The probability distribution of a discrete random variable $X$ is given as
Find the value of $P(2 < X < 6)$.
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To minimize arithmetic steps during an exam, keep your fractions written in terms of the common denominator $21$ until the final step. Summing the coefficients directly ($3 + 4 + 5 = 12$) before introducing the fraction constant saves time and prevents simple addition errors.
Step 1: Set up the distribution. The variable $X$ takes values $1$ to $6$ with probabilities $K, 2K, 3K, 4K, 5K, 6K$ in turn. Step 2: Use the total-probability law. All probabilities must sum to $1$: $K+2K+3K+4K+5K+6K = 1$. Step 3: Solve for $K$. $21K = 1$, so $K = \dfrac{1}{21}$. Step 4: Identify the required range. We want $P(2 < X < 5)$, the strict interval, which includes $X=3$ and $X=4$ only. Step 5: Add the relevant probabilities. $P(X=3) + P(X=4) = 3K + 4K = 7K$. Step 6: Substitute $K$. $7K = 7\cdot\dfrac{1}{21} = \dfrac{7}{21} = \dfrac{1}{3}$. Listing exactly which values fall inside the open interval avoids accidentally including an endpoint. \[ \boxed{P(2 < X < 5) = \dfrac{1}{3}} \]
