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02 Jun 2020, 08:44
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (02:24) correct
63%
(02:31)
wrong
based on 94
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The probability of a manufacturing company producing a defective item is 0.1. If 5 items are drawn at random from a set of 100 distinct items, without replacement, what is the probability that at least 2 items would be defective?
A. \(1-0.9^4 * 1.4\)
B. \(1-0.9^4\)
C. \(1-0.9^5\)
D. \(1-0.9^4 * 0.5\)
E. \(1-0.9^4* 0.1\)
A. \(1-0.9^4 * 1.4\)
B. \(1-0.9^4\)
C. \(1-0.9^5\)
D. \(1-0.9^4 * 0.5\)
E. \(1-0.9^4* 0.1\)
02 Jun 2020, 08:59
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Bunuel
Probability of a defective item = 0.1
Probability of a NON-defective item = 1-0.1 = 0.9
i.e. Total defective items in 100 items = 10
Probability of atleast 2 defective items = 1- (Probability of no item defective + Probability of 1 item defective)
Probability of atleast 2 defective items = = \(1-(0.9^5+5C4*0.9^4*0.1)\) = \(1-(0.9^4(0.9+0.5) - 1-0.9^4*1.4\)
Answer: Option A
02 Jun 2020, 09:33
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IMO A
Probability of defective item = 0.1
Probability of non-defective item = 1-0.1= 0.9
Probability of atleast 2 defective items = 1- (Probability of no item defective + Probability of 1 item defective)
= 1- ( (0.9)^5 + 5*(0.1) (0.9)^4) ---- Multiplied by 5 as (DNNNN, NDNNN, NNDNN, NNNDN, NNNND)
= 1- (0.9^4)* (0.9 + 0.5)
= 1-(0.9^4) * 1.4
Probability of defective item = 0.1
Probability of non-defective item = 1-0.1= 0.9
Probability of atleast 2 defective items = 1- (Probability of no item defective + Probability of 1 item defective)
= 1- ( (0.9)^5 + 5*(0.1) (0.9)^4) ---- Multiplied by 5 as (DNNNN, NDNNN, NNDNN, NNNDN, NNNND)
= 1- (0.9^4)* (0.9 + 0.5)
= 1-(0.9^4) * 1.4
