Question

A random variable X has the following probability distribution

Then P ( 3 < x $\le$ 6 ) =

• A. $\frac{3}{7}$
• B. $\frac{4}{7}$
• C. $\frac{13}{21}$
• D. $\frac{8}{21}$

Hint:

When counting the coefficients for total probability, group them symmetrically from the outside in: $(1+1+1+1) + (2+2) + (3+3) + (4+4) = 4 + 4 + 6 + 8 = 21$. This quick visual grouping prevents calculation mistakes when summing long strings of terms on scratch paper.

Answer 1

The Correct Option is A

Step 1: Recall the total-probability rule.
All the probabilities $K, 2K, 3K, 4K, 4K, 3K, 2K, K, K$ for $x = 0$ to $8$ must add to 1.
Step 2: Add the coefficients.
$1+2+3+4+4+3+2+1+1 = 21$, so $21K = 1$.
Step 3: Solve for K.
$K = \dfrac{1}{21}$.
Step 4: Identify the required interval.
We need $P(3 < X \le 6) = P(4) + P(5) + P(6)$, using the values for $x = 4, 5, 6$.
Step 5: Add those probabilities.
$P(4) = 4K$, $P(5) = 3K$, $P(6) = 2K$, so the sum is $4K + 3K + 2K = 9K$.
Step 6: Substitute K.
$9K = 9\cdot\dfrac{1}{21} = \dfrac{9}{21} = \dfrac{3}{7}$, which is option 1 and matches the key.
\[ \boxed{\dfrac{3}{7}} \]