Bombs pulled out to crack, do blast?

theCodeToGMAT wrote:Is it [spoiler]{A}[/spoiler]?

sanju09 wrote:

theCodeToGMAT wrote:Is it [spoiler]{A}[/spoiler]?

[spoiler]It doesn't match my answer. Can you tell me your approach?[/spoiler]

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

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

sanju09 wrote: How if we only try find the probability that 2 of the last 4 bombs do not blast?

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

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!