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Question
In a box of floppy discs, it is known that 95% will work. A sample of three of the discs is selected at random. Find the probability that exactly two floppy disc work.
Solution
Let X = number of working discs.
p = probability that a floppy disc works
∴ p = 95% = `95/100 = 19/20`
and q = 1 - p = `1 - 19/20 = 1/20`
Given: n = 3
∴ X ~ B`(3, 19/20)`
The p.m.f. of X is given by
P(X = x) = `"^nC_x p^x q^(n - x)`
i.e. p(x) = `"^3C_x (19/20)^x (1/20)^(3-x)`, x = 0, 1, 2, 3
P(exactly two floppy discs work) = P(X = 2)
= p(2) = `"^3C_2 (19/20)^2 (1/20)^(3 - 2)`
`= (3* 2!)/(2! * 1!) xx (19)^2/(20)^2 xx (1/20)`
`= 3(19^2/20^3)`
Hence, the probability that none of the floppy disc will work = 3`(19^2/20^3)`
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