A random variable X has the following probability distribution
Then P ( 3 < x $\le$ 6 ) =
Show 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.
Step 1: Understanding the Question:
We are given a discrete probability distribution table with an unknown parameter $K$. We need to compute the cumulative probability inside the specific interval $3 < X \le 6$.
Step 2: Key Formula or Approach:
1. The sum of all individual probabilities in a valid discrete distribution must equal 1:
$$\sum P(X = x_i) = 1$$
2. The interval probability $P(3 < X \le 6)$ sums the discrete values within that range, namely $X = 4, 5, 6$:
$$P(3 < X \le 6) = P(X = 4) + P(X = 5) + P(X = 6)$$
Step 3: Detailed Explanation:
The probability distribution from the reference data contains the following values:
$$\begin{array}{c|c|c|c|c|c|c|c|c|c}
x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \ \\
\hline
P(X=x) & K & 2K & 3K & 4K & 4K & 3K & 2K & K & K
\end{array}$$
3. Set up the equation for the sum of all probabilities to find $K$:
$$K + 2K + 3K + 4K + 4K + 3K + 2K + K + K = 1$$
Summing the coefficients:
$$21K = 1 \implies K = \frac{1}{21}$$
4. Identify the probabilities for the target interval $3 < X \le 6$:
For $X = 4$: $P(X = 4) = 4K$
For $X = 5$: $P(X = 5) = 3K$
For $X = 6$: $P(X = 6) = 2K$
3. Sum these components:
$$P(3 < X \le 6) = 4K + 3K + 2K = 9K$$
Substitute $K = \frac{1}{21}$ into the expression:
$$P(3 < X \le 6) = 9 \times \frac{1}{21} = \frac{9}{21}$$
Reduce the fraction by dividing the numerator and denominator by their greatest common divisor, 3:
$$P(3 < X \le 6) = \frac{3}{7}$$
Step 4: Final Answer:
The value of the probability is $\frac{3}{7}$, which corresponds to option (A).
