Finite Math Examples

$\begin{array}{c} x & P (x) \\ 0 & 0.643 \\ 1 & 0.224 \\ 2 & 0.088 \\ 3 & 0.023 \\ 4 & 0.014 \\ 5 & 0.009 \end{array}$

Step 1

A discrete random variable $x$ takes a set of separate values (such as $0$ , $1$ , $2$ ...). Its probability distribution assigns a probability $P (x)$ to each possible value $x$ . For each $x$ , the probability $P (x)$ falls between $0$ and $1$ inclusive and the sum of the probabilities for all the possible $x$ values equals to $1$ .

1. For each $x$ , $0 \leq P (x) \leq 1$ .

2. $P (x_{0}) + P (x_{1}) + P (x_{2}) + \dots + P (x_{n}) = 1$ .

Step 2

$0.643$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.643$ is between $0$ and $1$ inclusive

Step 3

$0.224$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.224$ is between $0$ and $1$ inclusive

Step 4

$0.088$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.088$ is between $0$ and $1$ inclusive

Step 5

$0.023$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.023$ is between $0$ and $1$ inclusive

Step 6

$0.014$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.014$ is between $0$ and $1$ inclusive

Step 7

$0.009$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.009$ is between $0$ and $1$ inclusive

Step 8

For each $x$ , the probability $P (x)$ falls between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0 \leq P (x) \leq 1$ for all x values

Step 9

Find the sum of the probabilities for all the possible $x$ values.

$0.643 + 0.224 + 0.088 + 0.023 + 0.014 + 0.009$

Step 10

The sum of the probabilities for all the possible $x$ values is $0.643 + 0.224 + 0.088 + 0.023 + 0.014 + 0.009 = 1.001$ .

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Step 10.1

Add $0.643$ and $0.224$ .

$0.867 + 0.088 + 0.023 + 0.014 + 0.009$

Step 10.2

Add $0.867$ and $0.088$ .

$0.955 + 0.023 + 0.014 + 0.009$

Step 10.3

Add $0.955$ and $0.023$ .

$0.978 + 0.014 + 0.009$

Step 10.4

Add $0.978$ and $0.014$ .

$0.992 + 0.009$

Step 10.5

Add $0.992$ and $0.009$ .

$1.001$

Step 11

The sum of the probabilities for all the possible $x$ values is not equal to $1$ , which does not meet the second property of the probability distribution.

$0.643 + 0.224 + 0.088 + 0.023 + 0.014 + 0.009 = 1.001 \neq 1$

Step 12

For each $x$ , the probability $P (x)$ falls between $0$ and $1$ inclusive. However, the sum of the probabilities for all the possible $x$ values is not equal to $1$ , which means that the table does not satisfy the two properties of a probability distribution.

The table does not satisfy the two properties of a probability distribution