Question

If the probability distribution is given by:

X	0	1	2	3	4	5	6	7
P(x)	0	k	2k	2k	3k	k²	2k²	7k² + k

Then find: \( P(3 < x \leq 6) \)

• A. 0.33
• B. 0.22
• C. 0.11
• D. 0.44

Hint:

The first step in any probability distribution problem with an unknown constant is to use the property that the sum of all probabilities is 1. Also, remember that all individual probabilities must be non-negative.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a probability distribution for a discrete random variable X. First, we need to find the value of the constant k. Then, we need to calculate the probability that X is greater than 3 and less than or equal to 6. 

Step 2: Key Property of Probability Distribution: 
For any probability distribution, the sum of all probabilities must be equal to 1. \[ \sum P(x_i) = 1 \] Applying this to the given distribution: \[ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1 \] \[ 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2+k) = 1 \] 
Step 3: Solving for k: 
Combine the terms with k and k\(^2\): \[ (k + 2k + 2k + 3k + k) + (k^2 + 2k^2 + 7k^2) = 1 \] \[ 9k + 10k^2 = 1 \] Rearrange into a standard quadratic equation: \[ 10k^2 + 9k - 1 = 0 \] Factor the quadratic equation: \[ 10k^2 + 10k - k - 1 = 0 \] \[ 10k(k+1) - 1(k+1) = 0 \] \[ (10k-1)(k+1) = 0 \] This gives two possible values for k: \(k = 1/10\) or \(k = -1\). 
Since probabilities cannot be negative (e.g., P(1) = k must be \(\geq 0\)), we must choose the positive value. \[ k = \frac{1}{10} = 0.1 \] 
Step 4: Calculating the Required Probability: 
We need to find P(3 \(<\) x \(\leq\) 6), which is the sum of probabilities for x=4, x=5, and x=6. \[ P(3<x \leq 6) = P(4) + P(5) + P(6) \] From the table: \[ P(3<x \leq 6) = 3k + k^2 + 2k^2 = 3k + 3k^2 \] Substitute the value of k = 0.1: \[ P(3<x \leq 6) = 3(0.1) + 3(0.1)^2 = 0.3 + 3(0.01) = 0.3 + 0.03 = 0.33 \] 
Step 5: Final Answer: 
The required probability is 0.33.