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Of the 20 lightbulbs in a box, 2 are defective. An inspector will sele

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Of the 20 lightbulbs in a box, 2 are defective. An inspector will sele  [#permalink]

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New post 28 Jan 2019, 01:19
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  45% (medium)

Question Stats:

50% (01:37) correct 50% (01:47) wrong based on 24 sessions

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Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously and at random from the box. What is the probability that neither of the lightbulbs selected will be defective?

A. 37/190
B. 17/19
C. 9/10
D. 153/190
E. 189/190

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Of the 20 lightbulbs in a box, 2 are defective. An inspector will sele  [#permalink]

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New post Updated on: 29 Jan 2019, 00:45
Bunuel wrote:
Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously and at random from the box. What is the probability that neither of the lightbulbs selected will be defective?

A. 37/190
B. 17/19
C. 9/10
D. 153/190
E. 189/190


P of picking bulbs not defective ;
18/20 * 17/19 = 153/190

IMO D
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Originally posted by Archit3110 on 28 Jan 2019, 03:24.
Last edited by Archit3110 on 29 Jan 2019, 00:45, edited 1 time in total.
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Re: Of the 20 lightbulbs in a box, 2 are defective. An inspector will sele  [#permalink]

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New post 28 Jan 2019, 18:40
1
Don't be confused by the wording. This is your basic –"red marbles in a jar" - probability question. Simultaneously is treated the same as take one, don't replace, then take another.

The probability space (the denominator) is 20·19. There are initially 20 lightbulbs to choose from. When one is taken, there are only 19.

The numerator is 18·17. Neither of the lightbulbs selected will be defective so we can pick any lightbulb bar the defective ones. 20-2=18. The defective ones stay. Now there are 19 lightbulbs and still two defective ones 19-2=17.

\(\frac{18}{20}·\frac{17}{19}=\frac{9}{10}·\frac{17}{19}=\frac{153}{190}\)=D
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Re: Of the 20 lightbulbs in a box, 2 are defective. An inspector will sele  [#permalink]

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New post 28 Jan 2019, 19:07
1
Leave out the two defective bulbs;
18C2/20C2=153/190

or

Deduct the case of “one defective+one non-defective” and “both defective”;
(20C2-18C1*2C1-2C2)/20C2=(190-36-1)/190=153/190
(The formar solution is way simpler)

D is the answer

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Re: Of the 20 lightbulbs in a box, 2 are defective. An inspector will sele   [#permalink] 28 Jan 2019, 19:07
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