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Question
The probability that a machine will produce all bolts in a production run within specification is 0.998. A sample of 8 machines is taken at random. Calculate the probability that 7 or 8 machines.
Solution
Let X = number of machines that produce the bolts within specification.
p = probability that a machine produce bolts within specification
p = 0.998
and q = 1 - p = 1 - 0.998 = 0.002
Given: n = 8
∴ X ~ B (8, 0.998)
The p.m.f. of X is given by
P(X = x) = `"^nC_x p^x q^(n - x)`
i.e. p(x) = `"^8C_x (0.998)^x (0.002)^(8 - x)`, x = 0, 1, 2,...,8
P(7 or 8 machines will produce all bolts within specification) = P(X = 7) + P(X = 8)
`= ""^8C_7 (0.998)^7 (0.002)^(8 -7) + "^8C_8 (0.998)^8 (0.002)^(8 -8)`
`= 8 xx (0.998)^7 (0.002)^1 + 1xx (0.998)^8 (0.002)^0`
`= (0.998)^7 [8(0.002) + 0.998]`
`= (0.016 + 0.998)(0.998)^7`
`= (1.014) xx (0.998)^7`
Hence, the probability that 7 or 8 machines produce all bolts within specification = `(1.014) xx (0.998)^7`
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