---
title: "Homework 2 Help: Conditional Probability Example"
author: "Tom Fletcher"
date: "September 7, 2017"
output: html_document
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE, comments = "")
```

When dealing with conditional probability, there are several questions you could answer with the same starting information. The starting point is always to draw the __conditional probability tree__ for the problem. Following that, you may be asked a question that involves __total probability, Bayes' rule, or independence.__ We will go through all of these possibilities in the following example.


# Problem Setup
_You have a system with a main power supply and auxillary power supply. The main power supply has a 10\% chance of failure. If the main power supply is running, the auxillary power supply also has a 10\% chance of failure. But if the main supply fails, the auxillary supply is more likely to be overloaded and has a 15\% chance to fail._

# 1. Draw a Conditional Probability Tree

You can draw a conditional probability tree in your favorite drawing software, or you can format the information into a table. First, be sure to define the notation you will use for events:

- $M$ =   "main power is working"
- $M^c$ = "main power fails"
- $A$ =   "auxillary power is working"
- $A^c$ = "auxillary power fails"

Now we can format the conditional probability tree as a table:

 First Events    | Second Events            | Joint Probabilities       |
-----------------|--------------------------|---------------------------|
 $P(M) = 0.9$    | $P(A \mid M) = 0.9$      | $P(A \cap M) = 0.81$      |
 <br>            | $P(A^c \mid M) = 0.1$    | $P(A^c \cap M) = 0.09$    |
 $P(M^c) = 0.1$  | $P(A \mid M^c) = 0.85$   | $P(A \cap M^c) = 0.085$   |
 <br>            | $P(A^c \mid M^c) = 0.15$ | $P(A^c \cap M^c) = 0.015$ |

Note above that you need to put an explicit line break `<br>` in a cell that you want to be blank.

# 2. Total Probability Question
_What is the probability of the auxillary power failing?_

### Solution
Here we just add all of the joint probabilities involving the event $A^c$.

$$P(A^c) = P(A^c \cap M) + P(A^c \cap M^c) = 0.09 + 0.015 = 0.105$$

# 3. Bayes' Rule Question
_If the auxillary power is working properly, what is the probability that the main power fails?_

### Solution
First, using the Complement Rule, we compute
$$P(A) = 1 - P(A^c) = 1 - 0.105 = 0.895$$

Now we can just plug the information we have into the formula for Bayes' Rule:
$$
\begin{align}
P(M^c \mid A) &= \frac{P(A \mid M^c) P(M^c)}{P(A)} & \text{Bayes' Rule}\\
              &= \frac{0.85 \times 0.1}{0.895}     & \text{Plugging in numbers}\\
              &= `r 0.85 * 0.1 / 0.895`            & \text{Answer!}\\
\end{align}
$$

Notice in the last line we used an inline R command to do the final calculation for us. (Be lazy, don't do your own arithmetic!)

# 4. Independence Question
_Is the auxillary power failing independent of the main power failing?_

There are many equivalent formulas you can check for independence. (See the lecture notes.) We will first try this one:

$$P(A^c | M^c) \stackrel{?}{=} P(A^c).$$

The left-hand side of this is from the tree, and the right-hand side is from the answer to the total probability question above. We have

$$P(A^c | M^c) = 0.15 \neq 0.105 = P(A^c).$$

This means that $A^c$ and $M^c$ are __not__ independent (in other words, they are dependent).