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Calculating theoretical probabilities

Calculating theoretical probabilities

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Calculating theoretical probabilities

​​In a nutshell

In order to calculate theoretical probabilities for events from experiments you need to know all the possible outcomes. An outcome of an experiment is itself an event, but it often is also one of many outcomes that can lead to a certain event occurring.

Theoretical probability


An activity or experiment.
A result of a trial.
A subset of all of the possible outcomes of a trial.

The theoretical probability of an event AA is 

P(A)=Number of times event occursTotal number of trialsP(A) = \dfrac{\text{Number of times event occurs}}{\text{Total number of trials}}

Example 1

Alan rolls a fair die. What is the probability that the die lands on a prime number?

As the outcomes of a fair die are all equal, the probability of landing on any number is 16\dfrac{1}{6}. The outcomes and their resepective probabilities can be show in a table:

X123456P(X=x)161616161616\begin{array}{c|cccccc} X &1 &2 &3 &4 &5 &6 \\ \hline P(X=x) &\frac 1 6 &\frac 1 6 &\frac 1 6 &\frac 1 6 &\frac 1 6 &\frac 1 6 \\ \end{array}​​

The outcomes associated with the event of landing on a prime number is if the die lands on 2,3 or 5. Hence,

P(prime)=P(2,3,5)=P(2)+P(3)+P(5)=16+16+16=12P(prime) = P(2 ,3,5) = P(2)+P(3)+P(5) = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} =\underline{\dfrac{1}{2}}

Counting outcomes

To find the probability of an event, we need to count the number of outcomes that lead to the event occurring. To find the total number of outcomes that lead to multiple mutually exclusive events, you can multiply the number of outcomes together.

Note: Mutually exclusive events are events that cannot occur at the same time. If two events AA and BB are mutually exclusive, then P(AB)=0P(A \cap B) = 0.

Example 2

A restaurant offers 33​ starters, 1010​ main courses and 55​ desserts. How many different outcomes are there for an entire sitting?

Obtaining a starter is an event, as are obtaining main courses and desserts. Hence there are 3×10×5=1503 \times10 \times 5 = \underline{150} different possible outcomes.

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FAQs - Frequently Asked Questions

Why is counting important in probability?

How do you count all possible outcomes?

How do you tell if events are mutually exclusive?


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