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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?
Solution
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
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