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Source: — Problem Solving |
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by cramya » Sat Nov 01, 2008 11:08 am
Prob is not my forte but is it 1/3


P(NOT BURNING IN FIRST 6 MONTHS) = 1/3

P(BURNING IN FIRST 6 MONTHS) = 1 - 1/3 = 2/3


P(BURNING 6 MONTHS - 1 YR) = 1/2*2/3 = 1/3
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by stop@800 » Sat Nov 01, 2008 11:12 am
first 6 months
burn 1/3
not burn 2/3

next 6 months
not burn = 1/2 prev not burn
= 1/2 * 2/3
= 1/3

so burn = 1-1/3
=2/3
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by cramya » Sat Nov 01, 2008 11:12 am
Juts realized my mistake in the post

P( BURNING IN FIRST 6 MONTHS) = 1/3

P( not BURNING IN FIRST 6 MONTHS) = 1 - 1/3 = 2/3


P(BURNING 6 MONTHS - 1 YR) = 1/2*2/3 = 1/3
p(not burning 6 MONTHS - 1 YR)) = 1-1/3 = 2/3


2/3 E) is my final vote(confirming the fact that prob is not my forte :D )

OA please?
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by cramya » Sat Nov 01, 2008 11:17 am
The first time I read it incorrectlty thinking the given data 1/3 was for not burning.

CARELESS ABD LETHARGIC reading can cost u heavily on GMAT: -) A lesson learnt.....

Good luck All!!!!
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by stubbornp » Sat Nov 01, 2008 1:34 pm
oa is not 2/3
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by stop@800 » Sat Nov 01, 2008 2:50 pm
stubbornp wrote:oa is not 2/3
it shud be 4/9.
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by stubbornp » Sun Nov 02, 2008 9:51 am
stop@800 wrote:
stubbornp wrote:oa is not 2/3
it shud be 4/9.
yup it is 4/9....Please explain
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by cramya » Sun Nov 02, 2008 12:14 pm
Ok Shooting in the dark with these prob problems continues.Stp@800 let me know if this is how u came to the soln 4/9


P(BURNING IN 6 MONTHS-1YR) =
P(NOT BURNING IN THE FIRST 6 MONTHS) * P(BURNING IN 6MONTHS-1YR)

P(NOT BURNING IN FIRST 6 MONTHS) = 1-1/3 = 2/3

P(NOT BURNING IN 6 MONTHS-1YR) = 1-2/3 = 1/3 (WHICH IS 1/2 *1/3 AS GIVEN IN THE PROBLEM)

P(BURNING IN 6 MONTHS-1YR) = 1-1/3 = 2/3

P(BURNING IN 6 MONTHS-1YR) = 2/3*2/3 = 4/9

Is this conditional probanility P(A) * P(B/A) WHERE P(A) is not burning in forst 6 months and P(b/a) is probab of burning in 6 months-1yr given the bulb has not got burnt in first 6 months


Suggestions welcome!!!
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by cramya » Sun Nov 02, 2008 12:18 pm
CORRECTION IN BOLD, Sorry guys I made a meal out of this problem:


Ok Shooting in the dark with these prob problems continues.Stop@800 let me know if this is how u came to the soln 4/9


P(BURNING IN 6 MONTHS-1YR) =
P(NOT BURNING IN THE FIRST 6 MONTHS) * P(BURNING IN 6MONTHS-1YR/NOT BURNING IN THE FIRST 6 MONTHS)

P(NOT BURNING IN FIRST 6 MONTHS) = 1-1/3 = 2/3

P(NOT BURNING IN 6 MONTHS-1YR) = 1/2 *2/3 = 1/3

P(BURNING IN 6 MONTHS-1YR/NOT BURNING IN THE FIRST 6 MONTHS) = 1-1/3 = 2/3


P(BURNING IN 6 MONTHS-1YR) = 2/3*2/3 = 4/9

Is this conditional probability P(A) * P(B/A) WHERE P(A) is not burning in forst 6 months and P(b/a) is probab of burning in 6 months-1yr given the bulb has not got burnt in first 6 months


Suggestions welcome!!!
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by luisengard » Mon Nov 03, 2008 11:10 am
cramya wrote: P(BURNING IN 6 MONTHS-1YR) =
P(NOT BURNING IN THE FIRST 6 MONTHS) * P(BURNING IN 6MONTHS-1YR/NOT BURNING IN THE FIRST 6 MONTHS)
Hi guys,

I could use some help, I'm stuck with probability, permutations and combinations.

I don't understant why you state that the answer is that if the question of the problem is exactly "P(BURNING IN 6MONTHS-1YR/NOT BURNING IN THE FIRST 6 MONTHS)", isn't it?

Thanks a lot!!

(BTW is my first post, thank you very much for the help and this lovely forum!)
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by cramya » Mon Nov 03, 2008 2:38 pm
I could be way off here but hope I am not.

I think since these are dependednt events i.e. P(NOT BURNING IN THE FIRST 6 MONTHS) * P(BURNING IN 6MONTHS-1YR/NOT BURNING IN THE FIRST 6 MONTHS)


To find the prob that abulb will burn in 6 months-1yr we have to take in to account the prob that it hasnot burt in the first 6 months. If it did then this probability would be 0

I am sure prob experts can jump in on luisengard's question. Thanks!
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