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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 12 of the travelers will not have a laptop?
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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(12 of the travels will not have a laptop)
= 1 – P(X = 12)
= `1 - 15"c"_12 (2/5)^12(3/5)^(15 - 12)`
= `1 - 15"c"_3 (2/5)^12 (3/5)^3`
= `1 - ((15 xx 14 xx 13)/(1 xx 2 xx 3)) [((2)^12 (3)^3)/(5)^15]`
= `1 - 455 ((4096 xx 27)/(3.055 xx 10^10))`
= `1 - 50319360/300550000000`
= `1 - 0.001647`
= 0.99835
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