Textbook

A probability describes how likely it is that an event may occur. You’ll see probabilities represented as percentages from 0% to 100%, which correspond to their decimal values from 0 to 1.

- If the event will never occur, the probability is $0$, i.e. a $0%$ chance
- If the event will always occur, the probability is $1$, i.e. a $100%$ chance
- If the event will occur 7 out of 10 times, the probability is $0.7$, i.e. a $70%$ chance

Since something will either happen or it won’t, the sum of these two probabilities will *always* add up to $1$, i.e. a $100%$ chance. For example, imagine we’re picking a marble from a bag, and there’s a $20%$ chance it will be blue. What is the probability it won’t be blue? That probability is $100%−20%=80%$.

A probability can be found by dividing the number of desired outcomes by the number of total possible outcomes. For example, if you and your brother are participating in a raffle, and you both would be happy with either of you winning, and there are 20 total people participating in the raffle, the chance of winning the raffle is 2/20 = .1 or 10%. There are two desired tickets to be drawn and 20 total tickets possible.

There are two rules that can be helpful when working with probabilities: *AND* and *OR*.

Let’s think about this question for a moment:

What is the chance of winning a coin flip twice in a row?

Coin flips are luck-based, and the chance of winning a single coin flip is $50%$, represented in decimal form as $0.50$. Because we are trying to win twice in a row (i.e. winning the first flip *and* the second flip), this is an *AND* situation.

$.50×.50=.25$

The probability of winning twice in a row is $25%$.

Let’s try another example.

You are tasked with randomly selecting criminals from a police lineup. There are two accomplices in the crime. What is the chance you overlook both criminals when randomly selecting two from a lineup that includes 7 suspects? Express the probability in terms of a fraction.

Try to solve it yourself, then check below!

(spoiler)

Answer: $10/21$

This question asks for the probability of selecting a non-criminal *and* then selecting a second non-criminal, so we have to multiply the two probabilities together.

There are $2$ criminals, so given our group has $7$ suspects, there must be $5$ non-criminals. The probability of selecting a non-criminal for the first pick is $5/7$. After making this choice that one non-criminal is no longer in the lineup, so the chance of selecting a second non-criminal is $4/6$.

$5/7×4/6=20/42=10/21$

Think about this question:

What is the chance of randomly selecting a consonant or the letter A from the alphabet?

There are 21 consonants in the alphabet and obviously, there is a single letter A. Because this is an *OR* situation, you should add the probabilities. There is a $21/26$ chance of selecting a consonant and a $1/26$ chance of selecting the letter A. The total probability of selecting either a consonant or A is:

$21/26+1/26=22/26=11/13$

Divided out, $11/13$ is approximately $85%$.

Let’s try another example.

Probabilities $p$, $q$, and $r$ are the chances of three mutually exclusive events occurring. No other possible events can occur beyond these three events.

Quantity A: The chance that $p$ or $r$ occurs

Quantity B: $1−q$

Give it a try!

(spoiler)

Answer: C. The quantities are equal

The probability that $p$ or $r$ could occur is equivalent to $p+r$.

In Quantity B, the $1$ represents all possible outcomes, i.e., there is a 100% chance that one of the three mutually exclusive events occurs. This can be represented as the equation $1=p+q+r$, which we can rearrange to get Quantity B’s value of $1−q$.

$11−q =p+q+r=p+r $

So, it turns out that both Quantity A and Quantity B are $p+r$.

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