Use of binomial probability distribution

2. Use of Binomial Probability Distribution
Example-1:
At a supermarket, 60% of customers pay by credit card. Find the
Probability that in a randomly selected sample of ten customers,
(a). exactly two pay by credit card.
(b). exactly eight not pay by credit card.
(c). at least one pay by credit card.
Solution :
X is the number of customers in a sample of 10, who pay by
credit card.
n = 10
P = 0.60 (paying by credit card as success)
q = 1 – P = 1 – 0.60 = 0.40
So X ̴ B(10,0.6)
(a). 𝑃( 𝑥 = 𝑥) = 𝑐 𝑥
𝑛
𝑝 𝑥
𝑞 𝑛−𝑥
𝑃( 𝑥 = 2) = 𝑐2
10
(0.6)2(0.4)10−2
= 0.0106

3. (b).
P(exactly eight not pay by credit card) = P(exactly two pay by credit card)
P(exactly eight not pay by credit card) = 0.0106 (from part (a))
(c). 𝑃( 𝑥 ≥ 1) = 1 − 𝑝( 𝑥 = 0)
= 1 − 𝑐0
10
(0.6)0(0.4)10−0
= 1 − 0.0001048
= 0.9998
Example-2:
Suppose a die is tossed 5 times. What is the probability of getting
exactly 2 fours?
Solution:
This is a binomial experiment in which the number of trials is equal to
5, the number of successes is equal to 2, and the probability of success
(four) on a single trial is 1/6 or about 0.167. Therefore, the binomial
probability is:
Where n=5
P = 1/6
q = 1 – p = 5/6
P(x = 2) = ?
P(x = 2) = 𝑐2
5 * (0.167)2
* (0.833)3
P(x = 2) = 0.161

4. Example-3:
If n = 20 and p = 0.8 then find mean and variance of binomial
distribution.
Solution:
As we know Mean = np = 20*0.8= 16
Variance = npq = np(1-p) = 20*0.8*0.2 = 3.2
Example-4:
If Mean and variance of a binomial distribution is 16 and 3.2,
Find n and p.
Solution:
Mean = 16
Variance = 3.2
We know that Mean = np = 16 ---------- (i)
Variance = npq = 3.2 ----- (ii)
Put the value of np = 16 in (ii)
Then 16q = 3.2
q = 3.2/ 16 = 0.2
Since p + q = 1
Then p + 0.2 = 1
P = 1- 0.2 = 0.8
Now put value of p = 0.8 in (i)
0.8n = 16
n = 16 / 0.8 = 20