Probability rules help us understand how likely different outcomes are. They are applied in a range of ways, like risk assessment and modelling. Having a good understanding of these basics will help you to predict and make sense of different possibilities.
When considering a set of all possible outcomes, an event is a particular outcome of interest. For example:
When tossing a coin, the event of interest might be "obtaining a head".
When considering the weather for Saturday, the event of interest might be "it doesnt rain".
When throwing a die, the event of interest might be "rolling a \(3\)."
The probability of an event \(E\) can be found with the formula:
\[\Pr(E)=\frac{\textrm{Number of ways $E$ can occur}}{\textrm{Total number of possible outcomes}}\]
This assumes all outcomes are equally likely.
\(\Pr(E)\) is expressed as a number between \(0\) and \(1\):
\[0\leq Pr(E)\leq1\]
If the event is impossible, \(\Pr(E)=0\).
If the event is certain, \(\Pr(E)=1\).
Example 1 – finding the probability of an outcome
If two coins are tossed, find the probability of obtaining two heads.
Let \(E\) be the event "two heads". If H = heads and T = tails, the total number of possible outcomes is \(4\): HH HT TH TT.
\[\begin{align*} \Pr(E) & = \frac{\textrm{Number of ways E can occur}}{\textrm{Total number of possible outcomes}}\\
& = \frac{1}{4}
\end{align*}\]
If a die is thrown, find the probability of obtaining an odd number.
Let \(E\) be the event "an odd number". The possible outcomes are: \(1, 3, 5\). The total number of possible outcomes is \(6\).
\[\begin{align*} \Pr(E) & =\frac{\textrm{Number of ways E can occur}}{\textrm{Total number of possible outcomes}}\\
& = \frac{3}{6}\\
& = \frac{1}{2}
\end{align*}\]
The multiplication principle
The multiplication principle is used to find the probability of two events happening at the same time by multiplying their probabilities—that is, the probability of \(A\) and \(B\) occurring.
We can only use the multiplication principle if the probability of event \(A\) does not affect the probability of event \(B\) occurring. If this is true, we say that \(A\) and \(B\) are independent events.
The probability of \(A\) and \(B\) occurring together is denoted by \(\Pr(A\cap B)\) where \(\cap\) is the symbol for intersection.
\[\Pr(A\textrm{ and }B)=\Pr(A\cap B)=\Pr(A)\times(B)\]
Example – using the multiplication principle
If two coins are tossed, find the probability of obtaining two heads.
This is the same question as Example 1 – finding the probability of an outcome. Because tosses of a coin are independent events, an alternate way of calculating the probability would be to use the multiplication principle.
Two coins are tossed, each representing an event. In both, the possibility of a heads is \(\dfrac{1}{2}\).
The probability of a head on the first toss (\(H_{1}\)) and a head on the second toss (\(H_{2}\)) is:
\[\begin{align*} \Pr(H_{1}\cap H_{2}) & = \Pr(H_{1})\times \Pr(H_{2})\\
& = \frac{1}{2}\times\frac{1}{2}\\
& = \frac{1}{4}
\end{align*}\]
The addition principle
The addition principle tells us how to find the probability of one event or another happening by adding their individual probabilities—that is, the probability of \(A\) or \(B\) occurring.
We denote this using the union symbol, \(\cup\).
\[\Pr(A\textrm{ or }B)=\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)\]
If \(A\) and \(B\) are mutually exclusive or cannot happen at the same time (i.e. \(\Pr(A\cap B)=0\), we use:
\[\Pr(A\textrm{ or }B)=\Pr(A\cup B)=\Pr(A)+\Pr(B)\]
Example – using the addition principle
If a die is tossed twice, find the probability that a \(6\) occurs.
We are looking at two events, the first die toss and the second die toss. The \(6\) could occur on the first toss (\(S_{1}\)) or the second toss (\(S_{2}\)).
In both events, the possibility of a \(6\) is \(\dfrac{1}{6}\).
\[\begin{align*} \Pr(S_{1}\cup S_{2}) & = \Pr(S_{1})+\Pr(S_{2})-\Pr(S_{1}\cap S_{2})\\
& = \frac{1}{6}+\frac{1}{6}-\frac{1}{36}\\
& = \frac{11}{36}
\end{align*}\]
Complementary events
Complementary events show us what happens when an event doesn't occur, helping to see the whole picture.
If \(E\) is an event, then "not \(E\)" is the complement of \(E\). The complement of \(E\) is denoted by \(\overline{E}\), \(E'\) or \(E^{c}\).
Some examples of complementary events are:
"winning the grand final" and "not winning the grand final"
"passing a test" and "failing a test"
"being left handed" and "being right handed".
To find the probability of a complementary event, we just minus the probability of the event occurring from \(1\).
\[\Pr(\overline{E})=1-\Pr(E)\]
Example – finding the probability of a complementary event
If a die is tossed twice, find the probability of not getting a \(6\) on either toss.
From Example – using the addition principle, we found that \(\Pr(S_{1}\cup S_{2})=\dfrac{11}{26}\).
\[\begin{align*} \Pr(\overline{S_{1}\cup S_{2}}) & = 1-\Pr(S_{1}\cup S_{2})\\
& = 1-\frac{11}{36}\\
& = \frac{25}{36}
\end{align*}\]
Exercise – using the probability rules
If \(1000\) tickets are sold in a raffle and one winning ticket is chosen at random, what is the probability of winning the raffle if \(5\) tickets are purchased?
If a die is rolled, what is the probability that the number is greater than \(4\)?
A bag contains \(6\) white marbles and \(4\) black marbles. A marble is chosen, the colour recorded and it is then returned to the bag. This is repeated three times. What is the probability that all three marbles are white?
The probability that phone A will continue to function in \(10\) years' time is \(0.7\). The probability that phone B will continue to function in \(10\) years' time is \(0.4\). Find the probability that:
both are still functioning in \(10\) years' time
neither are still functioning in \(10\) years' time
only one is still functioning in \(10\) years' time
at least one is still functioning in \(10\) years' time.
