The probability distribution of random variable X is given by: Let p = P(1

Q: Wait, why do we assume the sum of probabilities is 1? Is that always true?

That is the fundamental axiom of probability! In any sample space, the sum of probabilities of all mutually exclusive and exhaustive outcomes must be 1. If it weren't, the model wouldn't be a valid probability distribution.

Q: I'm confused about the intersection $A \cap B$. Why is it just $X=2$?

Great question. Event $A$ is $1 < X < 4$, so $X \in \{2, 3\}$. Event $B$ is $X < 3$, so $X \in \{1, 2\}$. The intersection is the set of values present in both lists. Only $X=2$ appears in both!

Q: Do I really need to find the value of $K$ to solve for $p$?

Mathematically, you might notice $K$ cancels out in the ratio $p = \frac{2K}{3K} = \frac{2}{3}$. However, you *do* need $K$ to solve the final equation $5p = \lambda K$. So, don't skip finding $K$!

Question

The probability distribution of random variable $X$ is given by: $$\begin{array}{|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 \ \hline P(X) & K & 2K & 2K & 3K & K \ \hline \end{array}$$ Let $p = P(1 < X < 4 | X < 3)$. If $5p = \lambda K$, then $\lambda$ equal to .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To begin, we must determine the value of the mystery constant $K$. For any valid discrete probability distribution, the sum of all probabilities must equal unity.

This normalization condition is expressed as:

$$\sum_{i=1}^{5} P(X_i) = 1$$

By summing the given terms, we obtain:

$$K + 2K + 2K + 3K + K = 1$$

Combining these coefficients, we arrive at $9K = 1$, which reveals that $K = \frac{1}{9}$. This value serves as the foundation for our entire calculation.

### The Conditional Universe: Defining Events A and B

We are tasked with finding $p = P(1 < X < 4 | X < 3)$. This notation restricts our sample space to the condition $X < 3$.

Let us define our events clearly. Event $A$ is defined by the inequality $1 < X < 4$. The values of $X$ that satisfy this are $X=2$ and $X=3$. Thus, $A = \{2, 3\}$.

Next, we define our condition, Event $B$, as $X < 3$. The values of $X$ that satisfy this are $X=1$ and $X=2$. Thus, $B = \{1, 2\}$.

### The Intersection: Finding the Overlap

The formula for conditional probability is defined as:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

To solve this, we identify the intersection $A \cap B$. Comparing our sets $A = \{2, 3\}$ and $B = \{1, 2\}$, the only common element is $X=2$.

Therefore, $A \cap B = \{2\}$. The numerator of our probability fraction is the probability of this intersection:

$$P(A \cap B) = P(X=2) = 2K$$

The denominator is the probability of the condition $B$:

$$P(B) = P(X=1) + P(X=2) = K + 2K = 3K$$

### The Final Synthesis: Solving for $\lambda$

With the components identified, we calculate $p$:

$$p = \frac{2K}{3K} = \frac{2}{3}$$

The problem provides the final equation $5p = \lambda K$. Substituting our known values $p = \frac{2}{3}$ and $K = \frac{1}{9}$, we get:

$$5 \left( \frac{2}{3} ight) = \lambda \left( \frac{1}{9} ight)$$

This simplifies to:

$$\frac{10}{3} = \frac{\lambda}{9}$$

Multiplying both sides by $9$, we find:

$$\lambda = \frac{10 	imes 9}{3} = 30$$

The final result is **$\lambda = 30$**.

Animated Solution for Mathematics - Probability: The probability distribution of random variable $X$ is given by: $X P (X) 1 K 2 2 K 3 2 K 4 3 K 5 K$ Let $p = P (1 < X < 4∣ X < 3)$ . If $5 p = λ K$ , then $λ$ equal to .

Enter Numerical Value:

Visualized Solution

Understanding the Distribution

Summing the Probabilities

Solving for $K$

Defining Conditional Probability

Identifying Event $A$

Identifying Event $B$

Finding the Intersection

Calculating Numerator and Denominator

Calculating $p$

Setting up the Final Equation

Solving for $λ$

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Conditional Universe

The Intersection

The Final Synthesis

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