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A manufacturing plant produces widgets. It is known that 10%

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A manufacturing plant produces widgets. It is known that 10% [#permalink] New post 11 Nov 2004, 14:36
A manufacturing plant produces widgets. It is known that 10% are defective. If 10 widgets are selected at random find the probability that the sample does not contain more than 3 defective widgets.
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 [#permalink] New post 11 Nov 2004, 15:31
An ugly solution here:

(1/10)^3 * (9/10)^7 * 10C3 --> 3 defective
+
(1/10)^2 * (9/10)^8 * 10C2 --> 2 defective
+
1/10 * (9/10)^9 * 10C1 --> 1 defective
+
(9/10)^10 --> no defective

Answer is the sum of the above
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 [#permalink] New post 12 Nov 2004, 11:08
need to apply binomial theorem here.
nCr*p^r*q^(n-r)

n=10
r=0,1,2,3.
p= probability of defect=1/10

q=1-p=9/10

required answer= 10C0(1/10)^0*(9/10)10+ 10C1(1/10)^1*(9/10)^9+
10C2(1/10)^2*(9/10)^8+ 10C3(1/10)^3*(9/10)^7.

i think the calculation is too big. this question shouldnt appear on GMAT. cant solve something so big in less than 2 mins.
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Let's get it right!!!!

  [#permalink] 12 Nov 2004, 11:08
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A manufacturing plant produces widgets. It is known that 10%

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