Step 1: Understanding the Concept:
First, we need to find the value of $k$ by using the property that the sum of all probabilities in a probability distribution equals 1.
Then, we calculate the conditional probability using the formula $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$. Step 2: Key Formula or Approach:
$\sum P(X) = 1$
Conditional Probability: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$
Let $A$ be the event $1 \le X<4$ and $B$ be the event $X \le 2$. Step 3: Detailed Explanation:
Sum of probabilities:
$P(0) + P(1) + P(2) + P(3) + P(4) = 1$
$k + 2k + 4k + 2k + k = 1$
$10k = 1 \Rightarrow k = 0.1$
However, we might not even need to substitute $k$'s numerical value if it cancels out.
We want to find $P(1 \le X<4 \mid X \le 2)$.
Using the conditional probability formula:
\[ P(1 \le X<4 \mid X \le 2) = \frac{P((1 \le X<4) \cap (X \le 2))}{P(X \le 2)} \]
The condition $1 \le X<4$ means $X$ can be $1, 2, \text{ or } 3$.
The condition $X \le 2$ means $X$ can be $0, 1, \text{ or } 2$.
The intersection of these two sets of outcomes is $\{1, 2, 3\} \cap \{0, 1, 2\} = \{1, 2\}$.
So, the numerator is $P(X = 1 \text{ or } X = 2) = P(X=1) + P(X=2)$.
Numerator $= 2k + 4k = 6k$.
The denominator is $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.
Denominator $= k + 2k + 4k = 7k$.
Now substitute these back into the fraction:
\[ P(1 \le X<4 \mid X \le 2) = \frac{6k}{7k} \]
Since $k \neq 0$, we can cancel it out:
\[ P(1 \le X<4 \mid X \le 2) = \frac{6}{7} \] Step 4: Final Answer:
The value is $\frac{6}{7}$.
