If a random variable x has the probability distribution (Table), then P(3

Q: Wait, why do we reject the negative root here? Can't $k$ be negative?

That is a fantastic question! In pure algebra, $k = -1$ is a perfectly valid solution to the quadratic equation. However, we are working with probability. By definition, the probability of any event $P(x)$ must be greater than or equal to 0. If you plug $k = -1$ into the expression for $P(1)$, which is $2k$, you get $-2$. A negative probability is physically impossible. Therefore, we must reject any value of $k$ that leads to a negative probability.

Q: Why can't we just use energy conservation or some other physics principle here?

I love the enthusiasm for physics! But here, we are strictly in the realm of statistics and probability. The 'conservation' principle we use is the conservation of total probability. Just as energy is conserved in a closed system, the total probability of all possible outcomes in a sample space must always be conserved at 1. It is the fundamental axiom that governs this entire problem.

Q: How do I know which values of $x$ to include in $P(3 < x \le 6)$?

Look closely at the inequality signs. The expression $3 < x \le 6$ means $x$ is strictly greater than 3 (so we exclude 3) and less than or equal to 6 (so we include 6). Since our random variable $x$ takes integer values, the integers that satisfy this condition are 4, 5, and 6. Always take a moment to draw a number line if you are ever unsure about inclusive versus exclusive boundaries!

Question

If a random variable $x$ has the probability distribution $$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \ \hline P(x) & 0 & 2k & k & 3k & 2k^2 & 2k & k^2 + k & 7k^2 \ \hline \end{array}$$ then $P(3 < x \le 6)$ is equal to :

Options:

(A) 0.64

(B) 0.22

(C) 0.33

(D) 0.34

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student. Today, we are going to peel back the layers of a classic probability problem. Often, when you see a table filled with variables like $k$, it is easy to feel a sense of dread.

But I want you to look at this table not as a source of confusion, but as a puzzle waiting to be solved. We have a discrete random variable $x$ and a set of probabilities defined by an unknown constant $k$. Our mission is to find the probability of a specific range of outcomes, but first, we must identify $k$.

### The Fundamental Axiom: The Conservation of Probability

Think of the entire probability distribution as a complete, closed system. If you were to list every single possible outcome of an experiment and add up their individual probabilities, the sum must equal 1.

This is the fundamental axiom of probability:

$$\sum P(x_i) = 1$$

It is the bedrock upon which all of statistics is built. Without this, the entire structure collapses. So, let us apply this to our table by setting the sum of all probabilities to 1:

$$0 + 2k + k + 3k + 2k^2 + 2k + (k^2 + k) + 7k^2 = 1$$

### The Quadratic Battle

Now, let us bring order to this chaos by grouping the terms by their powers of $k$. First, the quadratic terms: $2k^2$, $k^2$, and $7k^2$ sum to $10k^2$.

Next, the linear terms: $2k$, $k$, $3k$, $2k$, and the $k$ from the bracket $(k^2 + k)$ sum to $9k$. Our equation now stands as:

$$10k^2 + 9k = 1$$

To solve this, we bring the 1 to the left side, transforming it into the standard quadratic form:

$$10k^2 + 9k - 1 = 0$$

We need to split the middle term by finding two numbers that multiply to $-10$ and add to $9$. Those numbers are $10$ and $-1$. Thus, we rewrite the equation as:

$$10k^2 + 10k - k - 1 = 0$$

Factoring by grouping, we get $(10k - 1)(k + 1) = 0$. This yields two potential values for $k$:

$$k = \frac{1}{10} \quad 	ext{or} \quad k = -1$$

### The Reality Check

Here is where the true mathematician separates themselves from the calculator. We have two roots, but we must determine if they are physically meaningful.

Probability cannot be negative. If we test $k = -1$, we find that $P(1) = 2(-1) = -2$, which is an impossibility. We must reject $k = -1$ and embrace **$k = 0.1$**.

### The Final Target

The question asks for $P(3 < x \le 6)$. We must be precise with our boundaries; the inequality $3 < x \le 6$ tells us to exclude 3 but include 6.

Thus, we are looking for the sum of probabilities for $x = 4, 5, 6$:

$$P(4) + P(5) + P(6) = 2k^2 + 2k + (k^2 + k) = 3k^2 + 3k$$

Substituting our value $k = 0.1$ into the expression:

$$3(0.1)^2 + 3(0.1) = 3(0.01) + 0.3 = 0.03 + 0.3 = 0.33$$

And there you have it! Through careful summation, algebraic rigor, and a healthy dose of physical intuition, we have arrived at the final answer: **$0.33$**.

Animated Solution for Mathematics - Probability: If a random variable $x$ has the probability distribution $x P (x) 00 1 2 k 2 k 3 3 k 4 2 k^{2} 5 2 k 6 k^{2} + k 7 7 k^{2}$ then $P (3 < x \leq 6)$ is equal to :

Select Answer:

Visualized Solution

Understanding the Probability Distribution

The Fundamental Probability Axiom

Setting up the Equation

Grouping the $k^{2}$ Terms

Grouping the $k$ Terms

Forming the Quadratic Equation

Factorizing the Quadratic Equation

Validating the Value of $k$

Identifying the Target Interval

Substituting the Expressions

Simplifying the Target Expression

Executing the Final Substitution

Final Arithmetic Computation

Conclusion and Key Takeaway

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Fundamental Axiom

The Quadratic Battle

The Reality Check

The Final Target

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