Finite Math Examples

$\begin{array}{c} x & P (x) \\ 1 & 0.15 \\ 2 & 0.2 \\ 3 & 0.25 \\ 4 & 0.05 \\ 5 & 0.35 \end{array}$

Step 1

Prove that the given table satisfies the two properties needed for a probability distribution.

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Step 1.1

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 1.2

$0.15$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.15$ is between $0$ and $1$ inclusive

Step 1.3

$0.2$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.2$ is between $0$ and $1$ inclusive

Step 1.4

$0.25$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.25$ is between $0$ and $1$ inclusive

Step 1.5

$0.05$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.05$ is between $0$ and $1$ inclusive

Step 1.6

$0.35$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$0.35$ is between $0$ and $1$ inclusive

Step 1.7

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 1.8

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

$0.15 + 0.2 + 0.25 + 0.05 + 0.35$

Step 1.9

The sum of the probabilities for all the possible $x$ values is $0.15 + 0.2 + 0.25 + 0.05 + 0.35 = 1$ .

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Step 1.9.1

Add $0.15$ and $0.2$ .

$0.35 + 0.25 + 0.05 + 0.35$

Step 1.9.2

Add $0.35$ and $0.25$ .

$0.6 + 0.05 + 0.35$

Step 1.9.3

Add $0.6$ and $0.05$ .

$0.65 + 0.35$

Step 1.9.4

Add $0.65$ and $0.35$ .

$1$

Step 1.10

For each $x$ , the probability of $P (x)$ falls between $0$ and $1$ inclusive. In addition, the sum of the probabilities for all the possible $x$ equals $1$ , which means that the table satisfies the two properties of a probability distribution.

The table satisfies the two properties of a probability distribution:

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

Property 2: $0.15 + 0.2 + 0.25 + 0.05 + 0.35 = 1$

The table satisfies the two properties of a probability distribution:

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

Property 2: $0.15 + 0.2 + 0.25 + 0.05 + 0.35 = 1$

Step 2

The expectation mean of a distribution is the value expected if trials of the distribution could continue indefinitely. This is equal to each value multiplied by its discrete probability.

$Expectation = 1 \cdot 0.15 + 2 \cdot 0.2 + 3 \cdot 0.25 + 4 \cdot 0.05 + 5 \cdot 0.35$

Step 3

Simplify the expression.

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Step 3.1

Simplify each term.

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Step 3.1.1

Multiply $0.15$ by $1$ .

$Expectation = 0.15 + 2 \cdot 0.2 + 3 \cdot 0.25 + 4 \cdot 0.05 + 5 \cdot 0.35$

Step 3.1.2

Multiply $2$ by $0.2$ .

$Expectation = 0.15 + 0.4 + 3 \cdot 0.25 + 4 \cdot 0.05 + 5 \cdot 0.35$

Step 3.1.3

Multiply $3$ by $0.25$ .

$Expectation = 0.15 + 0.4 + 0.75 + 4 \cdot 0.05 + 5 \cdot 0.35$

Step 3.1.4

Multiply $4$ by $0.05$ .

$Expectation = 0.15 + 0.4 + 0.75 + 0.2 + 5 \cdot 0.35$

Step 3.1.5

Multiply $5$ by $0.35$ .

$Expectation = 0.15 + 0.4 + 0.75 + 0.2 + 1.75$

Step 3.2

Simplify by adding numbers.

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Step 3.2.1

Add $0.15$ and $0.4$ .

$Expectation = 0.55 + 0.75 + 0.2 + 1.75$

Step 3.2.2

Add $0.55$ and $0.75$ .

$Expectation = 1.3 + 0.2 + 1.75$

Step 3.2.3

Add $1.3$ and $0.2$ .

$Expectation = 1.5 + 1.75$

Step 3.2.4

Add $1.5$ and $1.75$ .

$Expectation = 3.25$