For the probability distribution, ∑p_i = 1
⇒ a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1
⇒81a = 1 ⇒ a = \(\frac{1}{81}\)
Also P(X< 3) = P(X = 0)+P(X = 1) + P(X = 2)
= a + 3a + 5a = 9a = 9(\(\frac{1}{81}\))
= \(\frac{1}{9}\)

Since the random variable. X is the probability mass function, Σp_i = 1
\(\Rightarrow \frac { k }{ 6 } +\frac { k }{ 4 } +\frac { k }{ 12 } =1\Rightarrow \frac { 2k+3k+k }{ 12 } =1\)
\(\Rightarrow \frac { 6k }{ 12 } =1\Rightarrow k=\frac { 12 }{ 6 } =2\quad \therefore k=2\)

A bag contains 2 bad eggs and 8 good eggs
∴ Total number of eggs = 10
We are going to select 3 eggs, out of that 2 must be bad eggs.
∴ Required probability \(=\frac { { 2C }_{ 2 }\times { 8C }_{ 1 } }{ 10{ C }_{ 3 } } =\frac { 1\times 8 }{ \frac { 10\times 9\times 8 }{ 3\times 2\times 1 } } \)
\(=\frac { 1\times 8\times 3\times 2\times 1 }{ 10\times 9\times 8 } =\frac { 1 }{ 15 } \)
∴ Probability of getting two bad eggs = \(\frac{1}{15}.\)

When a die is rolled, sample space 
S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6
∴P(X = 0) = Probability of getting an odd number = \(\frac{3}{6}\)[∵ Their are 3 favourable events]
= \(\frac{1}{2}\)
Thus, the probability distribution of the random variable X is given by

P(X = 0) + P(X = 1) + P(X = 2)
= 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Hence the given distribution of probabilities is not a probability distribution.