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Tagged: probability
This topic contains 15 replies, has 5 voices, and was last updated by harshita 5 years, 5 months ago.

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August 11, 2014 at 3:46 pm #531

August 11, 2014 at 4:51 pm #537
i’m not sure but i think the answer is 50c2/100c2+50c2/100c2
how i got here is tht if u observe the pattern then u will notice tht if both m and n are odd or both m and n are even then 7^m+7^n will be divisible by 5 so there are 2 cases
case 1 when both m and n are odd
case 2 whn both are even
and the prbability will be case1 + case 2
shruti290Quote 
August 11, 2014 at 5:28 pm #539

August 11, 2014 at 6:00 pm #541


August 12, 2014 at 1:39 pm #543
if we take m=1 and n=5 it is not divisble by 5 so we cant generalize m and nas odd will always yield numbers divisible by 5 .
rohit pandeyQuote
August 12, 2014 at 2:04 pm #545


August 12, 2014 at 5:16 pm #547

August 13, 2014 at 11:40 am #553
What will b the total outcome ???
I got stuck on this point. So someone help me out…..
sumit tripathiQuote 
August 13, 2014 at 1:29 pm #555
I think there will be 25 series that satisfy this condition e.g
(1,3)(2,4)(3,5)—–(98,100)=98 possibility
(1,7)(2,8)(3,9)—(94,100)=94 possibility
(1,11)(2,12)—–(90,100)=90
possibility
!
!
!
!
(1,99)(2,100)=2 possibilitythen the total possibility will be 1250.
Bt I got stuck on total outcome.
What’s total outcome???
M I right upto total possibility??
sumit tripathiQuote 
August 13, 2014 at 7:44 pm #558
Exactly shruti …I was just forgot to calculate its repetition.
so my method is right???
sumit tripathiQuote
August 13, 2014 at 8:12 pm #559


August 14, 2014 at 5:03 pm #564
guys, I’m unable to follow this discussion. apologies. but you’re getting the right answer…
the way i was thinking about this is: the ending numbers of powers of 7 cycle between 7, 9, 3, 1. So the remainders are 2, 4, 3, 1. Now pick any n. Suppose 7^n gives you a number ending in 7. ie.. a remainder of 2. So, you need 7^m to give you a remainder of 3. i.e. 7^m should end in 3. And this will happen for exactly 1/4th of the numbers. Which implies the probability should be 1/4 (since we picked a random n)
adminQuote 
August 14, 2014 at 5:17 pm #566
Sir i am not getting it… For divisibility by 5 we need the units digit to be either 0 or 5… So units digit in the power of 7 will follow the pattern 9,3,1,7 like this… N i think we have to divide the elements into some groups like (3,1) aise… But m not able to execute it properly… N sir how do we come to know that it is exactly for 1/4th of the nos?
harshitaQuote 
August 14, 2014 at 5:18 pm #567

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