Questions: Is the following table a valid discrete probability distribution? x 0 1 2 3 4 P(X=x) 0.111 0.214 0.312 0.163 0.159 No, since the probabilities do not add up to 1. No, since not all the probabilities are between 0 and 1. Yes, since all the probabilities are between 0 and 1 and the probabilities add up to one. Yes, since all the x values are greater than 0.

Is the following table a valid discrete probability distribution? x 0 1 2 3 4 P(X=x) 0.111 0.214 0.312 0.163 0.159 No, since the probabilities do not add up to 1. No, since not all the probabilities are between 0 and 1. Yes, since all the probabilities are between 0 and 1 and the probabilities add up to one. Yes, since all the x values are greater than 0.
Transcript text: Is the following table a valid discrete probability distribution? \begin{tabular}{|c|c|c|c|c|c|} \hline$x$ & 0 & 1 & 2 & 3 & 4 \\ \hline$P(X=x)$ & 0.111 & 0.214 & 0.312 & 0.163 & 0.159 \\ \hline \end{tabular} No, since the probabilities do not add up to 1 . No, since not all the probabilities are between 0 and 1 . Yes, since all the probabilities are between 0 and 1 and the probabilities add up to one. Yes, since all the x values are greater than 0 .
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Solution

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Solution Steps

To determine if the table is a valid discrete probability distribution, we need to check two conditions: (1) all probabilities must be between 0 and 1, and (2) the sum of all probabilities must equal 1.

Step 1: Verify Probabilities are Between 0 and 1

To determine if the given table is a valid discrete probability distribution, we first check if all probabilities \( P(X=x) \) are between 0 and 1. The probabilities given are:

\[ P(X=0) = 0.111, \quad P(X=1) = 0.214, \quad P(X=2) = 0.312, \quad P(X=3) = 0.163, \quad P(X=4) = 0.159 \]

All these values satisfy \( 0 \leq P(X=x) \leq 1 \).

Step 2: Check if the Sum of Probabilities is 1

Next, we calculate the sum of all probabilities:

\[ 0.111 + 0.214 + 0.312 + 0.163 + 0.159 = 0.959 \]

Since the sum is not equal to 1, the condition for a valid probability distribution is not met.

Final Answer

The table is not a valid discrete probability distribution because the sum of the probabilities is not equal to 1. Therefore, the answer is the first option: "No, since the probabilities do not add up to 1."

\[ \boxed{\text{No, since the probabilities do not add up to 1.}} \]

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