Probability Introduction: Articles and Videos with Solutions!

Dice rolling, finding a parking space, winning at cards; These are just some situations where you might want to find the odds of an event happening. If you want to know how to find probability, you first have to figure out what kind of question you have. For example, how you find the odds of an event happening is different from finding the odds of group members choosing the same thing.

Some probabilities are easy to identify, like finding dice rolling probabilities or picking from a deck of cards.

Questions that involve the binomial theorem are also easy to identify. In these types of experiment, the only possible outcomes of an event are “Success” or “Failure”: like yes/no, heads/tails or black/white.

Other question types you might come across involve people, like:

Events

Do you want to find the probability of a simple event happening? Like it raining, or finding a parking space downtown? See:
Probability of a simple event happening.
On the other hand, if you want the exact opposite, see:
How to find out the probability of an event NOT happening.

Is your question about one event happening given another event? Like the odds of finding a parking space, given that it’s game day, or finding a certain popular toy on Black Friday? If so, check out:

A frequency distribution table.

Frequency Distributions

Do you have a frequency distribution to work with? Or are you able to make a frequency distribution table with your given data? For example, you have x number of items with a certain traits. Probability frequency distribution.

Introduction to Probability: How-to Articles and Videos.

Introduction to Probability: Definitions

Calculators.

Online permutations calculator and combinations calculator.

Probability Formulas / Probability Rules

One of the most uncomfortable things that students don’t like about formulas is the lack of them in probability and statistics. There are a few staples, including some must-know notation:

Probability Range
0 ≤ P(A) ≤ 1
This states that the probability of an event is somewhere between zero and 100% (as a decimal, that’s 0 and 1). You’ll want to remember this rule when adding or multiplying probabilities of events. If your answer is over 100%, that’s a clue you might have done something wrong.

Rule of Complementary Events
P(A^C) + P(A) = 1
Complementary events happen when there are only two outcomes, like flipping a coin. Rolling a die to see if you would get six is also complementary; the only two outcomes are getting a six (a 1/6 chance) or not getting a six (5/6 chance). The two probabilities must add up to 1.
You may also see this formula written like this:
p(A) + p(A’) = 1
which can be rearranged algebraically to read:
p(A’) = 1 – p(A).
All three formulas are equivalent: which terminology (A’ or A^c) is used is up to the textbook author and teacher. Personally, I prefer A’, which I call “not A.” The probability of “not A”, I think, is easier to comprehend than “the complement” (either an event happens, or it does not happen).

Addition Rule
P(A∪B) = P(A)+P(B)-P(A∩B)
where ∪ is the “union” and ∩ is the intersection.
What this is saying (in English!) is the probability of event A OR event B happening (or both at the same time) is:

The probability of event A happening on its own,
Plus the probability of event B happening on its own,
Plus the probability of both events happening at the same time.

If you have mutually exclusive events, then P(A∩B) can’t happen (the events can’t happen together), so the formula becomes:
P(A∪B) = P(A)+P(B)- 0 = P(A)+P(B)

Disjoint Events
Events A and B are disjoint if:
P(A∩B) = 0
This is just another way of saying the events are mutually exclusive. They cannot happen at the same time.

A Related formula is P((A∪B)^c) or, equivalently ((A∪B)^‘). In English, this says “not the union.” To solve this, figure out the union and take the result away from 1 (because the probability of the events happening or not happening must add up to 1).

Conditional Probability
P(A|B) = P(A∩B) / P(B)
The “|” symbol means “given that”. In other words, the probability of event B happening, given that event A happens. For examples of how to use the formula, see: conditional probability.

Bayes Formula
P(A|B) = P(B|A) · P(A) / P(B)
Bayes’ theorem is a way to figure out conditional probability, although it is slightly more nuanced. In a nutshell, it gives you the actual probability of an event given information about tests. For example, what is the probability you have cancer if your medical test is positive (answer = a lot lower than you think). For some examples, see: Bayes Theorem Examples and Posterior Distributions / Posterior Probabilities.

Independent Events
Events A and B are independent if one of them doesn’t affect the probability of the other. Events are also said to be independent if the following equation holds true.
P(A∩B) = P(A) · P(B).
This equation is derived from the multiplication rule, which says that P(A∩B) = P(A) * P(B|A). As we know that P(B|A) = P(B) for independent events, we can substitute P(B|A) for P(B), giving us the formula.

Some Notes on Probability Rules

Statistics is about chance and ballparking, not absolutes and the “right answer.” Just look up any Gallup poll — they are rarely more than 90% confident that they have the “right answer.” There are techniques you can use to figure out odds (like multiplying two probabilities together, or adding them). However, other than perhaps the binomial distribution table and the formulas above (which in real life, aren’t used that much), there aren’t really any probability formulas you can easily apply. You have to revert back to that old elementary school staple, logic (you remember it…it was there right before they started standardized testing the logic out of you).

The three Rules of probability formulas:

There are no rules (well, very few, except those listed above).
Use logic, not equations.
There are many, many different ways to get to the answer — none of which really use formulas.

Here’s a question that arrived in my inbox this morning that tackles probabilities:

“If you were trying to collect 6 baseball cards that came in packets of cheese puffs, assuming they are distributed evenly how many packets of cheese puffs would you expect to buy before you have all 6 cards?”

Step 1 to solving this problem is realizing you can’t look up the answer in a table. In order to solve it, you have to think like a kid.

Card# 1:You’re 8-years-old again, walking into a store with enough money to buy one bag of cheese puffs. You hope to collect all 5 of the baseball cards but you don’t have any yet. What are the odds you buy a bag and get a card you want?

The answer, of course, is 100%. Buy your first bag, and you have 100% chance that there’s a card in it that you want.

Card# 2: Now it gets a little trickier. You return to the store to get card#2. But as you already got card #1, Mickey Mantle. You don’t want him again, but there’s a 1/5 chance you might get him (and therefore a 4/5 chance you won’t). How many bags of cheese puffs will you have to buy to get card #2 becomes a ratio problem. You could figure it out in your head, but if you want to model it mathematically you have to set up the equation. If one bag of cheese puffs gives you a 80% chance of getting the card you want, how many bags do you need to buy to get a 100% chance?

To get 100% you would need to purchase 1.25 bags.

Card# 3:The odds start to get tougher. You have a 60% chance of getting the third baseball card, and you’ll have to buy 1.667 bags to get that third card.

Card# 4:The odds start to get a little dismal. You have a 40% chance of getting the fourth baseball card, and you’ll have to buy 2.5 bags to get that third card.

Card# 5:The odds are against you. You have a 20% chance of getting the last card, and you’ll have to buy 5 bags to get that third card.

So the total amount of bags you’ll have to buy equals:
1 + 1.25 + 1.667 + 2.5 + 5
But wait! (And here’s where a little logic comes in). You can’t go into a store and buy 1.25 bags of cheese puffs, so you’ll have to round up. The equation becomes:
1 + 2 + 2 + 3 + 5 = 13 bags.

One important point for formulas: remember that nothing is absolute. I am reasonably confident that if you purchased 13 bags of cheese puffs and assuming the cards are evenly distributed, that you would get all 5 cards. But…probability is about chance, and it could be just your bad luck to get the same card 10 times in a row (I remember that happening to me as a kid). But there could be another explanation for your “bad luck,” and that is that companies want you to buy as many bags as possible, so they will try to make the odds work in their favor. Even if the cards are evenly distributed, the company could ship bags with cards #1, #2, and #3 to one store (therefore reeling you in by getting you to collect more than half the cards) and #4 and #5 to another store.

What other ways can you think of that manufacturers might turn the odds in their favor?

Probability of a Group Choosing the Same Thing

Probability questions can be broken down into different types. When you are asked to find the probability of a group choosing the same thing, you’re considering the actions of random members of a group (it could be as small as a committee or it could be as large as the population of the U.S.).
These probability questions give you a group, and ask you to calculate the probability of an event occurring for a certain number of random members within that group.

Probability of a Group Choosing the Same Thing : Steps

Sample Problem: There are 200 people at a book fair. 159 of them will buy at least one book. If you survey 5 random people coming out of the door, what is the probability they all will have purchased at least one book?

What is the probability of a group buying at least one book?

Step 1: Convert the data in the question to a fraction. For example, the phrase “159 people out of 200” can be converted to: 159/200.

Step 2: Multiply the fraction by itself. Repeat for however many random items (i.e. people) are chosen. In our example, we have 5 people surveyed, so we want:

159/200 x 159/200 x 159/200 x 159/200 x 159/200 = 0.3176

That’s how to find the Probability of a Group Choosing the Same Thing!

Tip: It may be easier to convert the fraction into a decimal before multiplying. In this case, 159/200 = 0.795.

Check out our YouTube channel for more stats help and tips! We’ve got videos for the most common problems you’ll be likely to come across. Plus videos for using Excel in stats > everything from making basic bar graphs to solving complex data analysis problems.

------------------------------------------------------------------------------

Need help with a homework or test question? Chegg offers 30 minutes of free tutoring, so you can try them out before committing to a subscription. Click here for more details.

If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track.

Comments? Need to post a correction? Please post a comment on our Facebook page.

Probability Introduction: Articles and Videos with Solutions! was last modified: June 18th, 2018 by Stephanie

Statistics How To

Statistics for the rest of us!