# 2.1.3.2.5.1 - Advanced Conditional Probability Applications

2.1.3.2.5.1 - Advanced Conditional Probability Applications## Advanced Formulas

Conditional probabilities can also be computed using the following formulas. Note that these two formulas are identical, but A and B are switched. Again, if the contingency table is available it is usually most efficient to take the appropriate values from the table, as shown above, as opposed to using these formulas.

- Conditional Probability of A Given B
- \(P(A\mid B)=\dfrac{P(A \: \cap\: B)}{P(B)}\)

- Conditional Probability of B Given A
- \(P(B\mid A)=\dfrac{P(A \: \cap\: B)}{P(A)}\)

## Example: Clubs

In a standard 52-card deck, there are 26 black cards including 13 clubs. All clubs are black, therefore there are 13 black clubs.

What is the probability that a randomly selected card is a club given that it is a black card?

We are given that \(P(club)=\frac{13}{52}=0.25\), \(P(black)=\frac{26}{52}=0.50\), and \(P(club \: \cap\: black)=\frac{13}{52}=0.25\)

\(P(club\mid black)=\dfrac{P(club \: \cap\: black)}{P(black)}=\dfrac{0.25}{0.50}=0.50\)

Given that a randomly selected card is black, there is a 50% chance that it's a club.

## Independent Events Written as Conditional Probabilities

If events A and B are independent then \(P(A) = P(A \mid B)\). In other words, whether or not event B occurs does not change the probability of event A occurring.

## Example: Checking for Independence, Aces and Hearts

A card is randomly drawn from a 52-card deck. Are the events of drawing an ace and drawing a heart independent?

In a standard 52-card deck, there are 4 aces and 13 hearts. Therefore \(P(ace)=\frac{4}{52}\) and \(P(heart)=\frac{13}{52}\). Out of 13 hearts, 1 is an ace, which translates to \(P(ace \mid heart) = \frac{1}{13}\).

To determine if these two events are independent we can compare \(P(A)\) to \(P(A\mid B)\). If we call being an ace event A and being a heart event B, then we're comparing \(P(ace)\) to \(P(ace \mid heart)\).

\(P(ace)=\frac{4}{52}=0.0769\)

\(P(ace \mid heart) = \frac{1}{13}=0.0769\)

These values are identical, therefore we can conclude that the events of drawing an ace and drawing a heart are independent.