Joint, Marginal, Conditional Probability

class: center, middle, inverse, title-slide

# Joint, Marginal, Conditional Probability
### Dr. Dogucu

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 Copyright &copy; <a href="https://mdogucu.ics.uci.edu">Dr. Mine Dogucu</a>. <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/">CC BY-NC-SA 4.0</a></div>

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## Data - GSS 2018

The General Social Survey (GSS) is a sociological survey that has been regularly conducted since 1972. It is a comprehensive survey that provides information on experiences of residents of the United States.

<table align = "center">
<thead>
  <tr>
    <th colspan="2" rowspan="2"></th>
    <th colspan="3">Belief in Life After Death</th>
  </tr>
  <tr>
    <td>Yes</td>
    <td>No<br></td>
    <td>Total<br></td>

</tr>
</thead>
<tbody>
  <tr>
    <td rowspan="3">College Science Course<br></td>
    <td>Yes</td>
    <td>375</td>
    <td>75</td>
    <td>450</td>

</tr>
  <tr>
    <td>No</td>
    <td>485</td>
    <td>115</td>
    <td>600</td>

</tr>
  
  <tr>
    <td>Total</td>
    <td>860</td>
    <td>190</td>
    <td>1050</td>

</tr>
</tbody>
</table>

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## Events

Let B represent an event that a randomly selected person in this sample believes in life after death.

Let C represent an event that a randomly selected person in this sample took a college level science course.

---

## Joint Probability

</tr>
</thead>
<tbody>
  <tr>
    <td rowspan="3">College Science Course<br></td>
    <td>Yes</td>
    <td>375</td>
    <td>75</td>
    <td>450</td>

</tr>
  <tr>
    <td>No</td>
    <td>485</td>
    <td>115</td>
    <td>600</td>

</tr>
  
  <tr>
    <td>Total</td>
    <td>860</td>
    <td>190</td>
    <td>1050</td>

</tr>
</tbody>
</table>

Note that events `\(B\)` and `\(C\)` are not mutually exclusive. A randomly selected person can believe in life after death and might have taken a college science course. `\(B \cap C \neq \emptyset\)`.

`\(P(B \cap C) = \frac{375}{1050}\)`

Note that `\(P(B\cap C) = P(C\cap B)\)`. Order does _not_ matter.

---
## Marginal Probability

</tr>
</thead>
<tbody>
  <tr>
    <td rowspan="3">College Science Course<br></td>
    <td>Yes</td>
    <td>375</td>
    <td>75</td>
    <td>450</td>

</tr>
  <tr>
    <td>No</td>
    <td>485</td>
    <td>115</td>
    <td>600</td>

</tr>
  
  <tr>
    <td>Total</td>
    <td>860</td>
    <td>190</td>
    <td>1050</td>

</tr>
</tbody>
</table>

`\(P(B)\)` represents a __marginal probability__. So do `\(P(C)\)`, `\(P(B^C)\)` and `\(P(C^C)\)`. In order to calculate these probabilities we could only use the values in the margins of the contingency table, hence the name.

`\(P(B) = \frac{860}{1050}\)`  
`\(P(C) = \frac{450}{1050}\)`

---

## Conditional Probability

</tr>
</thead>
<tbody>
  <tr>
    <td rowspan="3">College Science Course<br></td>
    <td>Yes</td>
    <td>375</td>
    <td>75</td>
    <td>450</td>

</tr>
  <tr>
    <td>No</td>
    <td>485</td>
    <td>115</td>
    <td>600</td>

</tr>
  
  <tr>
    <td>Total</td>
    <td>860</td>
    <td>190</td>
    <td>1050</td>

</tr>
</tbody>
</table>

`\(P(B|C)\)` represents a __conditional probability__. So do `\(P(B^c|C)\)`, `\(P(C| B)\)` and `\(P(C|B^c)\)`. To calculate the probabilities we focus on the row or the column of the given information. We _reduce_ the sample space to this given information.

Probability that a randomly selected person believes in life after death given that they have taken a college science course  
`\(P(B|C) = \frac{375}{450}\)`

---

## Conditional Probability

The order matters!

`\(P(\text{has a dog | like dogs}) \neq\)` `\(P(\text{like dogs | has a dog})\)`

---

## Addition Rule

</tr>
</thead>
<tbody>
  <tr>
    <td rowspan="3">College Science Course<br></td>
    <td>Yes</td>
    <td>375</td>
    <td>75</td>
    <td>450</td>

</tr>
  <tr>
    <td>No</td>
    <td>485</td>
    <td>115</td>
    <td>600</td>

</tr>
  
  <tr>
    <td>Total</td>
    <td>860</td>
    <td>190</td>
    <td>1050</td>

</tr>
</tbody>
</table>

`\(P(B\cup C) = P(B) + P(C) - P(B \cap C)\)`

`\(P(B\cup C) = \frac{860}{1050} + \frac{450}{1050} -\frac{375}{1050} = \frac{935}{1050}\)`

`\(P(B\cup C) = \frac{375}{1050} + \frac{75}{1050} + \frac{485}{1050} = \frac{935}{1050}\)`