Probability is a mathematical tool used to study randomness and provide predictions about the likelihood of something happening. There are several basic rules of probability that can be used to help determine the probability of multiple events happening together, separately, or sequentially. This article covers the fundamentals of probability, which are important both when conducting or interpreting the results of clinical trials, and when making clinical decisions for patients Patients Individuals participating in the health care system for the purpose of receiving therapeutic, diagnostic, or preventive procedures. Clinician–Patient Relationship based on the probability of different outcomes.

Last updated: Sep 12, 2022

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%.

- 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
Mean
Mean is the sum of all measurements in a data set divided by the number of measurements in that data set.
Measures of Central Tendency and Dispersion that over a large number of coin tosses, the frequency of getting heads will be close to 50%.

- 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 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

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.

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.

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
Product
A molecule created by the enzymatic reaction.
Basics of Enzymes 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.

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).

- 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.

- 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%

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