Two events are independent if the outcome of one doesn't affect the outcome of the other. Otherwise they are dependent.

Examples:

• When tossing a fair coin twice, the result of the first toss doesn't affect the probability of the outcome of the second toss, and vice versa. The event 'getting a head' in the second toss is independent of the event 'getting a head' in the first toss.
• When drawing two cards from a deck of 52 playing card, the event 'getting a King' on the first card and the event 'getting a black card' are not independent. The probability of the second card change after the first card is drawn. The two events would be independent if after drawing the first card, the card is returned to the deck (thus the deck is complete 52 again).

For two independent events, $A$ and $B$, the probability of both occuring, $\mathbb{P}\left(A\cap B\right)$, is the product of the probability of each event. $\cap$ is the mathematical symbol for and or intersection.

$\mathbb{P}\left(A\cap B\right)=\mathbb{P}\left(A\right)×\mathbb{P}\left(B\right)$

For example, when tossing a fair coin twice, the probability of getting a 'Head' (let's call it event $H$) on the first and then getting a 'Tail' (let's call it event $T$) on the second is

$\begin{array}{rl}\mathbb{P}\left(H\cap T\right)& =\mathbb{P}\left(H\right)×\mathbb{P}\left(T\right)\\ & =0.5×0.5\\ & =0.25\end{array}$

Two events are mutually exclusive if they cannot occur at the same time.

Examples:

• When tossing a fair coin, the event 'getting a head' and the event 'getting a tail' are mutually exclusive because they can't occur at the same time.
• When throwing a fair die, the event 'getting a 1' and the event 'getting a 4' are mutually exclusive because they can't occur at the same time. But the event 'getting a 3' and the event 'getting an odd number' are not mutually exclusive since it can happen at the same time (i.e. if you get 3)

For two mutually exclusive events, $A$ and $B$, the probability of either one occuring, $\mathbb{P}\left(A\cup B\right)$, is the sum of the probability of both events. $\cup$ is the mathematical symbol for union. $\mathbb{P}\left(A\cup B\right)=\mathbb{P}\left(A\right)+\mathbb{P}\left(B\right)$

For example, when choosing a ball at random from a bag containing 3 blue balls, 2 green balls, and 5 red balls, the probability of getting a blue (let's call it event $B$) or red ball (let's call it event $R$) is

$\begin{array}{rl}\mathbb{P}\left(B\cup R\right)& =\mathbb{P}\left(B\right)+\mathbb{P}\left(R\right)\\ & =\frac{3}{10}+\frac{5}{10}\\ & =\frac{8}{10}\\ & =0.8\end{array}$

For non mutually exclusive events the probability of either one or both occuring is

$\mathbb{P}\left(A\cup B\right)=\mathbb{P}\left(A\right)+\mathbb{P}\left(B\right)-\mathbb{P}\left(A\cap B\right)$

where $\mathbb{P}\left(A\cap B\right)$ is the probability of event $A$ and event $B$ happening at the same time.

For example, when drawing a card from a deck of 52 playing cards, the probability of getting a red card (let's call it event $R$) or a King (let's call it event $K$) is

$\begin{array}{rl}\mathbb{P}\left(R\cup K\right)& =\mathbb{P}\left(R\right)+\mathbb{P}\left(K\right)-\mathbb{P}\left(R\cap K\right)\\ & =\frac{26}{52}+\frac{4}{52}-\frac{2}{52}\\ & =\frac{28}{52}\\ & =\frac{7}{13}\end{array}$

This is so because a card can either be red, king, or both (i.e. red king). So that's why we need to subtract the probability of a card being both red and king because it has already been accounted for in the probability of the card being red and the probability of the card being king.