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22 Sep 2020, 22:59
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A
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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. \(\frac{8}{105}\)
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. \(\frac{8}{105}\)
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
Arun Kumar
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
globaldesi
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23 Sep 2020, 20:43
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CrackVerbalGMAT
HI Team,
Quote:
why did we say 6 places remaining, as of now only 3 places are left .
Can you explain the part how you got 20
