The key point to solve the problem:

If a probability distribution is given then as per its definition, Sum of probabilities associated with each value of a random variable of given distribution is equal to 1

i.e. ∑(p_i) = 1

∴ Sum of probabilities = 1

∴ a+3a+5a+7a+9a+11a+13a+15a+17a = 1

a(1+3+5+7+9+11+13+15+17) = 1

Thus, a = ………..ans (i)

P(X<3) = P(X = 0) + P(X = 1) + P(X = 2)

= a+3a+5a

= 9a =

P(X≥3) = 1 - P(X<3) {∵ sum of probabilities in distribution is 1}

=

P(0<X<5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 3a + 5a + 7a + 9a

= 24a

= ……..ans (ii)