# APPLICATION PROBLEMS ON BINOMIAL DISTRIBUTION

﻿Application of binomial distribution :

﻿Binomial distribution is applicable when the trails are independent and each trail has just two outcomes success and failure. It is applied in coin tossing experiments, sampling inspection plan, genetic experiments and so on.

A discrete random variable x is defined to follow binomial distribution with parameters n and p to be denoted by

x ~ B(n, p)

If the probability of mass function of x is given by

Problem 1 :

A coin is tossed 10 times. Assuming the coin to be unbiased, what is the probability of getting

i) 4 heads ?

ii) at least 4 heads ?

iii) at most 3 heads ?

Solution :

Let X be the random variable representing number of heads.

Number of trails (n) = 10

probability of getting head when a coin is tossed = 1/2

probability of not getting head when a coin is tossed = 1/2

p = 1/2 and q = 1/2

i)

ii) at least 4 heads :

P(x  ≥ 4) = 1 - P(x < 4)

= 1 - [P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)]

iii)  at most 3 heads :

Problem 2 :

If 15 dates are selected at random, what is the probability of getting two Sundays ?

Solution :

Number of days selecting (n) = 15

Probability of getting two Sundays (p) = 1/7

Probability of getting failure (q) = 6/7

Problem 3 :

The incidence of occupational disease in an industry is such that the workmen have a 10% chance of suffering from it. What is the probability that out of 5 workmen 3 or more will contract the disease ?

Solution :

Probability that the workmen suffers from the occupational disease p = 0.1 = 1/10

Probability does not suffer q = 0.9 = 9/10

Number of workmen choosing n = 5

= P(x ≥ 3)

Problem 4 :

6 coins are tossed 512 times. Find the expected frequency of heads. Also compute the mean and standard deviation of the number number of heads.

Solution :

Number of coins tossed (n) = 6

Probability of getting heads = 1/2

not getting heads q = 1/2

Getting 0 heads :

Getting 1 head :

Getting 2 heads :

Getting 3 heads :

Getting 4 heads :

Getting 5 heads :

Getting 6 heads :

Expected Frequency :

Mean :

Mean = np

= 6 (1/2)

= 3

Standard deviation :

Variance = npq

Standard deviation = √npq

√6x(1/2)x(1/2)

= √1.5

= 1.22

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