Question

twenty percent of the applications recieved for a particular position are rejected. what is the probability that among the next 14 applicants
1a. no more than four will be rejected?
1b. determine the expected number of rejected applications and its variance

Answer 1

The given problem is solved using Binomial distribution;

Binomial Distribution

If 'X' is the random variable representing the number of successes, the probability of getting ‘r’ successes and ‘n-r’ failures, in 'n' trails, ‘p’ probability of success ‘q’=(1-p) is given by the probability function

P(X r)=() п-т рФ

Mean of X : Expected value of X : E(X) =np

Variance of X = npq

Probability of an application is rejected : p =20/100=0.20

n : Number of applications

X : number of applications that are rejected in the next 14 applicants.

X follows binomial distribution with n=14 and p= 0.20 ; (q=1-p=1-0.20=0.80)

Probability mass function of X

P(X r)() x 0.2 x 0.814- r = 0, 1, 2,14

1a. Probability that among the next 14 applicants no more than four will be rejected

P(X$\small \leq$4) = P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)

x	P(x)	P(x)
0	14 P(X 0) x 0.20 x 0.814= 1 x 1 x 0.044 0.044	0.0440
1	14 x 0.21 x 0.813 = 14 x 0.2 x 0.055 0.1539 1 P(X 1)	0.1539
2	14 0.22 x 0.812 2 P(X 2) 91 x 0.04 x 0.0687 0.2501 X	0.2501
3	14 0.23 x 0.811 3 P(X 3) 364 x 0.008 x 0.0859 0.2501	0.2501
4	14 x 0.24 x 0.810 4 P(X =4) 1001 x 0.0016 x 0.1074 0.172 X	0.1720
	Total	0.8702

P(X$\small \leq$4) = P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4) = 0.8702

Probability that among the next 14 applicants no more than four will be rejected = 0.8702

1b. determine the expected number of rejected applications and its variance

E(X) = np =14*0.2 = 2.8

expected number of rejected applications = 2.8

variance = npq = 14*0.2*0.8=2.24