Question

which of the following represents a valid probability distribution?
probability distribution d:

x	p(x)
2	0.56
3	0
4	1.38

probability distribution a:

x	p(x)
2	0.6
3	0.25
4	0.29

probability distribution b:

x	p(x)
2	0.45
3	0.16
4	0.39

probability distribution c:

x	p(x)
2	1.23
3	-0.87
4	0.19

Explanation:

To determine a valid probability distribution, we use two rules:

1. Each probability \( P(x) \) must satisfy \( 0 \leq P(x) \leq 1 \).
2. The sum of all probabilities must equal \( 1 \).

Step 1: Check Probability Distribution A

• For \( x = 1 \), \( P(1) = -0.14 \), which is less than \( 0 \). So, it violates the first rule.

Step 2: Check Probability Distribution B

• For \( x = 1 \), \( P(1) = 0 \); \( x = 2 \), \( P(2) = 0.45 \); \( x = 3 \), \( P(3) = 0.16 \); \( x = 4 \), \( P(4) = 0.39 \).
• Check individual probabilities: All \( P(x) \) values are between \( 0 \) and \( 1 \).
• Sum the probabilities: \( 0 + 0.45 + 0.16 + 0.39 = 1 \). It satisfies both rules.

Step 3: Check Probability Distribution C

• For \( x = 2 \), \( P(2) = 1.23 \) (greater than \( 1 \)) and \( x = 3 \), \( P(3) = -0.87 \) (less than \( 0 \)). Violates the first rule.

Step 4: Check Probability Distribution D

• Sum the probabilities: \( 0.87 + 0.56 + 0 + 1.38 = 2.81 \), which is not equal to \( 1 \). Also, \( P(4) = 1.38 > 1 \). Violates both rules.

Answer:

Probability Distribution B (where \( X \) has \( P(x) \) values \( 0 \) (for \( x = 1 \)), \( 0.45 \) (for \( x = 2 \)), \( 0.16 \) (for \( x = 3 \)), and \( 0.39 \) (for \( x = 4 \))) is the valid probability distribution.