Finite Math Examples

$\begin{array}{c} x & P (x) \\ 1 & \frac{1}{36} \\ 2 & \frac{2}{36} \\ 3 & \frac{3}{36} \\ 4 & \frac{4}{36} \\ 5 & \frac{5}{36} \end{array}$

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

$\frac{1}{36}$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$\frac{1}{36}$ is between $0$ and $1$ inclusive

Step 1.3

$\frac{2}{36}$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$\frac{2}{36}$ is between $0$ and $1$ inclusive

Step 1.4

$\frac{3}{36}$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$\frac{3}{36}$ is between $0$ and $1$ inclusive

Step 1.5

$\frac{4}{36}$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$\frac{4}{36}$ is between $0$ and $1$ inclusive

Step 1.6

$\frac{5}{36}$ is between $0$ and $1$ inclusive, which meets the first property of the probability distribution.

$\frac{5}{36}$ 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.

$\frac{1}{36} + \frac{2}{36} + \frac{3}{36} + \frac{4}{36} + \frac{5}{36}$

Step 1.9

The sum of the probabilities for all the possible $x$ values is $\frac{1}{36} + \frac{2}{36} + \frac{3}{36} + \frac{4}{36} + \frac{5}{36} = \frac{5}{12}$ .

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Step 1.9.1

Combine the numerators over the common denominator.

$\frac{1 + 2 + 3 + 4 + 5}{36}$

Step 1.9.2

Simplify by adding numbers.

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Step 1.9.2.1

Add $1$ and $2$ .

$\frac{3 + 3 + 4 + 5}{36}$

Step 1.9.2.2

Add $3$ and $3$ .

$\frac{6 + 4 + 5}{36}$

Step 1.9.2.3

Add $6$ and $4$ .

$\frac{10 + 5}{36}$

Step 1.9.2.4

Add $10$ and $5$ .

$\frac{15}{36}$

Step 1.9.3

Cancel the common factor of $15$ and $36$ .

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Step 1.9.3.1

Factor $3$ out of $15$ .

$\frac{3 (5)}{36}$

Step 1.9.3.2

Cancel the common factors.

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Step 1.9.3.2.1

Factor $3$ out of $36$ .

$\frac{3 \cdot 5}{3 \cdot 12}$

Step 1.9.3.2.2

Cancel the common factor.

$\frac{3 \cdot 5}{3 \cdot 12}$

Step 1.9.3.2.3

Rewrite the expression.

$\frac{5}{12}$

Step 1.10

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.

$\frac{1}{36} + \frac{2}{36} + \frac{3}{36} + \frac{4}{36} + \frac{5}{36} = \frac{5}{12} \neq 1$

Step 1.11

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

Step 2

The table does not satisfy the two properties of a probability distribution, which means that the expectation mean can't be found using the given table.

Can't find the expectation mean