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Three of 12 remote controls in a box are defective. The remote control

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Three of 12 remote controls in a box are defective. The remote control  [#permalink]

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New post 29 Mar 2018, 23:54
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Three of 12 remote controls in a box are defective. The remote controls are tested one at a time and not replaced in the box after testing. What is the probability that neither of the first two remote controls tested is defective?

(A) 1/6
(B) 2/9
(C) 6/11
(D) 9/16
(E) 3/4

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Re: Three of 12 remote controls in a box are defective. The remote control  [#permalink]

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New post 29 Mar 2018, 23:59
Bunuel wrote:
Three of 12 remote controls in a box are defective. The remote controls are tested one at a time and not replaced in the box after testing. What is the probability that neither of the first two remote controls tested is defective?

(A) 1/6
(B) 2/9
(C) 6/11
(D) 9/16
(E) 3/4


3 defective
12 total
9 non defective


neither of first two defective = 9/12 x 8/11 = 6/11
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Re: Three of 12 remote controls in a box are defective. The remote control  [#permalink]

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New post 30 Mar 2018, 00:38
3 defective
12 total
9 non defective


neither of first two defective = 9/12 x 9/12(since not replaced) = 9/16
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Re: Three of 12 remote controls in a box are defective. The remote control  [#permalink]

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New post 30 Mar 2018, 01:38
kunalcvrce wrote:
3 defective
12 total
9 non defective


neither of first two defective = 9/12 x 9/12(since not replaced) = 9/16




„Not replaced“ means that in the second choice you only have 11 left to choose from, doesn’t it?

Hence -> 6/11
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Re: Three of 12 remote controls in a box are defective. The remote control  [#permalink]

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New post 30 Mar 2018, 01:41
kunalcvrce wrote:
3 defective
12 total
9 non defective


neither of first two defective = 9/12 x 9/12(since not replaced) = 9/16


your answer is for replaced

not please solve not replaced

9/12 now 1 ball is removed

8/11

this becomes the new fraction 6/11 would be the answer IMO
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Re: Three of 12 remote controls in a box are defective. The remote control  [#permalink]

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New post 30 Mar 2018, 01:53
JaBre wrote:
kunalcvrce wrote:
3 defective
12 total
9 non defective


neither of first two defective = 9/12 x 9/12(since not replaced) = 9/16




„Not replaced“ means that in the second choice you only have 11 left to choose from, doesn’t it?

Hence -> 6/11


Correct my mistake..Thanks
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Re: Three of 12 remote controls in a box are defective. The remote control  [#permalink]

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New post 04 Apr 2018, 17:30
Bunuel wrote:
Three of 12 remote controls in a box are defective. The remote controls are tested one at a time and not replaced in the box after testing. What is the probability that neither of the first two remote controls tested is defective?

(A) 1/6
(B) 2/9
(C) 6/11
(D) 9/16
(E) 3/4


Since there are 9 non-defective remotes out of 12 in the box, the probability that the first remote selected is not defective is 9/12, or 3/4. We don’t put the first selected remote control back in the box. Thus, there are now 11 remotes in the box, and 8 of them are not defective. Thus, the probability that the second remote control is not defective is 8/11.

Thus, the probability that the first two remote controls are not defective is 9/12 x 8/11 = 3/4 x 8/11 = 24/44 = 6/11.

Answer: C
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Re: Three of 12 remote controls in a box are defective. The remote control &nbs [#permalink] 04 Apr 2018, 17:30
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