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: Find k.  
Sum of probabilities $\sum P(x) = 1$. $(k + 2k + 2k + 3k + k) + (k^2 + 2k^2 + 7k^2) = 1$. $9k + 10k^2 = 1 \implies 10k^2 + 9k - 1 = 0$. $(10k - 1)(k + 1) = 0$. Since $P(x) \ge 0$, $k = 1/10$. 
Step 2: Calculate target probability. 
$P(3<x \le 6) = P(4) + P(5) + P(6)$. $= 3k + k^2 + 2k^2 = 3k + 3k^2$. Substitute $k=0.1$: $3(0.1) + 3(0.01) = 0.3 + 0.03 = 0.33$.