Math Problem Statement

Determine the value of k for the given probability distribution of the discrete random variable X, based on the table provided.

Solution

To solve this problem, we need to determine the value of kk by using the properties of a probability distribution. The given problem provides a discrete random variable XX and its probability distribution P(X)P(X).

Key Property:

The sum of all probabilities in a probability distribution must equal 1: P(X)=1\sum P(X) = 1

Data from the Table:

The probabilities for each value of XX are:

  • P(0)=0P(0) = 0
  • P(1)=kP(1) = k
  • P(2)=2kP(2) = 2k
  • P(3)=2kP(3) = 2k
  • P(4)=3kP(4) = 3k
  • P(5)=k2P(5) = k^2
  • P(6)=2k2P(6) = 2k^2
  • P(7)=7k2+kP(7) = 7k^2 + k

Step 1: Write the Total Probability Equation

P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)=1P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1 Substitute the probabilities: 0+k+2k+2k+3k+k2+2k2+(7k2+k)=10 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1

Step 2: Combine Like Terms

Combine all kk terms and k2k^2 terms: k+2k+2k+3k+k+7k=16kk + 2k + 2k + 3k + k + 7k = 16k k2+2k2+7k2=10k2k^2 + 2k^2 + 7k^2 = 10k^2 Thus, the equation becomes: 16k+10k2=116k + 10k^2 = 1

Step 3: Solve the Quadratic Equation

Rearrange the equation: 10k2+16k1=010k^2 + 16k - 1 = 0

Use the quadratic formula to solve for kk: k=b±b24ac2ak = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Here, a=10a = 10, b=16b = 16, and c=1c = -1. Substituting these values: k=16±1624(10)(1)2(10)k = \frac{-16 \pm \sqrt{16^2 - 4(10)(-1)}}{2(10)} k=16±256+4020k = \frac{-16 \pm \sqrt{256 + 40}}{20} k=16±29620k = \frac{-16 \pm \sqrt{296}}{20} k=16±27420k = \frac{-16 \pm 2\sqrt{74}}{20} k=8±7410k = \frac{-8 \pm \sqrt{74}}{10}

Step 4: Select the Positive Solution

Since kk represents a probability scaling factor, it must be positive. Thus: k=8+7410k = \frac{-8 + \sqrt{74}}{10}

This is the final value of kk.


Do you want me to simplify further, explain this in detail, or confirm with calculations? Let me know!

Here are 5 related questions for practice:

  1. What happens if one of the probabilities is negative?
  2. How can you verify if a given distribution is valid?
  3. How does the quadratic formula work in solving such problems?
  4. What is the significance of each k2k^2 term in the distribution?
  5. How would the solution change if additional constraints were given for kk?

Tip: Always check that all probabilities sum to 1 and are non-negative in any discrete probability distribution.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Discrete Random Variables
Quadratic Equations

Formulas

Sum of probabilities in a distribution: ∑P(X) = 1
Quadratic equation: ax^2 + bx + c = 0
Quadratic formula: k = (-b ± √(b^2 - 4ac)) / 2a

Theorems

The sum of all probabilities in a valid probability distribution equals 1.
Quadratic equation solutions using the quadratic formula.

Suitable Grade Level

Grades 10-12