Three distinct numbers a, b and c are chosen at random from the numbers 1 ,2, ..., 100. The probability that

n(s)=100C3a. 2b = a + c = evenThis means that a and c are both even or both odd.n(E)=50C2+50C2=50×49b. Taking r = 2, 3, . .., 10,a, b, c can be in GP in 53 ways.c. 1a,1b,1c are in GP = a, b, c are in GPd. P(a+b+c is even )= 50C3+50C1×50C2 100C3=12