Maharashtra State BoardHSC Arts 12th Board Exam
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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. - Mathematics and Statistics

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Sum

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.

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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`

Concept: Binomial Distribution
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 8 Binomial Distribution
Miscellaneous exercise 8 | Q 11.2 | Page 524
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