Question

A random variable $X$ has the following probability distribution:
{|c|c|c|c|c|c|c|} $X = x$ & 1 & 2 & 3 & 4 & 5 & 6 $P(X = x)$ & $k$ & $3k$ & $5k$ & $7k$ & $8k$ & $k$ Then $P(2 \le X < 5) =$

• A. $\frac{7}{25}$
• B. $\frac{3}{5}$
• C. $\frac{24}{25}$
• D. $\frac{23}{25}$

Hint:

Keep a sharp eye on the inequality signs! Notice that $2 \le X < 5$ includes $2$ because of the 'greater than or equal to' sign ($\le$), but excludes $5$ entirely due to the strict 'less than' sign ($<$). Double-checking the endpoints prevents simple counting errors.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a discrete probability distribution table with an unknown constant $k$. We need to find the value of $k$ first, and then calculate the cumulative probability for the interval $2 \le X < 5$.

Step 2: Key Formula or Approach:
1. The sum of all individual probabilities in a valid probability distribution must always equal 1: $$\sum P(X = x) = 1$$ 2. Identify the target values that satisfy the inequality $2 \le X < 5$. Since $X$ is a discrete integer value, the valid values are $X = 2, 3, \text{ and } 4$. $$P(2 \le X < 5) = P(X = 2) + P(X = 3) + P(X = 4)$$

Step 3: Detailed Explanation:
Let's sum all the probabilities from the given table and set them equal to 1: $$k + 3k + 5k + 7k + 8k + k = 1$$ $$25k = 1 \implies k = \frac{1}{25}$$ Now, let's target the exact terms needed for the inequality $2 \le X < 5$: $$P(2 \le X < 5) = P(X = 2) + P(X = 3) + P(X = 4)$$ $$P(2 \le X < 5) = 3k + 5k + 7k = 15k$$ Substitute the value of $k = \frac{1}{25}$ into the expression: $$P(2 \le X < 5) = 15 \times \frac{1}{25} = \frac{15}{25} = \frac{3}{5}$$

Step 4: Final Answer:
The value of the required probability is $\frac{3}{5}$, which matches option (B).