Question

identifying a valid probability distribution
which of the following represents a valid probability distribution?
probability
distribution b
x & p(x)\\
0 & 0.2\\
1 & 1.2\\
2 & -0.8\\
3 & -0.6\\\\
probability
distribution a
x & p(x)\\
0 & 0\\
1 & 0\\
2 & 0\\
3 & 1\\\\
probability
distribution c
x & p(x)\\
0 & 0.23\\
1 & 0.46\\
2 & 1.1\\
3 & 0.21\\\\
probability
distribution d
x & p(x)\\
0 & 0.23\\
1 & 0.4\\
2 & 0.15\\
3 & -0.2

Explanation:

To determine a valid probability distribution, we check two conditions:

1. Each probability \( p(x) \) must be between 0 and 1 (inclusive).
2. The sum of all probabilities must equal 1.

Step 1: Analyze Probability Distribution B

• \( p(0) = 0.2 \), \( p(1) = 1.2 \) (greater than 1), \( p(2) = -0.8 \) (negative), \( p(3) = -0.6 \) (negative).

Fails condition 1.

Step 2: Analyze Probability Distribution A

• \( p(0) = 0 \), \( p(1) = 0 \), \( p(2) = 0 \), \( p(3) = 1 \).

Sum: \( 0 + 0 + 0 + 1 = 1 \). All probabilities are between 0 and 1.
Passes both conditions? Wait, but let's check others too.

Step 3: Analyze Probability Distribution C

• \( p(2) = 1.1 \) (greater than 1).

Fails condition 1.

Step 4: Analyze Probability Distribution D

• \( p(3) = -0.2 \) (negative).

Fails condition 1.

Wait, recheck Distribution A: All \( p(x) \) are 0 or 1 (within [0,1]), and sum is 1. But wait, another check: for a discrete probability distribution, the sum of \( p(x) \) over all \( x \) must be 1, and each \( p(x) \in [0,1] \). Distribution A: \( X \) takes 0,1,2,3. \( p(0)=0 \), \( p(1)=0 \), \( p(2)=0 \), \( p(3)=1 \). Sum is 1, and all \( p(x) \) are in [0,1]. But wait, is there a mistake? Wait, maybe I missed. Wait, let's re-express:

Wait, maybe I made a mistake. Wait, let's check again:

Distribution B: negative probabilities and \( p(1)=1.2>1 \) → invalid.

Distribution A: \( p(x) \) values: 0,0,0,1. Sum is 1. Each \( p(x) \) is between 0 and 1. So valid?

Wait, but let's check Distribution A again. The table is:

\( X \): 0,1,2,3; \( p(x) \): 0,0,0,1.

Yes, sum is 1, and each \( p(x) \in [0,1] \). So it's valid.

Wait, but maybe the intended answer is Distribution A? Wait, no, wait: Wait, maybe I messed up. Wait, let's check Distribution A again. Wait, the problem is to identify which is valid. Let's confirm the two conditions:

1. For all \( x \), \( 0 \leq p(x) \leq 1 \).

1. \( \sum p(x) = 1 \).

Distribution A:

• \( p(0)=0 \) (0 ≤ 0 ≤ 1), \( p(1)=0 \) (0 ≤ 0 ≤ 1), \( p(2)=0 \) (0 ≤ 0 ≤ 1), \( p(3)=1 \) (0 ≤ 1 ≤ 1).
• Sum: \( 0 + 0 + 0 + 1 = 1 \).

So it satisfies both conditions.

Distribution B: \( p(1)=1.2>1 \), \( p(2)=-0.8<0 \), \( p(3)=-0.6<0 \) → invalid.

Distribution C: \( p(2)=1.1>1 \) → invalid.

Distribution D: \( p(3)=-0.2<0 \) → invalid.

So the valid one is Probability Distribution A.

Answer:

Probability Distribution A (with \( X \) values 0,1,2,3 and \( p(x) \) values 0,0,0,1) is the valid probability distribution.