2. www.sanjivanimba.org.in
302-DECISION SCIENCE
Unit No.5. Probability
5.2.6 Case 1: Poisson
Probability Distribution
Presented By:
Dr. V. M. Tidake
Ph. D (Financial Management), MBA(FM), MBA(HRM) BE(Chem)
Dean EDP & Associate Professor MBA
2
Sanjivani College of Engineering, Kopargaon
Department of MBA
www.sanjivanimba.org.in
4. www.sanjivanimba.org.in
Probability Distribution
If 5% of electric bulbs manufactured by a company are
defective,, use poisson distribution to find the probability
that in a box of 100 bulbs-
i. None is defective
ii. 3 bulbs are defective
iii. More than 3 bulbs are defective
Given: e-5 = 0.007
5. www.sanjivanimba.org.in
Case 1: Poisson Distribution
We have, p – Probability that a bulb is defective = 5% i.e. p = 0.05
Number of bulbs in a box = 100
According to poisson distribution, m=np
m = 100*0.05 = 5
Probability of getting r defective bulbs in a box of n=100 bulbs is-
𝑒 − 𝑚 𝑚 𝑟
p(r) = __________
r!
(0.007)5 𝑟
p(r) = __________
r!
6. www.sanjivanimba.org.in
Case 1: Poisson Distribution
i. Probability that no bulb is defective-
𝑒 − 𝑚 𝑚 𝑟
p(r) = __________
r!
(0.007)50
p(0) = __________
0!
0.007 ∗ 1
p(0) = __________
1
p(0) = 0.007
7. www.sanjivanimba.org.in
Case 1: Poisson Distribution
ii. Probability that 3 bulbs are defective-
𝑒 − 𝑚 𝑚 𝑟
p(r) = __________
r!
(0.007)53
p(3) = __________
3!
0.007 ∗ 125
p(3) = __________
3*2*1
p(3) = 0.146
8. www.sanjivanimba.org.in
Case 1: Poisson Distribution
iii. Probability that more than 3 bulbs are defective-
p(r>3) = 1- p(r≤3)
p(r>3) = 1- [P(0)+P(1)+P(2)+P(3)]
p(r>3) = 1- [
𝟎.𝟎𝟎𝟕 ∗𝟓𝟎
𝟎!
+
𝟎.𝟎𝟎𝟕 ∗𝟓𝟏
𝟏!
+
𝟎.𝟎𝟎𝟕 ∗𝟓𝟐
𝟐!
+
𝟎.𝟎𝟎𝟕 ∗𝟓𝟑
𝟑!
]
p(r>3) = 0.725