19-02-09 Divisible by 12

what is the probability that the product of two randomly chosen integers is divisible by 12?

39 Answers

1
neil.dhruva ·

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

33
Abhishek Priyam ·

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

13
Двҥїяuρ now in medical c ·

is the ans 5/24

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

62
Lokesh Verma ·

yes abhirup.. I got the same answer :)

21
tapanmast Vora ·

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

21
tapanmast Vora ·

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

11
Sunil Kumar ·

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

1
Terminator ·

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

11
rkrish ·

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 n→∞ (n2C1.n6C1 + n3C1.n4C1 + n12C1.2nC1) / 2nC2

= Lt n→∞ (n2.n6 + n3.n4 + n12.2n) / n(2n-1)

= Lt n→∞ ( (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 !!!

21
tapanmast Vora ·

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

13
Двҥїяuρ now in medical c ·

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

11
Sunil Kumar ·

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

21
tapanmast Vora ·

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]

11
Mani Pal Singh ·

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

11
rkrish ·

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

62
Lokesh Verma ·

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

where are the cases 12n+0?

21
tapanmast Vora ·

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

11
rkrish ·

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

62
Lokesh Verma ·

@ 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!

11
rkrish ·

thnx bhaiyya.

11
Mani Pal Singh ·

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

21
tapanmast Vora ·

sir,

is any of our answers corrct?

11
Mani Pal Singh ·

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

3
iitimcomin ·

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

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

21
tapanmast Vora ·

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

1
neil.dhruva ·

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

21
tapanmast Vora ·

allllllllll integers.....

11
Mani Pal Singh ·

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

33
Abhishek Priyam ·

hmmm.... [12] sochte hainn.

21
tapanmast Vora ·

HUA KYA?

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