Question Of the Day - 19-02-09 Divisible by 12

1

neil.dhruva ·2009-02-22 01:35:20

i am getting the same as tapan...5/18

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33

Abhishek Priyam ·2009-02-21 23:09:32

nishant bhaiya.. jaldi... instant chat dekhiye..

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-02-21 23:11:16

is the ans 5/24

i m not sure about being multiple of 4...is it 5/16[7]

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62

Lokesh Verma ·2009-02-21 23:30:02

yes abhirup.. I got the same answer :)

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21

tapanmast Vora ·2009-02-21 23:34:09

@abhi can u pl. xplain ur soln!!!

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21

tapanmast Vora ·2009-02-21 23:37:07

these + all the cases wer 12n + 12 is one of the 2 nos...

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11

Sunil Kumar ·2009-02-21 23:39:07

bhai log explain the solution yaar.............
Probability aur algebra me haath thoda dheela hai............

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1

Terminator ·2009-02-22 00:09:36

5/24.........itz the correct answer.......[1][1][1][1][1][1][1][1]..............

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11

rkrish ·2009-02-22 00:16:49

Let there be 2n integers i.e. -n to n
Out of these ,
2k type : 2[n/2] + 1 = n₂
3k type : 2[n/3] + 1 = n₃
4k type : 2[n/4] + 1 = n₄
6k type : 2[n/6] + 1 = n₆
12k type : 2[n/12] + 1 = n₁₂

where [.] denotes G.I.F.

Let Required Probability = P

P = Lt _{nâ†’âˆž} (^n₂C₁.^n₆C₁ + ^n₃C₁.^n₄C₁ + ^n₁₂C₁.²ⁿC₁) / ²ⁿC₂

= Lt _{nâ†’âˆž} (n₂.n₆ + n₃.n₄ + n₁₂.2n) / n(2n-1)

= Lt _{nâ†’âˆž} ( (2[n/2]+1).(2[n/6]+1) + (2[n/3]+1).(2[n/4]+1) + (2[n/12]+1).(2n) ) / n(2n-1)

Someone Plz. tell me how to proceed after this !!!

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21

tapanmast Vora ·2009-02-22 01:23:59

pl. xplain ur ans : terminator!!!!!!

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-02-21 23:08:08

the probability that the product is multiple of 3 is 2/3[7]

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11

Sunil Kumar ·2009-02-22 02:08:48

@rkrish. yaar kucch samajh nahi aaya.........

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21

tapanmast Vora ·2009-02-22 03:29:58

Sir, can u pl. tell me wer hav I gone wrong in my approach????? [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7] [7]

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11

Mani Pal Singh ·2009-02-22 03:51:40

i think if we consider integers upto 12 we could get requisite answer
the total cases will be 11
and possible outcome is 1
so answer is 1/11 as per me
ISSKO DHYAAN SE DEKHO

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11

rkrish ·2009-02-22 04:16:37

@sunil....which part samajh nahi aaya???

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62

Lokesh Verma ·2009-02-22 06:40:13

@tapan.. your method is correct.. but you have not counted the cases correctly!

where are the cases 12n+0?

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21

tapanmast Vora ·2009-02-22 06:52:03

NO sir i've addded them in the factor : "12 + 11" 11 bcoz (12n + 0)*(12n + 0) is same so ek bar kum.....

otherwise 17 + 12 + 11 = 40

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11

rkrish ·2009-02-22 07:25:06

@nishant bhaiyya..
is my approach correct???...if yes pls help to proceed.

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62

Lokesh Verma ·2009-02-22 07:30:57

@ rkrish

note that you have brought down an infinite set to a set of 12 sets given by

12n, 12n+1, 12n+2, 12n+3 and so on..

So you should look at these 12 sets only and forget that n is infinite!

From here your approach should help a lot!

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11

rkrish ·2009-02-22 07:32:53

thnx bhaiyya.

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11

Mani Pal Singh ·2009-02-21 11:34:46

kya hua tapan mera sol pasand nahin aaya kya??????

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21

tapanmast Vora ·2009-02-21 06:02:01

sir,

is any of our answers corrct?

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11

Mani Pal Singh ·2009-02-21 06:04:36

sir is ur ques complete???????????????

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3

iitimcomin ·2009-02-21 06:45:20

1/12??????????

sorry im pathetic at probability!!!!![2]

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21

tapanmast Vora ·2009-02-21 07:23:52

que 2
wat r the odds in favour of one of the above answers being correct!!!

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1

neil.dhruva ·2009-02-21 10:04:42

wht's the limit between which we have to select the integers?

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21

tapanmast Vora ·2009-02-21 10:08:14

allllllllll integers.....

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11

Mani Pal Singh ·2009-02-21 10:16:21

i think if we consider integers upto 12 we could get requisite answer
the total cases will be 11
and possible outcome is 1
so answer is 1/11 as per me

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33

Abhishek Priyam ·2009-02-21 10:16:27

hmmm.... [12] sochte hainn.

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21

tapanmast Vora ·2009-02-21 11:02:44

HUA KYA?

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19-02-09 Divisible by 12

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