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805+ (Hard)|   Probability|      
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mdhyan2012
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A bag has 10 bulbs of which 4 are known to be fused, but are virtually 22 Sep 2020, 22:59
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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:

39% (03:01) correct 61% (03:02) wrong based on 56 sessions

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A bag has 10 bulbs of which 4 are known to be fused, but are virtually indistinguishable from the working ones. Bulbs are drawn out one by one, tested and kept aside, until all fused bulbs have been taken out of the bag.

What is the probability that separation process stops exactly at the 7th bulb?

A. \(\frac{2}{21}\)
B. \(\frac{5}{42}\)
C. \(\frac{3}{14}\)
D. \(\frac{4}{63}\)
E. \(\frac{8}{105}\)
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A
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A bag has 10 bulbs of which 4 are known to be fused, but are virtually 23 Sep 2020, 00:21
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\(Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}\)

This is based on picking without replacement, so the favorable outcomes and total outcomes keep reducing by 1

There are 6 Good and 4 Bad Bulbs

We know that the 7th Ball is a defective Bulb always.

Lets assume one of the possible Arrangements as G G G B B B B

P(1st Bulb is good) = \(\frac{6}{10}\)

P(2nd Bulb is good) = \(\frac{5}{9}\)

P(3rd Bulb is good) = \(\frac{4}{8}\)

P(4th Bulb is Bad) = \(\frac{4}{7}\)

P(5th Bulb is Bad) = \(\frac{3}{6}\)

P(6th Bulb is Bad) = \(\frac{2}{5}\)

P(7th Bulb is Bad) = \(\frac{1}{4}\)


The probability for this particular arrangement = \(\frac{6}{10} \space * \frac{5}{9} \space * \frac{4}{8} \space * \frac{4}{7} \space * \frac{3}{6} \space * \frac{2}{5} \space * \frac{1}{4} = \frac{1}{210}\)

We now have to see the number of possible arrangements. The last place is fixed by a bad bulb.

So the remaining 6 places will have 3 good and 3 bad bulbs as \(\frac{6!}{3!3!} = 20\).


Each of the 20 options will have the same probability of occurring i.e \(\frac{1}{210}\)


Therefore the required probability = \(20 * \frac{1}{210} = \frac{2}{21}\)


Option A

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A bag has 10 bulbs of which 4 are known to be fused, but are virtually 23 Sep 2020, 20:43
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CrackVerbalGMAT
\(Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}\)

This is based on picking without replacement, so the favorable outcomes and total outcomes keep reducing by 1

There are 6 Good and 4 Bad Bulbs

We know that the 7th Ball is a defective Bulb always.

Lets assume one of the possible Arrangements as G G G B B B B

P(1st Bulb is good) = \(\frac{6}{10}\)

P(2nd Bulb is good) = \(\frac{5}{9}\)

P(3rd Bulb is good) = \(\frac{4}{8}\)

P(4th Bulb is Bad) = \(\frac{4}{7}\)

P(5th Bulb is Bad) = \(\frac{3}{6}\)

P(6th Bulb is Bad) = \(\frac{2}{5}\)

P(7th Bulb is Bad) = \(\frac{1}{4}\)


The probability for this particular arrangement = \(\frac{6}{10} \space * \frac{5}{9} \space * \frac{4}{8} \space * \frac{4}{7} \space * \frac{3}{6} \space * \frac{2}{5} \space * \frac{1}{4} = \frac{1}{210}\)

We now have to see the number of possible arrangements. The last place is fixed by a bad bulb.

So the remaining 6 places will have 3 good and 3 bad bulbs as \(\frac{6!}{3!3!} = 20\).


Each of the 20 options will have the same probability of occurring i.e \(\frac{1}{210}\)


Therefore the required probability = \(20 * \frac{1}{210} = \frac{2}{21}\)


Option A

Arun Kumar
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HI Team,

Quote:
So the remaining 6 places will have 3 good and 3 bad bulbs as \(\frac{6!}{3!3!} = 20\).
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why did we say 6 places remaining, as of now only 3 places are left .
Can you explain the part how you got 20
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A bag has 10 bulbs of which 4 are known to be fused, but are virtually 23 Sep 2020, 22:32
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globaldesi
by fixing the last place for bad bulb we are left with 6 places out of which 3 can be filled by good bulbs and 3 by bad bulbs
our target is to remove the bulb by 7th place which is why 6!/3!*3! = 20 is the ways in which first 6 places can be arranged.


globaldesi
CrackVerbalGMAT
\(Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}\)

This is based on picking without replacement, so the favorable outcomes and total outcomes keep reducing by 1

There are 6 Good and 4 Bad Bulbs

We know that the 7th Ball is a defective Bulb always.

Lets assume one of the possible Arrangements as G G G B B B B

P(1st Bulb is good) = \(\frac{6}{10}\)

P(2nd Bulb is good) = \(\frac{5}{9}\)

P(3rd Bulb is good) = \(\frac{4}{8}\)

P(4th Bulb is Bad) = \(\frac{4}{7}\)

P(5th Bulb is Bad) = \(\frac{3}{6}\)

P(6th Bulb is Bad) = \(\frac{2}{5}\)

P(7th Bulb is Bad) = \(\frac{1}{4}\)


The probability for this particular arrangement = \(\frac{6}{10} \space * \frac{5}{9} \space * \frac{4}{8} \space * \frac{4}{7} \space * \frac{3}{6} \space * \frac{2}{5} \space * \frac{1}{4} = \frac{1}{210}\)

We now have to see the number of possible arrangements. The last place is fixed by a bad bulb.

So the remaining 6 places will have 3 good and 3 bad bulbs as \(\frac{6!}{3!3!} = 20\).


Each of the 20 options will have the same probability of occurring i.e \(\frac{1}{210}\)


Therefore the required probability = \(20 * \frac{1}{210} = \frac{2}{21}\)


Option A

Arun Kumar


HI Team,

Quote:
So the remaining 6 places will have 3 good and 3 bad bulbs as \(\frac{6!}{3!3!} = 20\).
why did we say 6 places remaining, as of now only 3 places are left .
Can you explain the part how you got 20
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A bag has 10 bulbs of which 4 are known to be fused, but are virtually 24 Sep 2020, 03:02
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The seventh one to be taken out will be faulty one . Hence its position is fixed . Assuming rest three faulty bulbs are taken out in first go

The probability will be 6C3 * 4/10* 3/9* 2/8 * 1/7 = 2/21

IMO A
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A bag has 10 bulbs of which 4 are known to be fused, but are virtually 28 Mar 2023, 17:23
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