We have, first 30 natural numbers.
Two number are selected a and b such that
a2−b2 is divisible by 2
n(S)=30C2∵a2−b2 is divisible by 3
∴(a+b)(a−b)=3mIf
(a−b) is multiple of 3, then
a2−b2 is multiple of 3
∴ Such numbers are
(1,4,7,....,28),(2,5,....,29),(3,6,9,...,30) are possible are
∴10C2+10C2+10C2=45×3=135o c 5) is multiple of 3 then
a2−b2 is multiple of 3
Such numbers are possible
{(1,2),(1,5)...1,29},{(4,2)(4,5)...}...{28,2)...28,29}
Total
=10×10=100Total numbers are
135+100=235∴ Required probability
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