Bombs pulled out to crack, do blast?

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Bombs pulled out to crack, do blast?

by sanju09 » Wed Nov 06, 2013 1:09 am
Records suggest that out of every 10 Cracker Bombs manufactured by The Fluffy Fireworks, 2 do not blast when cracked. A sample of 10 Cracker Bombs manufactured by The Fluffy Fireworks is taken to crack them one by one. What is the probability that all of the first six Cracker Bombs pulled out to crack, do blast?
A. (4/5)^6
B. 96/625
C. 16/625
D. (1/5)^6
E. (1/5)^6 (4/5)^2



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by theCodeToGMAT » Wed Nov 06, 2013 2:11 am
Is it [spoiler]{A}[/spoiler]?
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by sanju09 » Wed Nov 06, 2013 2:15 am
theCodeToGMAT wrote:Is it [spoiler]{A}[/spoiler]?
[spoiler]It doesn't match my answer. Can you tell me your approach?[/spoiler]
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by theCodeToGMAT » Wed Nov 06, 2013 2:36 am
sanju09 wrote:
theCodeToGMAT wrote:Is it [spoiler]{A}[/spoiler]?
[spoiler]It doesn't match my answer. Can you tell me your approach?[/spoiler]
Actually, i solved by: 8C6/10C6 =

8!/6!2! / 10!/6!*4! = 8*7/2! / 10*9*8*7/4*3*2 = 28 / 10*3*7 = 28/210 = 4/30 = 2/15

I dint get {A} :P .. I was just guessing..
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by mevicks » Wed Nov 06, 2013 2:47 am
sanju09 wrote:Records suggest that out of every 10 Cracker Bombs manufactured by The Fluffy Fireworks, 2 do not blast when cracked. A sample of 10 Cracker Bombs manufactured by The Fluffy Fireworks is taken to crack them one by one. What is the probability that all of the first six Cracker Bombs pulled out to crack, do blast?
A. (4/5)^6
B. 96/625
C. 16/625
D. (1/5)^6
E. (1/5)^6 (4/5)^2
P (defective) = 1/5
P (not defective) = 4/5

P (first six are not defective) = First is not defective & Second is not defective ... = (4/5)^6

Answer seems to be A

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by sanju09 » Wed Nov 06, 2013 2:53 am
mevicks wrote:
sanju09 wrote:Records suggest that out of every 10 Cracker Bombs manufactured by The Fluffy Fireworks, 2 do not blast when cracked. A sample of 10 Cracker Bombs manufactured by The Fluffy Fireworks is taken to crack them one by one. What is the probability that all of the first six Cracker Bombs pulled out to crack, do blast?
A. (4/5)^6
B. 96/625
C. 16/625
D. (1/5)^6
E. (1/5)^6 (4/5)^2
P (defective) = 1/5
P (not defective) = 4/5

P (first six are not defective) = First is not defective & Second is not defective ... = (4/5)^6

Answer seems to be A
How if we only try find the probability that 2 of the last 4 bombs do not blast?
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by mevicks » Wed Nov 06, 2013 3:04 am
sanju09 wrote: How if we only try find the probability that 2 of the last 4 bombs do not blast?
The two bombs dont blast implying that 20% are always defective. Since in our case first six bombs do blast these 20% defective ones fall in the last 4. So imo we can use the P(non defective) probability for first six.

If 2 of the last 4 bombs blast we can use the same logic as all are independent events.
2 blast --> 1/5 * 1/5
2 dont --> 4/5 * 4/5
in all --> 16/5^4

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by sanju09 » Wed Nov 06, 2013 3:06 am
mevicks wrote:
sanju09 wrote: How if we only try find the probability that 2 of the last 4 bombs do not blast?
The two bombs dont blast implying that 20% are always defective. Since in our case first six bombs do blast these 20% defective ones fall in the last 4. So imo we can use the P(non defective) probability for first six.

If 2 of the last 4 bombs blast we can use the same logic as all are independent events.
2 blast --> 1/5 * 1/5
2 dont --> 4/5 * 4/5
in all --> 16/5^4
Just one more step and you are done!
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by mevicks » Wed Nov 06, 2013 3:17 am
is it to do with 5^4 conversion... but I think if we are considering even the first 6 we should reach the same answer? please correct me if wrong!

Edit : Since we have a best case scenario for first six why do should one consider the last 4?

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by sanju09 » Wed Nov 06, 2013 3:25 am
mevicks wrote:is it to do with 5^4 conversion... but I think if we are considering even the first 6 we should reach the same answer? please correct me if wrong!
I mean, consider the last four bombs only. If B for Blast and F for Flop, then the following six outcomes are possible:

BBFF or BFBF or BFFB or FFBB or FBFB or FBBF, and each with a probability 16/625 as mentioned by you. Hence, the correct answer to this question would be

[spoiler]6 (16/625) = 96/625.


My answer is B.
[/spoiler]
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