Question

It is known that 60% of all women are doing online business from home. In a sample of 10 women, compute the probability that.

(i). exactly 5 are doing online business.

(ii). at least 8 are doing online business.

(iii). at most 2 are not doing online business.

Answer 1

solution:

Given data

Percent of women doing online business (p) = 60% = 0.60

sample size (n) = 10

Let X be the random variable representing no.of women doing online business

Here,X~B(n,p)

~B(10 , 0.60)

P(X=x) = nCx * (p)^x * (1-p)^(n-x)

i) Probability that exactly 5 women are doing online business = P(X=5)

= 10C5 * (0.60)^5 * (0.40)^5

= 252 * (0.60)^5 * (0.40)^5

= 0.2007

Probability that exactly 5 women are doing online business = 0.2007

ii) Probability that at least 8 are doing online business = P(X>= 8)

= P(X=8) + P(X=9) + P(X=10)

= 10C8 * (0.60)^8 * (0.40)^2 + 10C9 * (0.60)^9 * (0.40)^1 + 10C10 * (0.60)^10 * (0.40)^0

= 45 * (0.60)^8 * (0.40)^2 + 10 * (0.60)^9 * (0.40)^1 + 1 *   (0.60)^10 * (0.40)^0

= 0.12093 + 0.04031 + 0.00605

= 0.16729

Probability that at least 8 are doing online business =~ 0.1673

iii) Let Y be the binomial random variable representing no.of women not doing online business

Here p = 0.40

Probability that at most 2 are not doing online business = P(Y<=2)

= P(Y=0) + P(Y=1) + P(Y=2)

= 10C0 * (0.40)^0 * (0.60)^10 + 10C1 * (0.40)^1 * (0.60)^9 + 10C2 * (0.40)^2 * (0.60)^8

= 1 * (0.40)^0 * (0.60)^10 + 10 * (0.40)^1 * (0.60)^9 + 45 * (0.40)^2 * (0.60)^8

=  0.00605 + 0.04031 + 0.12093

= 0.16729

Probability that at most 2 are not doing online business =~ 0.1673