19-02-09 Divisible by 12

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

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

is the ans 5/24

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

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

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

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

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 !!!

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

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]

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

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

sir,

is any of our answers corrct?

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

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

allllllllll integers.....

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