Probability of Dependent Events
Let’s go back to the deck of cards. We know that the probability of a randomly chosen card being a diamond is \(\frac{1}{4}\) or \(25\%\). If we choose a diamond and don’t put it back in the deck, is our probability of choosing a spade for the second card also \(25\%\)? Although it will be close, it is not the same because even though there are \(13\) spades left in the deck, there is now a total of only $\(51\) cards left. When the outcome of one event affects the outcome of another event, we say these are dependent events.
Let’s go back to the deck of cards. We know that the probability of a randomly chosen card being a diamond is \(\frac{1}{4}\) or \(25\%\). If we choose a diamond and don’t put it back in the deck, is our probability of choosing a spade for the second card also \(25\%\)? Although it will be close, it is not the same because even though there are \(13\) spades left in the deck, there is now a total of only $\(51\) cards left. When the outcome of one event affects the outcome of another event, we say these are dependent events.
If two events are dependent, then the probability that both events will happen is \(P\left(A\ \text{and}\ B\right)=P\left(A\right)\cdot P\left(B\ |\
A\right)\).
\(P\left(B\ |\ A\right)\) is the probability that the second event (\(B\)) occurs given that the first event (\(A\)) has already occurred.
A\right)\).
\(P\left(B\ |\ A\right)\) is the probability that the second event (\(B\)) occurs given that the first event (\(A\)) has already occurred.
In the scenario above, event \(A\) would be choosing a diamond and event \(B\) would be choosing a spade. The probability would look as follows:
\(P\left(\text{diamond and spade}\right)=P\left(\text{diamond}\right)\cdot P\left(\text{spade}\ | \text{diamond}\right)\)
\(\large\frac{13}{52}\cdot\frac{13}{51}=\frac{13}{204}\ \normalsize\text{or}\ 6.37\%\)
Example 1: There are \(40\) employees at a company in Naperville. Only \(13\) of those employees live in Naperville. Two people are randomly selected to go on a company trip.
a) Find the probability that the first person chosen is a Naperville resident but the second person is not a Naperville resident.
b) P(resident|resident)
\(P\left(\text{diamond and spade}\right)=P\left(\text{diamond}\right)\cdot P\left(\text{spade}\ | \text{diamond}\right)\)
\(\large\frac{13}{52}\cdot\frac{13}{51}=\frac{13}{204}\ \normalsize\text{or}\ 6.37\%\)
Example 1: There are \(40\) employees at a company in Naperville. Only \(13\) of those employees live in Naperville. Two people are randomly selected to go on a company trip.
a) Find the probability that the first person chosen is a Naperville resident but the second person is not a Naperville resident.
b) P(resident|resident)
Example 2: A study finds that \(52\%\) of seniors at a local high school have their own car. If a senior has their own car, there is a \(95\%\) chance that they are involved in some type of extra-curricular activity. If a senior does not have their own car, there is a \(72\%\) chance that they are involved in some type of extra-curricular activity. What is the probability that a random senior from that high school is involved in some type of extra-curricular activity?