## Probability and Random Phenomena

### Probability

Probability** **is a mathematical tool used to study randomness and provide predictions about the likelihood of something happening.

**Abbreviations:**- Probability is abbreviated as P(event).
- Example: P(A) refers to the probability of event A happening.

**Types of probability**include:- Theoretical probability: mathematical models and not observations
- Relative frequency: observations and/or measurements
- Personal or subjective probability: nothing other than personal feelings

**Example of probability:**

If you toss a fair coin, the theoretical probability of heads on any given toss is 50%.

### Law of large numbers (LLN)

- When an event is repeated over and over again, the proportion of times that the event results in a particular outcome will settle around a particular number.
- That particular number is the
**probability**of the outcome. - Refers to long-term outcomes (rather than short-term outcomes)

**Example of short-term vs long-term outcomes:**

If you toss a fair coin, there is a 50% chance of getting heads on any given toss. If you toss 10 heads in a row, this does **not** increase the chance of getting heads on the next toss (a short-term outcome). The LLN does mean that over a large number of coin tosses, the frequency of getting heads will be close to 50%.

### Probability assignment

- If the probability of an event = 0, the event will never occur.
- If the probability of an event = 1, the event will always occur.
- The probability of an event happening can be distributed between the possible outcomes associated with it.
- The probability of all outcomes adds up to 1.

The probability of an event (A) happening is between 0 (absolute certainty of it not occurring) and 1 (absolute certainty of it occurring):

$$ 0\leq P(A)\leq 1 $$Summation of all the probabilities of all possible outcomes in a sample space add up to 1:

$$ P(S) = 1 $$### Random phenomena

Random phenomena are situations where the possible outcomes are known, but the one that will happen is unknown.

- They can be observed during what are known as trials, where outcomes occur.
**Sample space:**the set of all possible outcomes

### Related videos

## Basic Rules of Probability

### The complement rule

Some events only have 2 possible outcomes: event A occurs, or event A does not occur. The complement of event A is that event A does not occur, and is represented as A^{C}. The probability of A^{C} occurring equals 1 minus the probability of the event itself (A).

**Example:** You have a 1-in-4 chance of drawing a club from a standard deck of cards. What is the probability that you will **not** draw a club?

**Answer:** In this example, drawing a club is event A and not drawing a club is A^{C}. If the chance of drawing a club is 0.25, then A^{C} = 1 ‒ 0.25, which is 0.75.

### Rule of disjoint events

If 2 or more events cannot occur simultaneously, they are called mutually exclusive or **disjoint events. **While it is not possible for the 2 disjoint events to occur simultaneously, it **is **possible for **neither **of them to occur.

When 2 events (A and B) are mutually exclusive (or disjoint), the probability that A or B will occur is the sum of the probability of each event.

$$ P(A\cup B) = P(A) + P(B) $$This rule can be applied to any number of disjoint events. For instance, to find the probability of either A, B, or C occurring, you can simply add P(A) + P(B) + P(C), assuming all 3 are completely mutually exclusive events.

**Example 1:**

- You come to a stop light. There is a 35% chance you will pull up to the light while it is green, a 5% chance you will pull up to the light while it is yellow, and a 60% chance you will pull up to the light while it is red. When you pull up to a stop light, what is the chance that the light will be green
**or**yellow? **Answer:**Since the light cannot be green and yellow at the same time in this example, there are only 3 possibilities: P(green) can simply be added to P(yellow), which is 0.35 + 0.05 = 0.4. Thus, the chance that the light will be either green or yellow is 40%.

**Example 2:**

- You have a standard deck of cards. What is the probability of drawing either a club, spade, or heart?
**Answer:**The card you draw will only have 1 of the 4 suits at a time; thus, these are all mutually exclusive events. Therefore, P(club) + P(spade) + P(heart) = 0.25 + 0.25 + 0.25 = 0.75. There is a 75% chance you will draw a club, spade, or heart.

### The multiplication rule for independent events

Events are independent when the probability of one does not affect that of the other (Note: Disjoint events **cannot** be independent events (Example 2)). The probability of 2 independent events **both** occurring equals the product of the probabilities of events A and B.

**Example 1:**

- Event A is the probability of drawing a club from a deck of cards, which is 13 / 52, or 0.25. Event B is the probability of drawing a face card, which is 12 / 52, or 0.23. What is the probability of drawing a club that is also a face card?
**Answer:**These 2 events are independent of each other (the probability of drawing a club has no effect on the probability of drawing a face card); thus, the probabilities can simply be multiplied together. 0.25 x 0.23 = 0.057 or 5.7% (which is 3 / 52, representing the King, Queen, and Jack of clubs).

**Example 2: Disjoint events cannot be independent of each other**

- Disjoint events are 2 events that cannot happen at the same time: e.g, a stop light cannot be both red and green at the same time. If the light is green, it cannot also be red.
- Independent events
**can**happen simultaneously: e.g., a card can be both a club and a face card.

### General addition rule

The probability of an event (A), or another event (B), or both happening is given by the equation:

$$ P(A\cup B) = P(A) + P(B) – P(A\cap B) $$**Example:** You have a stack of money with 4 bill denominations: $1, $5, $10, and $20. Event A represents drawing an odd-numbered bill; event B represents drawing a bill between $4 and $12. What is the probability of A or B happening?

**Answer: **Note that $5 is in both events; therefore, they are not disjoint or mutually exclusive. Thus, we cannot simply add P(A) + P(B), because we will have counted the probability of a $5 bill being drawn (P($5)) twice. We, therefore, must subtract P($5) so that it is only counted once in the end. If the chances of drawing each bill are the same, then the probability of drawing each individual bill is 1 in 4, or 25%.

So, to answer our question, 1st we can calculate P(A), which equals P($1) + P($5) = 0.25 + 0.25 = 0.5. Similarly, P(B) = P($5) + P($10) = 0.25 + 0.25 = 0.5. We know that P($5) by itself is 0.25. So, overall, 0.5 + 0.5 ‒ 0.25 = 0.75, which represents 3 out of the 4 bills (the $1, $5, and $10 bills, which are all included in either event A or B).

### Pitfalls

- Beware of probabilities that do not add up to 1.
- Do not add probabilities of events if they are not disjoint.
- Do not multiply probabilities of events if they are not independent.
- Disjoint events cannot be independent.
- Do not use LLN to describe short-term events.
- Consider whether assuming events to be independent is reasonable.

### Related videos

## Conditional Probability

- Conditional probability of event B is the probability that B will occur knowing that event A has already occurred.
- Notated as P(B|A)
- Conditional probability of
**independent**events is simply the probability of event B, meaning P(B|A) = P(B).**Example:**A person wishes to draw 2 clubs in a row from a standard deck. Assuming their 1st card is a club; what is the probability that the 2nd card will also be a club?**Answer:**Each draw is independent of the last draw, so the conditional probability, in this case, can be described as: P(Drawing a 2nd club|1st card is clubs). As there are 13 cards in each suit, and 1 has already been drawn (“event A”), this leaves 12 clubs out of a total of 51 cards. Therefore, the answer is 12 / 51 or 23.5%.

- Conditional probability of events that are not independent represents the probability of both events occurring, meaning P(B|A) = P(A)*P(B).
**Example:**A student applying to college has an 80% chance of being accepted. On-campus housing is available for 60% of accepted students. What is the chance of acceptance and getting on-campus housing?**Answer:**0.8 x 0.6 = 0.48 or 48%

- Conditional probability of more than 2 events requires consideration of all preceding events.
**Example:**The same student above knows that of students who get on-campus housing, 90% have at least 1 roommate. What is the chance this student gets accepted, gets on-campus housing, and has at least 1 roommate?**Answer:**0.8 x 0.6 x 0.9 = 0.432 or 43.2%

## References

- Haidich, A.B. (2010). Meta-analysis in medical research. Hippokratia, 14 (Suppl 1): pp. 29–37.
- Smith, V., Devane, D., Begley, C.M., Clarke, M. (2011). Methodology in conducting a systematic review of systematic reviews of healthcare interventions. BMC Medical Research Methodology, 11 (1).
- Rind, D. (2019). Proof, p-values, and hypothesis testing. UpToDate. Retrieved May 25, 2021, from https://www.uptodate.com/contents/proof-p-values-and-hypothesis-testing
- Mahutte, N., Duleba, A. (2021). Evaluating diagnostic tests Diagnostic tests Diagnostic tests are important aspects in making a diagnosis. Some of the most important epidemiological values of diagnostic tests include sensitivity and specificity, false positives and false negatives, positive and negative predictive values, likelihood ratios, and pre-test and post-test probabilities. Epidemiological Values of Diagnostic Tests. UpToDate. Retreived May 25, 2021, from https://www.uptodate.com/contents/evaluating-diagnostic-tests