Step 1: Understanding the Concept:
For any valid probability distribution of a discrete random variable, the sum of all individual probabilities must be exactly equal to 1. By applying this rule, we can determine the unknown constant $k$. Afterward, we compute the required probabilities 'a' and 'b' by summing the relevant individual probabilities and comparing them.
Step 2: Key Formula or Approach:
1. Normalization property: $\sum P(X=x_i) = 1$.
2. Finding interval probabilities: $P(c \le X<d) = \sum_{c \le x_i<d} P(X=x_i)$.
Step 3: Detailed Explanation:
First, find the value of $k$ by summing all probabilities in the table and setting the sum to 1:
\[ P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1 \]
\[ 2k + k + 2k + 4k + k = 1 \]
Combine the terms:
\[ 10k = 1 \implies k = 0.1 \]
Now, calculate the value of $\text{a} = P(x<3)$:
The condition $x<3$ means we include $x=0, x=1$, and $x=2$.
\[ \text{a} = P(X=0) + P(X=1) + P(X=2) \]
\[ \text{a} = 2k + k + 2k = 5k \]
Since $k = 0.1$, $\text{a} = 5 \times 0.1 = 0.5$.
Next, calculate the value of $\text{b} = P(2 \le X<4)$:
The condition $2 \le X<4$ means we include $X=2$ and $X=3$.
\[ \text{b} = P(X=2) + P(X=3) \]
\[ \text{b} = 2k + 4k = 6k \]
Since $k = 0.1$, $\text{b} = 6 \times 0.1 = 0.6$.
Finally, compare the calculated values of $\text{a}$ and $\text{b}$:
We have $\text{a} = 0.5$ and $\text{b} = 0.6$.
Clearly, $0.5<0.6$, which means $\text{a}<\text{b}$.
(Note: You can also directly compare $5k$ and $6k$. Since probabilities must be positive, $k>0$, hence $5k<6k$).
Step 4: Final Answer:
The relationship between the probabilities is $\text{a}<\text{b}$.
