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12.1 Discrete Probability Distributions
(Binomial Distribution)
1
At the end of the lesson, students will be able to
(c)Understand the Binomial distribution B(n,p)
(d) find the mean and variance of Binomial
distribution.
2
BERNOULLI TRIAL

Bernoulli trial is a trial that has 2 possible
outcomes
or
3

The Bernoulli trial is repeated n times and
each trial is independent. The probability of
a success stays the same for each trial.

The number of successes (X) are recorded
as a random variable.
Intro Bernoulli
4
Probability distribution for X is a binomial
distribution
 Denoted by X ~ B( n, p ) where
n = the number of independent trials
p = the probability of success

X~B(n,p) read as ‘X is binomially distributed
where n is the number of trials and p is the
probability of success’
5

The binomial probability distribution is given by
n
x n- x
P(X = x) = Cxp q
x = 0, 1, 2, 3, … , n
Where
x = the number of success
n = the number of trials
p = the probability of success
q=1–p
Binomial Example
6
EXAMPLE 1
A fair coin is tossed three times. Find the
probability of getting
(a) no head,
(b) exactly one head,
(c) exactly two heads,
(d) exactly three heads.
7
SOLUTION
If X represents the number of heads
obtained, then X~B(3, 0.5)
P(X=x) =
n
x
Cxp (1 - p)
(a) P(X = 0) =
(b) P(X = 1) =
3
n- x
0
3- 0
= 0.125
1
3- 1
= 0.375
C0 (0.5) (0.5)
3
C1 (0.5) (0.5)
8
(c) P(X = 2) =
3
2
3- 2
= 0.375
3
3- 3
= 0.125
C2 (0.5) (0.5)
3
(d) P(X = 3) = C3 (0.5) (0.5)
9
EXAMPLE 2
In a box, 20% of bulbs are damaged. If a
random sample of 10 bulbs is taken, find the
probability that
(a) exactly 3 bulbs are damaged.
(b) all bulbs are good.
(c) not more three bulbs are damaged.
10
SOLUTION
If X represents the number of damaged
bulbs, then X~B(10, 0.2)
(a) P(X = 3) =
10
10
3
7
C3 (0.2) (0.8) = 0.2013
0
10
(b) P(X = 0) = C0 (0.2) (0.8)
= 0.1074
11
(c) P(X ≤ 3) = P(X = 0) + P(X = 1) +
P(X = 2) + P(X = 3)
=
10
C0 (0.2)0 (0.8)10 +
+ 10 C2 (0.2)2 (0.8)8 +
10
10
C1 (0.2)1 (0.8)9
C3 (0.2)3 (0.8)7
= 0.1074 + 0.2684 +
0.3020 + 0.2013
= 0.8791
12
EXAMPLE 3
The probability that students watch a certain
TV program is 0.4. If eight students are
selected at random, find the probability that
(a) 5 students watch the TV program.
(b) less than 4 students watch the TV
program.
(c) at least one student watches the TV
program.
13
SOLUTION
If X represents the number of students watch
a certain TV program, then X~B(8, 0.4) and
8
x
8- x
P(X = x) = Cx (0.4) (0.6)
(a) P(X = 5) =
8
5
8- 5
5
3
C5 (0.4) (0.6)
8
= C5 (0.4) (0.6)
= 0.1239
14
(b) P(X < 4) = P(X = 0) + P(X = 1) +
P(X = 2) + P(X = 3)
8
0
8
8
1
= C0 (0.4) (0.6) + C1 (0.4) (0.6)
8
2
6
8
3
7
+ C2 (0.4) (0.6) + C3 (0.4) (0.6)
5
= 0.01679 + 0.08957 + 0.20901
+ 0.27869
= 0.5941
15
(c) P(X ≥ 1) = 1 – P(X < 1)
= 1 – P(X = 0)
= 1-
8
0
C0 (0.4) (0.6)
8
= 0.9832
16
MEAN AND VARIANCE OF BINOMIAL
DISTRIBUTION
If X ~ B(n, p), then
(i) mean, μ = E(X) = np,
(ii) variances, σ2 = Var(X) = npq
*Notes: q = 1 – p
17
EXAMPLE 4
A fair die is rolled 300 times. Let X
represents the number of times the number
‘6’ is observed. Find the mean and standard
deviation.
18
SOLUTION
If X represents the number of times the
number 6 is observed, then
æ
X ~ B çç300,
çè
ö
1÷
÷
ø
6÷
æ1 ö
÷
= 50
Mean, μ = np = (300) ç
çç ÷
è6 ÷
ø
19
æ1 öæ5 ö
÷
= 41.67
Variance, σ2 = npq = (300) ççç ÷ç
֍
÷
è 6 ÷ç
øè 6 ÷
ø
Standard Deviation, σ =
=
Var(X)
41.67
= 6.45
20
EXAMPLE 5
A study revealed that two of five families
have a video recorder.
(a) If there are 4 families, find the
probability
that two or more families have
video recorders.
(b) If there are 900 families, calculate the
mean and standard deviation of the
numbers of families who have video
recorders
21
SOLUTION
If X represents the number of families who
have video recorders, then X~B(4, 0.4) and
P(X = x) = 4 Cx (0.4)x (0.6)44
x
2
2
4
3
1
(a) P(X ≥ 2) = C2 (0.4) (0.6) + C3 (0.4) (0.6)
4
4
+ C4 (0.4) (0.6)
0
= 0.5248
22
(b)
X~B(900, 0.4)
μ = np = (900)(0.4) = 360
σ=
npq =
(900)(0.4)(0.6) = 14.70
23
BINOMIAL PROBABILITY TABLES
To find probability for the value of
p from 0.001 to 0.5
n
P(X ³ r) =
å
n
Crp (1 - p)
r
n- r
X= r
P(X ≥ r) is ‘Probability of X is larger and
Equals to r.
24
Binomial Cumulative Distribution
F(x)  P(X  r)
 1  P(X  r)
 1  P(X  r  1)
25
EXAMPLE 6
Let X be a random such that X~B(5, 0.3). By
using the binomial table, find
(a) P(X ≥ 3)
(b) P(X > 3)
(c) P(X ≤ 3)
(d) P(X < 3)
(e) P(X = 3)
26
SOLUTION
(Turn to page 5 from Statistical Tables)
n
p 0.10 0.15 0.20
0.25
0.30
0.35
r
5
0
1
2
3
4
1.000 1.000 1.000
0.4095 0.5563 0.6723
0.0815 0.1648 0.2627
0.0086 0.0266 0.0579
0.0005 0.0022 0.0067
(a) P(X ≥ 3) = 0.1631
1.000 1.000 1.000
0.7627 0.8319 0.8840
0.3672 0.4718 0.5716
0.1035 0.1631 0.2352
0.0156 0.0308 0.0540
Check Table
27
(b) P(X > 3) = P(X ≥ 4) =
0.0308
(c) P(X ≤ 3) = 1 – P(X ≥ 4) = 0.9692
(d) P(X < 3) = 1 – P(X ≥ 3) = 0.8369
(e) P(X = 3) = P(X ≥ 3) – P(X ≥ 4)
= 0.1631 – 0.0308
= 0.1323
Check Table
28
TO FIND PROBABILITY FOR THE VALUE
OF P > 0.5
(i) P[ X  r / X
 ii  P[ X  r / X
 iii  P[ X  r / X
B(n, p)]  P[ X  n  r / X
B(n,1  p)]
B(n, p )]  P[ X  n  r / X
B (n,1  p )]
B(n, p)]  P[ X  n  r / X
B (n,1  p )]
29
EXAMPLE 7
Let X be a random variable such that X~B(10,
0.8). By using the binomial table, find
(a) P(X ≥ 4)
(b) P(X > 4)
(c) P(X ≤ 4)
(d) P(X < 4)
30
SOLUTION
(a)
P(X ≥ 4) 
P(X ≤ 6) 
= 1 - P(X ≥ 7)
= 1 - 0.0009
= 0.9991
(b) P (X > 4) = P(X ≥ 5)
P (X ≤ 5)
= 1 - P(X ≥ 6)
= 1 - 0.0064
= 0.9936
X ~ B(10,0.8)
X ~ B(10,0.2)


X ~ B(10,0.8)
X ~ B(10,0.2)
31
(c)
P(X ≤ 4) 
P(X ≥ 6) 
= 0.0064
X ~ B(10,0.8)
X ~ B(10,0.2)
(d) P(X < 4) = P (X ≤ 3 ) 
P(X ≥ 7)

X ~ B(10,0.8)
X ~ B(10,0.2)
= 0.0009
32
EXAMPLE 8
For a random variable X with a binomial
distribution B(10, 0.45), find a when
(a) P(X ≥ a) = 0.4956
(b) P(X < a) = 0.89801
(c) P(X = a) = 0.23836
33
SOLUTION
(a) P(X ≥ a) = 0.4956
from the binomial probability tables,
P(X ≥ 5) = 0.4956
Thus, a = 5
34
(b) P(X < a) = 0.89801
1 – P(X ≥ a) = 0.89801
P(X ≥ a) = 1 – 0.89801 = 0.1020
From the binomial probability tables,
P(X ≥ 7) = 0.1020
Thus, a = 7
35
(c) P(X = a) = 0.23836
P(X ≥ a) – P(X ≥ a + 1) = 0.23836
From the binomial probability tables,
P(X ≥ 4) – P(X ≥ 5) = 0.23836
Thus, a = 4
36
CONCLUSION

The binomial probability distribution is given
n
x n- x
by
P(X = x) = Cxp q
x = 0, 1, 2, 3, … , n
To find probability for the value of
p > 0.5
(i) P[ X  r / X
B(n, p)]  P[ X  n  r / X
B(n,1  P)
 ii  P[ X  r / X
 iii  P[ X  r / X
B(n, p)]  P[ X  n  r / X
B(n,1  P)
B(n, p)]  P[ X  n  r / X
B(n,1  P)
37
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