Question

Forty percent of business travellers carry a laptop. In a sample of 15 business travelers 

(i) what is the probability that 3 will have a laptop? 

(ii) what is the probability that 12 of the travelers will not have a laptop? 

(iii) what is the probability that atleast three of the travelers have a laptop?

Answer 1

Let X be the binomial variables that denote the number of business travelers having a laptop. 

Given that n = 15 and P = 40% = 0.4. So q = 1 – 0.4 = 0.6. Thus X ~ B (15, 0.4). 

The p.m.f of X is given by P (X = x) = ¹⁵C_x (0.4)^x(0.6)^{15 - x}

(i) P(3 travellers will have a laptop) = P (X = 3)

Note: The calculation can be done by the method of logarithms also. 

P(X = 3) = 455 (0.064) (0.002177) = 0.0634 

(ii) P(12 of the travellers will not have a laptop) 

= P(15 – 12 = 3 will have a laptop) 

= P(X = 3) = 0.0634 (from the previous subdivision) 

(iii) P(atleast three of the travellers have a laptop)

Thus the probability that at least 3 of the 5 mice recover is 0.8743.