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Question
One dialing certain telephone numbers assume that on an average, one telephone number out of five is busy, Ten telephone numbers are randomly selected and dialed. Find the probability that at least three of them will be busy.
Solution
`p = (1)/(10), q = (9)/(10), n = 10`
Required probability ( at least three phones busy)
= 1 - ( Probability maximum two phones are busy)
= `1 - [ ""^10C_0 (1/10)^0 (9/10)^0 + ""^10C_1 (1/10)^1 (9/10)^9 + ""^10C_2 (1/10)^2 (9/10)^8 ] `
= `1 - [ 1 xx 1 xx(9/10)^10 + 10 xx (1)/(10) xx (9/10)^9 + 45 xx (1)/(100) xx (9/10)^8 ]`
= `1 - [ (9/10)^9 (9/10 + 1 + 1/2) ]`
= `1 - [ (9/10)^9 ((9 + 10 + 5)/10) ]`
= `1 - [ (9/10)^9 (12/5) ]`
= `1 - [ 0.3874 xx (12)/(5) ]`
= `1 - [0.93] = 0.07`
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