𝑩 𝟐𝟓,
𝟔




There is a fixed number of trials
Each trial has the same two possible outcomes
The outcomes of the trials are independent of one another
The probability of ‘success’ remains constant
(f) A pair of dice is thrown 25 times.
X = average score of the sum of the numbers showing.
NOT binomial
Different outcomes – not counting how many
times something happens.




There is a fixed number of trials
Each trial has the same two possible outcomes
The outcomes of the trials are independent of one another
The probability of ‘success’ remains constant
A vet thinks that the number of male puppies in litters of a
Given size will follow a binomial distribution with p = 0.5.
(a) In litters of six puppies, what would be the mean and
variance of the number of males if the distribution is
binomial?
The vet records the number of males in 82 litters of six
Puppies and the results are summarised in the table:
Males
Frequency
0
8
1
10
2
16
3
15
4
14
5
12
6
7
(b) Calculate the mean and variance of the number of
males in litters of six puppies.
(c) Do you think the binomial distribution is a good model
for the number of males in a litter of
puppies?
A vet thinks that the number of male puppies in litters of a
Given size will follow a binomial distribution with p = 0.5.
(a) In litters of six puppies, what would be the mean and
variance of the number of males if the distribution is
binomial?
𝑿 ~ 𝑩 𝟔, 𝟎. 𝟓
E(X) = np = 6 x 0.5
=3
Var(X) = np(1 – p)
= 6 x 0.5 x 0.5
= 1.5
The vet records the number of males in 82 litters of six
Puppies and the results are summarised in the table:
Males
Frequency
0
8
1
10
2
16
3
15
4
14
5
12
6
7
(b) Calculate the mean and variance of the number of
males in litters of six puppies.
245
𝑓𝑥
=
𝑀𝑒𝑎𝑛 =
= 2.99
82
𝑓
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 =
𝑓𝑥 2
𝑓
−
𝑓𝑥
𝑓
2
985
245
=
−
82
82
= 3.09
2
(c) Do you think the binomial distribution is a good model
for the number of males in a litter of puppies?
Mean
Variance
Model
Actual
3
2.99
1.5
3.09
NO – Mean is a good match but variance is
significantly bigger than model predicts.
If X ~ B(40, p) and Var(X) = 9.6
(a) Find the two possible values of p
(b) For each of the values of p find P(X < μ – σ)
(a) 40p(1 – p) = 9.6
40p – 40p2 = 9.6
40p2 – 40p + 9.6 = 0
40 ± (−40)2 −4(40)(9.6)
𝑝=
80
𝑝 = 0.4 𝑜𝑟 0.6
If X ~ B(40, p) and Var(X) = 9.6
(a) Find the two possible values of p
(b) For each of the values of p find P(X < μ – σ)
(b)
Var(X) = 9.6
σ = 3.098
𝑋~𝐵(40, 0.4)
𝑝 = 0.4
𝑃 𝑋 < 16 − 3.098
𝑝 = 0.6
𝜇 = 40 × 0.4 = 16
= 𝑃 𝑋 < 12.902
𝑋~𝐵(40, 0.6)
= 𝑃 𝑋 ≤ 12
= 0.1285
𝜇 = 40 × 0.6 = 24
𝑃 𝑋 < 24 − 3.098 = 𝑃 𝑋 < 20.902 = 𝑃 𝑋 ≤ 20
𝑋~𝐵(40, 0.6) 𝑌~𝐵(40, 0.4)
𝑃 𝑋 ≤ 20
𝑃 𝑌 ≥ 20 = 1 − 𝑃 𝑌 ≤ 19
= 1 − 0.8702 = 0.1298