# probability

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• #531

If the integers m and n are chosen at random from 1 to 100 then the probability that a number of the form 7^m+7^n is divisible by 5 is??

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• #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

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• #539

No,this is not the ans… Ans is 1/4

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• #541

after making the correction i get is 98/4950
case 1 {(1,3) (3,5) (5,7) and so on} they make a total of 49 outcomes
case 2 {(2,4) (4,6) and so on} make a total of 49 outcomes

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• #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 .

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• #545

Rhoit I got tht but if u take the pattern tht I have taken in the correction it gets divisible by 5

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• #547

where is this qn from?

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• #553

What will b the total outcome ???
I got stuck on this point. So someone help me out…..

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• #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 possibility

then the total possibility will be 1250.
Bt I got stuck on total outcome.
What’s total outcome???
M I right upto total possibility??

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• #556

Total outcomes are 100c2

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• #557

If order is imp and repetition is allowed theb total outcomes would bcome 100*100 and fav events 2*1250 and hence the ans becomes 1/4

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• #558

Exactly shruti …I was just forgot to calculate its repetition.
so my method is right???

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• #559

i think sahi hai because we are getting the right answer

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• #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)

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• #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?

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• #567

And i saw this que online.

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