# MDM4U Dependent and Independent Events 2016-03-21

We talked about compound events, which are made up of a sequence of events. For example, if each trial in your experiment involved flipping a coin and then rolling a d6 die, each outcome would be a pair of results: the result of the coin flip followed by the result of the die roll.

A compound event is usually described in terms of “smaller” events for each stage of a trial. In the example above you might have event $A$ be the event that you flip heads on the coin, and event $B$ be the event that you roll a 5 or 6 on the die.

In this case, these events are independent: whether $A$ is achieved or not does not affect the probability that $B$ occurs (the coin flip doesn’t affect the die roll).

In general, when two events $A$ and $B$ are independent,

$P(A \text{ and } B) = P(A) \times P(B)$

For this example, $P(A) \times P(B) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$

So the probability of flipping heads and rolling a 5 or 6 is $\frac{1}{6}$.

For dependent events, the probability of the second event depends on the outcome of the first stage in the trial. For example, suppose the Steelhawks have a 70% chance of winning the first basketball game of the season. Suppose also that they have an 85% chance of winning the second game if they win the first game (perhaps their chance of winning the second game would be less if they lost the first game). This is a dependent situation – the probability of the second event occurring depends on the result of the first event.

Let’s let $A$ be the event that we win the first game, and let $B$ be the event that we win the second game.

In this case, we know two things: the probability of winning the first game, which is $P(A)=0.7$, and $P(B|A)=0.85$.

This is new notation for us: $B|A$ is the event that $B$ occurs given that $A$ has occurred. We say “$B$ given $A$“.

Together these give us what we want: the probability that $A$ and $B$ both occur is the probability that $A$ occurs times the probability that $B$ occurs given that $A$ has already occurred:

$P(A \text{ and } B) = P(A) \times P(B|A)$

In this case we have

$latex P(A) \times P(B|A) = 0.7 \times 0.85 = 0.595$

There is a 59.5% chance that the Steelhawks will win both games.

It’s important to not that this does not tell us the probability of the Steelhawks winning one game, or losing both games. We can figure out that there must be a $70%-59.5%=10.5%$ chance that we’ll win the first game but lose the second, but that’s all. If we lose the first game we know nothing about the probability for the second game based on what we’ve been given.

Your homework is page 334 #3, 4, 6.