The p.m.f of a d.r.v. X is P(X = x) `= {{:(((. ^(5)C_(x)))/(2^(5))", for","x = 0,1,2,3,4,5"),(" 0,"," otherwise"):}`
If `a = P(X le 2)` and `b = P(X ge 3)` then the relation between `a` and `b` is

Options

A. `a lt b`
B. `a gt b`
C. `a = b`
D. `a+ b =2`

Answer

To solve the problem, we need to find the probabilities \( a = P(X \leq 2) \) and \( b = P(X \geq 3) \) using the given probability mass function (PMF) of the discrete random variable \( X \).

### Step-by-Step Solution:

1. **Identify the PMF**:
The PMF of the discrete random variable \( X \) is given by:
\[ P(X = x) = \frac{5C_x}{2^5} \quad \text{for } x = 0, 1, 2, 3, 4, 5 \]
and \( P(X = x) = 0 \) otherwise.

2. **Calculate \( a = P(X \leq 2) \)**:
We need to calculate the probabilities for \( X = 0, 1, 2 \):
\[ P(X = 0) = \frac{5C_0}{2^5} = \frac{1}{32} \]
\[ P(X = 1) = \frac{5C_1}{2^5} = \frac{5}{32} \]
\[ P(X = 2) = \frac{5C_2}{2^5} = \frac{10}{32} \]

Now, we sum these probabilities:
\[ a = P(X = 0) + P(X = 1) + P(X = 2) = \frac{1}{32} + \frac{5}{32} + \frac{10}{32} = \frac{16}{32} = \frac{1}{2} \]

3. **Calculate \( b = P(X \geq 3) \)**:
We need to calculate the probabilities for \( X = 3, 4, 5 \):
\[ P(X = 3) = \frac{5C_3}{2^5} = \frac{10}{32} \]
\[ P(X = 4) = \frac{5C_4}{2^5} = \frac{5}{32} \]
\[ P(X = 5) = \frac{5C_5}{2^5} = \frac{1}{32} \]

Now, we sum these probabilities:
\[ b = P(X = 3) + P(X = 4) + P(X = 5) = \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2} \]

4. **Compare \( a \) and \( b \)**:
We have found that:
\[ a = \frac{1}{2} \quad \text{and} \quad b = \frac{1}{2} \]
Therefore, we conclude that:
\[ a = b \]

### Final Answer:
The relation between \( a \) and \( b \) is:
\[ a = b \]