If a random variable $x$ has the probability distribution
| $x$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| $P(x)$ | 0 | $2k$ | $k$ | $3k$ | $2k^2$ | $2k$ | $k^2 + k$ |
then $P(3 < x \leq 6)$ is equal to
✓
Solution Steps
1
Set up the constraint equation using probability axiom
Since the sum of all probabilities must equal 1:
$P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1$
$0 + 2k + k + 3k + 2k^2 + 2k + k^2 + k = 1$
2
Combine like terms
Group constant terms and quadratic terms:
$(2k + k + 3k + 2k + k) + (2k^2 + k^2) = 1$
$9k + 3k^2 = 1$
$3k^2 + 9k - 1 = 0$
3
Apply the quadratic formula
For equation $3k^2 + 9k - 1 = 0$, use $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$k = \frac{-9 \pm \sqrt{81 + 12}}{6} = \frac{-9 \pm \sqrt{93}}{6}$
$k = \frac{-9 \pm 9.644}{6}$
$k_1 = \frac{0.644}{6} \approx 0.107$ or $k_2 = \frac{-18.644}{6} \approx -3.107$
4
Choose valid value of k
Since all probabilities must be non-negative:
$k = -3.107$ makes $P(x)$ negative (invalid)
$k \approx 0.1$ makes all $P(x) \geq 0$ (valid)
Use $k = 0.1$ for exact calculation
5
Calculate the required probability
$P(3 < x \leq 6)$ includes $x = 4, 5, 6$ (exclude 3, include 6)
$P(3 < x \leq 6) = P(4) + P(5) + P(6)$
$= 2k^2 + 2k + k^2 + k$
$= 3k^2 + 3k$
$= 3(0.1)^2 + 3(0.1)$
6
Compute the final answer
$P(3 < x \leq 6) = 3(0.01) + 3(0.1)$
$= 0.03 + 0.30$
$= 0.33$
7
Verify the solution
Check with $k = 0.1$:
Sum: $0 + 0.2 + 0.1 + 0.3 + 0.02 + 0.2 + 0.11 = 0.93 \approx 1$ ✓
All probabilities positive ✓
$P(3 < x \leq 6) = 0.33$ ✓
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Key Insight
The fundamental property: all probabilities must sum to 1. This constraint uniquely determines k, allowing calculation of any required probability.
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Probability Theory
1. Axioms of Probability
Probability is a mathematical framework based on three fundamental axioms: (1) Non-negativity: $P(A) \geq 0$ for any event A. No probability can be negative. (2) Certainty: $P(\text{sample space}) = 1$. The total probability of all possible outcomes equals 1. (3) Additivity: For mutually exclusive events, $P(A \cup B) = P(A) + P(B)$. These axioms form the foundation of all probability calculations. In discrete probability distributions, these principles ensure the sum of all individual probabilities equals unity, providing a constraint to solve for unknown parameters like k.
2. Discrete Probability Distributions
A discrete probability distribution assigns non-negative probabilities to each possible value of a random variable. For a random variable x taking values $x_1, x_2, \ldots, x_n$, we have $P(x_i) \geq 0$ for all i and $\sum_{i=1}^{n} P(x_i) = 1$. This sum condition is crucial because it normalizes the distribution, ensuring probabilities are valid. Each value of x has an associated probability that represents the likelihood of that outcome. The distribution completely characterizes the random variable's behavior. In problems with unknown parameters like k, we use this summation constraint to set up equations and determine the parameters.
3. Calculating Compound Probabilities
When finding $P(a < x \leq b)$ for a discrete random variable, identify all values of x that satisfy the inequality condition. Include endpoints carefully: the notation $a < x$ excludes a while $x \leq b$ includes b. Sum the individual probabilities for all qualifying values. For example, $P(3 < x \leq 6)$ includes $x = 4, 5, 6$ but excludes $x = 3$. Always verify the interval notation: strict inequalities (<, >) exclude endpoints while non-strict inequalities (≤, ≥) include them. This precision is essential for obtaining correct answers.
4. Solving for Parameters in Distributions
When a probability distribution contains unknown parameters (like k), use the constraint $\sum P(x) = 1$ to establish equations. Combine like terms carefully, grouping constants separately from variables and higher powers. This typically yields polynomial equations—linear equations have one solution, quadratic equations may have two solutions requiring selection based on validity constraints. Check that all probabilities remain non-negative: this eliminates extraneous solutions. Once the parameter is determined, substitute back into probability expressions to calculate specific values. Always verify your solution by confirming the sum equals 1 and all individual probabilities are valid.
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Frequently Asked Questions
Q
What is a probability distribution?
⌄
A probability distribution assigns probabilities to each value of a random variable. For discrete distributions, P(x) ≥ 0 for all x and the sum of all probabilities equals 1.
Q
How do I find k using the constraint?
⌄
Sum all probabilities: 0 + 2k + k + 3k + 2k² + 2k + k² + k = 1. This gives 3k² + 9k - 1 = 0. Solve using quadratic formula to get k ≈ 0.1.
Q
What does P(3 < x ≤ 6) mean?
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It means probability that x is strictly greater than 3 AND less than or equal to 6. Include x = 4, 5, 6 but exclude x = 3.
Q
Why must probabilities be non-negative?
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Probability represents likelihood, which cannot be negative. It's one of the fundamental axioms of probability theory. Any negative result indicates an invalid solution.
Q
How to apply the quadratic formula?
⌄
For ax² + bx + c = 0, use k = [-b ± √(b² - 4ac)] / (2a). For 3k² + 9k - 1 = 0: a=3, b=9, c=-1, giving k = [-9 ± √93] / 6.
Q
How to verify if k is valid?
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Check that all P(x) values are non-negative. For k = 0.1, calculate each probability and confirm sum ≈ 1. If any probability is negative, reject that k value.
Q
Why sum probabilities from 4 to 6 only?
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The condition 3 < x ≤ 6 specifies a range. Strict inequality excludes 3, non-strict includes 6. So x can be 4, 5, or 6 only—add their probabilities.
Q
Can a probability equal 0?
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Yes. P(0) = 0 in this problem means event x=0 is impossible. Probabilities range from 0 (impossible) to 1 (certain), inclusive.
Q
What if quadratic has no real solutions?
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If discriminant (b² - 4ac) is negative, no real k exists. This means the probability distribution is impossible—constraints cannot be satisfied.
Q
How to express answer as decimal?
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P(3 < x ≤ 6) = 3k² + 3k = 3(0.1)² + 3(0.1) = 0.03 + 0.30 = 0.33. Convert to decimal form for final answer.
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