Question

Forty percent of business travellers carry a laptop. In a sample of 15 business travelers, what is the probability that atleast three of the travelers have a laptop?

Answer 1

Given n = 5

p = `40/10 = 2/5`

q = 1 – p = `1 - 2/5 = (5-2)/5 = 3/5`

The binomial distribution P(X = x) = `15"c"_x (2/5)^x (3/5)^(15 - x)`

P(at least three of the travelers have a laptop)

= P(X ≥ 3)

P(X ≥ 3) = 1 – P(X < 3)

= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= `15"c"_0 (2/5)^0 (3/5)^(15 - 0) + 15"c"_1 (2/5)^1 (3/5)^(15 - 1) + 15"c"_2 (2/5)^2 (3/5)^(15 - 2)`

= `(1)(1)(3/5)^15 + 15(2/5)(3/5)^14 + 105(2/5)^2(3/5)^13`

= `13^15/5^15 +(15 xx 2 xx 3^14)/5^5 +(105 xx 4 xx 3^13)/5^15`

= `(3^13 [3^2 + (30 xx 3) + (105 xx 4 xx 9)])/5^15`

= `(1.593 xx 10^6 xx (9 + 90 + 3780))/(3.055 xx 10^10)`

= `(1.593 xx 3879)/(3.055 xx 10^4)`

= `6179.247/30550`

= 0.20226

∴ P(X ≥ 3) = `1 - 020226`

= 0.79774