Problem Solving

orps1

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

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

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 )

OA please?

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!!!!

oa is not 2/3

stubbornp wrote:oa is not 2/3

it shud be 4/9.

stop@800 wrote:

stubbornp wrote:oa is not 2/3

it shud be 4/9.

yup it is 4/9....Please explain

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!!!

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!!!

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!)

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!