Consider the probability distribution:
Then the value of $P(X > 2)$ is ______.
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To save time on calculation, you can also compute $P(X > 2)$ using the complement rule: $P(X > 2) = 1 - [P(X=1) + P(X=2)] = 1 - (K + 2K) = 1 - 3K = 1 - 3(1/6) = 1 - 1/2 = 1/2$.
Step 1: Understanding the Question:
We are given a discrete probability distribution containing an unknown constant $K$. We must first find the legitimate value of $K$ and then calculate a specific cumulative probability. Step 2: Key Formula or Approach:
The fundamental rule of probability distributions states that the sum of all individual probabilities must exactly equal 1:
$$\sum P(X=x_i) = 1$$
Additionally, every individual probability must be non-negative: $P(X=x) \ge 0$, meaning $K$ cannot be negative. Step 3: Detailed Explanation:
1. Sum all the given probabilities and set equal to 1:
$$K + 2K + K^2 + 2K + 5K^2 = 1$$
Combine like terms to form a quadratic equation:
$$6K^2 + 5K - 1 = 0$$
Factor the quadratic by splitting the middle term:
$$6K^2 + 6K - K - 1 = 0$$
$$6K(K + 1) - 1(K + 1) = 0$$
$$(6K - 1)(K + 1) = 0$$
This gives roots $K = 1/6$ or $K = -1$.
Since probability cannot be negative, we reject $K = -1$. Thus, $K = 1/6$.
2. Calculate the required probability $P(X > 2)$:
$P(X > 2)$ means the probability of $X$ being 3, 4, or 5.
$$P(X > 2) = P(X=3) + P(X=4) + P(X=5)$$
Substitute their respective formulas from the table:
$$P(X > 2) = K^2 + 2K + 5K^2 = 6K^2 + 2K$$
Substitute $K = 1/6$:
$$P(X > 2) = 6\left(\frac{1}{6}\right)^2 + 2\left(\frac{1}{6}\right)$$
$$P(X > 2) = 6\left(\frac{1}{36}\right) + \frac{2}{6}$$
$$P(X > 2) = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$$ Step 4: Final Answer:
The value of $P(X > 2)$ is 1/2, matching option (c).
