Week 10 - Probability

# Week 10 - Probability
## Brief Review, Bayes’ Theorem
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### Danilo Freire
### 3 April 2019

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# Today's Agenda

.font150[
* Brief review:
  - Permutations
  - Combinations
  - Conditional probabilities
  
* Bayes' theorem/rule
]
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# Review: Permutation

* Example: in how many ways can we arrange 3 people?
]
--
.font140[
* We can count:
* ABC / ACB / BAC / BCA / CAB / CBA -- _6 ways_
]
--
.font140[
* Or we can think like this: 
  - How many people can we pick for the first seat? 3
  - How many people we can pick for the second seat? 2
  - And for the last one? 1
  
* Therefore, we have `$3 \times 2 \times 1 = 3! = 6$` 
]
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# Review: Permutation

.font150[
* In how many ways can 5 people sit in 3 chairs?
]
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.font150[
* 5 in the first chair, 4 in the second, 3 in the third... _and that's all_: `$5 \times 4 \times 3$`
 `$$\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{5!}{2!}$$`
 
* Formula = `$\frac{n!}{(n - k)!}$`, where `$n$` is the number of people and `$k$` is the number of places/seats 
]
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# Review: Combinations

* So ABC, ACB, BAC, BCA, CAB, CBA are 6 permutations but _only one combination_

* To calculate how many combinations one can get, we can .orange[take the number of permutations then divide by the possible combinations]

* We do that to _because we're counting the same group in different ways_
]
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# Example: Combinations

.font150[
* We have 6 people and 3 chairs. How many permutations are there? And how many combinations?
]
--
.font150[
* Permutations: `$6 \times 5 \times 4 = 120$`
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.font150[
* Combinations: We know can arrange the same 3 people in 6 different ways
* ABC / ACB / BAC / BCA / CAB / CBA
* So we can just divide the total of permutations to the ways we can pick 3 people to discount the excess counting: `$\frac{120}{6} = 20$` combinations
]
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# Review: Combinations

* We take `$\frac{n!}{(n-k)!}$`, the permutations, then divide by `$k!$`, the number of repeated permutations (the formula multiplies the fraction by `$k!$` to cancel the denominator)

* In `R`:

```r
choose(6, 3)
```

```
## [1] 20
```
]
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# Review: Conditional Probability

* Many real world events are conditional to one another

`$$P(A|B) = \frac{P(A \operatorname{and} B)}{P(B)}$$`

* `$P(A \operatorname{and} B)$` probability of `$A$` and `$B$` happening together

* `$P(B)$` marginal probability of `$B$`
]
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# Example: Conditional Probability

.font120[
* Researchers surveyed 100 students on which superpower they would most like to have. This two-way table displays data for the respondents:

| Superpower   | Male | Female | **Total** |
| :----------- | :--- | :----- | :-------- |
| Fly          | 26   | 12     | 38        |
| Invisibility | 12   | 32     | 44        |
| Other        | 10   | 8      | 18        |
| **Total**    | 48   | 52     | 100       |

* Find the probability that the student was male, given the student chose to fly as their superpower]
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.font120[
* `$P(Male|Fly) = \frac{P(MaleAndFly)}{P(Fly)} = \frac{26}{38} \approx 0.68$`
]
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# A Slightly Harder Example

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* John's two favourite foods are bagels and pizza. `$A$` represents the event he eats bagel for breakfast and `$B$` represents the event that he eats pizza for lunch.

* On a random day, the probability John will eat a bagel, `$P(A)$`, is `$0.6$`, the probability he will eat pizza is `$P(B) = 0.5$`, and the conditional probability that he eats a bagel for breakfast, given that he eats pizza for lunch is `$P(A|B) = 0.7$`

* Based on this information, what is `$P(B|A)$`, that is, the probability that John will eats pizza for lunch given that he eats a bagel for breakfast?
]
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# Step-by-step

.font150[
* P(A) = 0.6; P(B) = 0.5; P(A|B) = 0.7
* P(B|A)?
]
--
.font110[$$P(A|B) = \frac{P(A \operatorname{and} B)}{P(B)}$$]
--
.font110[$$P(B|A) = \frac{P(A \operatorname{and} B)}{P(A)}$$]
--
.font110[$$P(A|B) \times P(B) = P(A \operatorname{and} B)$$]
--
.font110[$$P(B|A) \times P(A) = P(A \operatorname{and} B)$$]
--
.font110[$$P(A|B) \times P(B) = P(A \operatorname{and} B) = P(B|A) \times P(A)$$]
--
.font110[$$0.7 \times 0.5 = 0.35 = P(A \operatorname{and} B) = P(B|A) \times 0.6$$]
--
.font110[$$\frac{0.35}{0.6} = P(B|A) \approx 0.58$$]
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# Let's See that Again

.font150[
`$$P(A|B) = \frac{P(A \operatorname{and} B)}{P(B)}$$`
`$$P(B|A) = \frac{P(A \operatorname{and} B)}{P(A)}$$`
`$$P(A|B) \times P(B) = P(A \operatorname{and} B) = P(B|A) \times P(A)$$`
`$$P(A|B) = \frac{P(B|A) \times P(A)}{P(B)}$$`
]
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# Bayes' Theorem

# Bayes' Theorem

* `$P(A | B)$`: posterior probability of event `$A$` given observed data `$B$`

* `$P(B | A)$`: probability of observing `$B$` given `$A$`

* `$P(B)$`: probability of observing `$B$`.... including both true and false positives!
]
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# False Positive

# Bayes' Theorem

# Medical Test

.font150[
* Imagine you have a serious disease. It is a rare disease, it happens only to 0.1% of the population. The test identifies the disease correctly in 99% of the cases, but incorrectly in 1% of them. If your test is positive, what is the probability you actually have the disease?
]
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# Medical Test

.font150[
`$$P(H|E) = \frac{P(E|H) \times P(H)}{P(E)}$$`
`$$P(H|E) = \frac{P(E|H) \times P(H)}{P(E|H) \times P(H) + P(E|H^c) \times P(H^c)}$$`
`$$P(disease|test+) = \frac{P(test+|disease) \times Prior(disease)}{P(test+) = \operatorname{true and false positives}}$$`
]
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# Medical Test

`$$P(H|E) = \frac{.99 \times .001}{.001 \times .99 + .999 \times .01} \approx .09 \approx 9\%$$`
]
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# Medical Test

* But 10 people will be tested positive _and will not have the disease_

* Which means that 11 people will be tested positive overall

* So we have `$\frac{1}{11} \approx .09 \approx 9\%$`
]
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# Will the Sun Rise Tomorrow?

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# Bayesian Updating

# xkcd

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# Questions?

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# See you next week!