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The probability of a manufacturing company producing a defective item

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The probability of a manufacturing company producing a defective item  [#permalink]

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New post 02 Jun 2020, 07:44
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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. \(1-0.9^4* 0.1\)

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The probability of a manufacturing company producing a defective item  [#permalink]

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New post 02 Jun 2020, 07:59
Bunuel wrote:
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. \(1-0.9^4* 0.1\)


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
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Re: The probability of a manufacturing company producing a defective item  [#permalink]

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New post 02 Jun 2020, 08:33
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
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The probability of a manufacturing company producing a defective item  [#permalink]

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New post 02 Jun 2020, 08:53
Bunuel wrote:
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. \(1-0.9^4* 0.1\)


Given: The probability of a manufacturing company producing a defective item is 0.1.
Asked: 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?

The probability of a manufacturing company producing a defective item is 0.1
The probability of a manufacturing company producing a non-defective item is 0.9

The probability that at least 2 items would be defective = 1- Probability of (0 item defective) - Probability of(1 item defective) = 1 - .9^5 - 5C1*(.1)(.9)^4 = 1 - .9^4 (.9 + .5) = 1 - .9^4*1.4

IMO A
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The probability of a manufacturing company producing a defective item   [#permalink] 02 Jun 2020, 08:53

The probability of a manufacturing company producing a defective item

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