If a and b are +ve integers.......find the probabilty that \frac{1}{2}(a^{2}+b^{2}) is a +ve integer
62if that is the case then you have to look at the integers a, b
they are of the form 5n or 5n+1 or 5n+4 or 5n+4 or 5n+1 depending on a/b being of the form 5n or 5n+1 or 5n+2 or 5n+3 or 5n+4
Then, to get the sum as a multiple of 5, we have to add particular kind of numbers..
Can you think the rest?
1i am extremely sorry i found out the answer for expression \frac{a^5 +b^5}{5}
if we proceed by same way
the possible combinations are
(5k,5l),(5k+1,5k+2),(5l+1,5k+2),(5k+1,5l+3),(5l+1,5k+3),(5k+4,5l+3),(5l+1,5k+3),(5k+2,5l+4),(5l+2,5k+4).
so 9 favourable cases
total combination is 25
so answer is
9/25
62this much is corect.. but then are you adding the numbers or their squares?
the squares are never of the form 5k+2 and 5k+3
62the answer is indeed 9/25
you should check for some missed cases..
62the answer is indeed 9/25
you should check for some missed cases..
3oh my bad....................SO soryy for that ya its 5 IN THE DENOMINATOR
1what is the correct answer.........& how are you not sure that 1/2 is not the correct answer?
