Key Concepts and Formulas

This problem requires a strong understanding of discrete probability distributions and conditional probability.

1. Probability Mass Function (PMF) Property: For any discrete random variable $X$, the sum of probabilities for all possible outcomes must equal 1. This ensures a complete and valid probability distribution. $\sum_{\text{all } x_i} P(X=x_i) = 1$

2. Conditional Probability: The probability of an event A occurring, given that another event B has already occurred, is defined as: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ where $P(A \cap B)$ is the probability that both events A and B occur, and $P(B)$ is the probability of event B occurring. This formula is applicable only when $P(B) > 0$.

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Step-by-Step Solution

Step 1: Determine the Value of the Constant $k$

• Why this step? The very first step in working with any probability distribution involving an unknown constant is to find that constant. This is achieved by using the fundamental property that the sum of all probabilities in a probability mass function (PMF) must equal 1. This ensures the distribution is valid.
• Working: We are given the probabilities for $X=0, 1, 2, 3, 4$ as $k, 2k, 4k, 6k, 8k$ respectively. Summing these probabilities and equating them to 1: $P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1$ $k + 2k + 4k + 6k + 8k = 1$ Combining the terms: $(1+2+4+6+8)k = 1$ $21k = 1$
• Result: Solving for $k$: $k = \frac{1}{21}$

Step 2: Define the Events for Conditional Probability

• Why this step? The problem asks for $P(1 < X < 4 | X \le 2)$. To apply the conditional probability formula, we must clearly identify the two events involved. Let's call the event we are interested in 'A' and the given condition 'B'.
• Working:
  • Let Event A be $1 < X < 4$.
    • Since X is a discrete random variable, the integer values strictly greater than 1 and strictly less than 4 are $X=2$ and $X=3$.
    • So, Event A corresponds to the outcomes $\{X=2, X=3\}$.
  • Let Event B be $X \le 2$.
    • This means $X$ can take any value from its defined domain that is less than or equal to 2. From the given distribution, these are $X=0, X=1,$ and $X=2$.
    • So, Event B corresponds to the outcomes $\{X=0, X=1, X=2\}$.

Step 3: Calculate the Probability of Event B, $P(B)$

• Why this step? The denominator of the conditional probability formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$ is $P(B)$. We need to calculate the probability that event B ($X \le 2$) occurs.
• Working: We sum the probabilities of the outcomes that constitute Event B: $P(B) = P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$ Using the probabilities given in terms of $k$: $P(B) = k + 2k + 4k$ $P(B) = 7k$ Now, substitute the value of $k = \frac{1}{21}$ found in Step 1: $P(B) = 7 \times \frac{1}{21} = \frac{7}{21} = \frac{1}{3}$

Step 4: Determine the Intersection of Events A and B, $P(A \cap B)$

• Why this step? The numerator of the conditional probability formula is $P(A \cap B)$. This represents the probability that both event A ($1 < X < 4$) and event B ($X \le 2$) occur simultaneously. We need to find the values of $X$ that satisfy both conditions.
• Working:
  • Event A includes $X \in \{2, 3\}$.
  • Event B includes $X \in \{0, 1, 2\}$.
  • The intersection $A \cap B$ consists of the values of $X$ common to both sets. The only common value is $X=2$.
  • Therefore, $A \cap B = \{X=2\}$.
  • Now, we calculate the probability of this intersection: $P(A \cap B) = P(X=2)$ From the given distribution, $P(X=2) = 4k$. $P(A \cap B) = 4k$ Substitute the value of $k = \frac{1}{21}$: $P(A \cap B) = 4 \times \frac{1}{21} = \frac{4}{21}$

Step 5: Apply the Conditional Probability Formula

• Why this step? With $P(A \cap B)$ and $P(B)$ calculated, we can now use the definition of conditional probability to find the desired value, $P(A|B)$.
• Working: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ Substitute the numerical values we found: $P(1 < X < 4 | X \le 2) = \frac{\frac{4}{21}}{\frac{1}{3}}$ To simplify, multiply the numerator by the reciprocal of the denominator: $P(1 < X < 4 | X \le 2) = \frac{4}{21} \times \frac{3}{1} = \frac{12}{21}$ Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 3: $P(1 < X < 4 | X \le 2) = \frac{4}{7}$

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Common Mistakes & Tips

1. Always Find k First: Neglecting to find the constant k or making an error in its calculation will propagate through the entire problem. This is a crucial first step.
2. Careful with Inequalities: For discrete random variables, be very precise when converting inequalities ($<, >, \le, \ge$) into the specific set of values $X$ can take. A common mistake is confusing strict inequalities ($<, >$) with non-strict ones ($\le, \ge$).
3. Clearly Define Events: Explicitly writing out the outcomes for events A, B, and their intersection $A \cap B$ helps avoid errors and provides clarity, especially under exam pressure.
4. k Cancellation: In many conditional probability problems involving an unknown constant like k, the constant often cancels out when forming the ratio $\frac{P(A \cap B)}{P(B)}$. This can serve as a quick check for your calculations.

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Summary

To find the conditional probability $P(1 < X < 4 | X \le 2)$, we first determined the value of the constant $k$ by using the property that the sum of all probabilities in a discrete distribution must equal 1. Then, we clearly defined the events A ($1 < X < 4$) and B ($X \le 2$), listing the possible outcomes for each. We calculated the probability of event B, $P(B)$, and the probability of the intersection of events A and B, $P(A \cap B)$. Finally, we applied the conditional probability formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$ to arrive at the result.

The final answer is $\boxed{\frac{4}{7}}$, which corresponds to option (A).