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\((a^2+b^2)/10\)
is integer

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[0] means ?

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Is it clear?

trying to understand...seeing this type of approach for first time...

how did u get a,b belongs t0 {......}mod 5 ?

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i think i am getting this a bit.....need to think

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can u expalin more on why there is just one option ?

a and b are independent so P(a,b)=P(a).P(b)

i think i got other parts, except that there is only 1 option

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ohh....right...i'll go through this again. thanks!

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Also, any alternative approach will be appreciated....

for some reason, i only have p = 0.10

me too, but thats incorrect.

i found my error. it is as what mahmit gave. but i have a different proof.

plz share.

sorry,didn't get u...

still polishing my argument.

1/6

oops, i got 0.18 this time. hahaha. do you have the correct answer?

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no, i don't have correct answer

for a^2, probability is 1/6, i get that
what about a^2+b^2

i got it, 0.18

Now what is probability:|dw:1352033286386:dw|

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are both of you doing same thing ?

Yes.

Can you write a table for \(a^2+b^2\) ? With probabilities?

it will also contain 10 columns,right ?
trying....

Right.

If you want to get it, better to draw the whole table for \(a^2+b^2\).

why is the answer drastically different from 2% ?

here's the first half of (a,b) mod 10
(0,0), (1,3), (1,7), (2,4), (2,6), (3,1), (3,9), (4,2),(4,8)

Answer is \(\frac{14}{100}=0.14\).

the 2% only counted 5,5 and 0,0.

I used Octave

for i=1:10;for j=1:10;A(i,j)=mod(i^2+j^2,10);endfor;endfor

here's the other nine
(5,5)
(6,2), (6,8)
(7,1), (7,9)
(8,4), (8,6)
(9,3), (9,7)

Sorry, my fault. Really - \(\frac{18}{100}\)

i got now, how we got 18 pairs for 100....

finally, we are all in agreement. hurrah.

why consider only 100?
will it make any difference, if we consider 1000 numbers ?

because if we consider only 10 numbers, we get different answer....

0.2

aah! okk, got that....

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0.18 is not true.
0.02 is true. Be sure.

where's the fault in our discussion ?

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