Probability plays an important part of our everyday lives. Will I get in a car accident on my way home from work? Will FC Barcelona win their match today? For events like this, where there are so many unknowns, it is impossible to say for certain what the future holds. However, with a little statistical analysis and understanding of “chance”, it is amazing how close we can come to predicting the future.

It is useful to remember that the total probability must sum to $1$. Thus, given an event $\rm A$, the probability of $\rm A$, $\rm P(A)$ plus the probability of NOT $\rm A$, $\rm P(A′)$ is one. Mathematically, $\rm P(A)+P(A′)=1.$ After all, something will happen or it will not happen.

Many times, we are concerned with the probability of an event occurring, given something else just occurred. For example, what is the probability that I earn a passing grade on a $10$ question exam, if someone told me I answered $3$ of the questions correctly and they are unsure about the remaining questions? Or, what is the probability it will rain tonight, given that it rained this afternoon? If we know certain events have already occurred, this changes the probability of future events to occur. In probability theory, given events $\rm A$ and $\rm B$, the probability of $\rm A$ given $\rm B$, $\rm P(A|B)$ is calculated as:

$\rm P(A|B)=P(A\cap B)P(B)$.

Example 1: A fair coin is flipped and a spinner is spun with six equally likely outcomes. Determine the probability that a head and the number five occur.

Explanation: These two events are said to be independent since one event does not affect the other. Therefore, $\rm P(H\bigcap 5)=\dfrac{1}{2}\cdot \dfrac{1}{6}=\dfrac{1}{12}$.

Note: The intersection of two events is denoted $\cap$ and is used when we multiply probabilities.

Example 2: A drawer contains $12$ socks of which $3$ are red, $4$ are blue, and $5$ are green. A sock is selected at random. Determine the probability that a red or green sock is selected.

Explanation: $\rm P(Red\cup Green)=\dfrac{3}{12}+\dfrac{5}{12}=\dfrac{8}{12}=\dfrac{2}{3}$

Note: The union of two events is denoted ∪ and is used when we add probabilities.

Example 3: An energy company makes batteries using $2$ different machines, $\rm A$ and $\rm B$. $65\%$ of the batteries are made from machine $\rm A$. $2\%$ of the batteries made by machine $\rm A$ fail and $3\%$ of the batteries made by machine $\rm B$ fail. A battery is selected at random. Given the battery failed, determine the probability it came from machine $\rm B$.

Explanation: It is important to note that the battery could come from machine $\rm A$ or $\rm B$.

$\rm P(B|F)=\dfrac{P(B\cap F)}{P(F)}$
$\rm P(B|F)=\dfrac{\frac{35}{100}\cdot \frac{3}{100}}{\frac{65}{100} \cdot \frac{2}{100} + \frac{35}{100} \cdot \frac{3}{100}}$
$\rm P(B|F)=\dfrac{35(3)}{35(3)+65(2)}$
$\rm P(B|F)=\dfrac{21}{47}=0.447$