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Question
Among 28 professors of a certain department, 18 drive foreign cars and 10 drive local made cars. If 5 of these professors are selected at random, what is the probability that atleast 3 of them drive foreign cars?
Solution
Let p be the probability of foreign cars driven by the professors
p = `18/28 = 9/14`
Let q be the probability of local made cars driv¬en by the professors
q = `10/28 = 5/14`
and n = 5
The binomial distribution p(x = x) = ncxpxqn-x
`"p"(x = x) = 5"c"_x (9/14)^x (5/14)^(5 - x)`
p(altleast 3 of them drive foreign cars) = p(x ≥ 3)
p(x ≥ 3) = p(x = 3) + p(x = 4)+ p(x = 5)
= `5"c"_3 (9/14)^3(5/1)^(5-3) + 5"c" (9/14)^4 (5/14)^(5 - 4) + 5"c"_5 (9/14)^5 (5/14)^(5 - 5)`
= `5"c"_2 (9/14)^3 (5/14)^2 + 5"c"_1 (9/14)^4 (5/14)^0 + 5"c"_0 (9/14)^5 (5/14)^0`
= `((5 xx 4)/(1 xx 2)) xx ((9^3 xx 5^2)/(14^5)) + 5 xx ((9^4 xx 5^1)/14) + (9^5/14^5)`
= `((10 xx 9^3 xx 5^2) + (5 xx 9^4 xx 5) + (9^5))/14^5`
= `((10 xx 729 xx 25) + (5 xx 9^4 xx 5) - (9^5))/537824`
= `405324/537824` .......`{"nc"_"r" = "nc"_("n" - "r")}`
= 0.7536
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