Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 10.7

p(x)= l0Cx(l/5)x(4/5)1–x

(i) Probability that all 10 are defective is

p(10)=l0C 10 (l/5)^10 (4/5)°= 5110

(ii) Probability that all 10 are good

= 1– P (all are defective) =^1 − 5110

. (iii) Probability that at least one is defective is given by the sum of probabilities,

viz. p(1) + p(2) + p(3)+...........+p(10)
or 1 – p(0) = 1– l0C 10 (1/5)^0 (4/5)^10 = 1– (4/5)^10
(iv) Probability of at the most three defective items is
p(X < 3) = p(X = 0)+ p(X = 1) + (X = 2) + p(X = 3)
= p(0)+p(1)+p(2)+p(3)
= l0C 0 (1/5)^0 (4/5)^10 + l0C 1 (1/5)l (4/5)^9 + l0C 2 (1/5)^2 (4/5)^8 + l0C 3 (1/5)^3 (4/5)^7
= 1(.107)+10(.026)+45(.0067)+120(.0016)
= 0.107 + 0.26 + 0.30 + 0.192 = 0.859
Example 6 :
The incidence of occupational disease in an industry is such that the workmen have a 20% chance of
suffering from it. What is the probability that out of six workmen, 4 or more will contact the disease?

Solution :
Let X represent the number of workers suffering from the disease. Then the possible values of X are 0, 1, 2, ... 6.

p = p ( worker suffer from a disease)

= 20/100 = 1/5

q= 1 – (1/5) = 4/5, n = 6

Using the formula for binomial distribution, we have

p (X) =^6 Cx(1/5)x(4/5)6–x

the probability that 4 or more workers contact the disease is

p(X>4) = p(4) + p(5) + p(6)

=^6 C 4 (1/5)^4 (4/5)^2 +^6 C 5 (1/5)^5 (4/5)+^6 C 6 (1/5)^6

= 15 16 6 415625 15625 15625 15625××++=^1265

= 0.016