Comprehensive theory, key formulas, diagrams, and memory aids for Conditional Probability.
In the real world, events rarely happen in a vacuum. Often, we receive new information that changes the likelihood of an event occurring. Conditional probability is the mathematical way of updating our probabilities based on new, given information.
It measures the probability of an event happening, under the condition that another specific event has already happened.
The conditional probability of an event $A$ occurring, given that event $B$ has already occurred, is written as $P(A|B)$. * This is read aloud as "The probability of $A$ given $B$". * In this scenario, $B$ represents the new information or the condition. Because $B$ has already happened, it essentially becomes our new, restricted sample space. We no longer care about the entire original sample space; we only care about the outcomes where $B$ is true.
If $P(B) > 0$, the conditional probability of $A$ given $B$ is defined as:
$$ P(A|B) = \frac{P(A \cap B)}{P(B)} $$
Alternatively, if dealing with classical counts of outcomes: $$ P(A|B) = \frac{n(A \cap B)}{n(B)} $$
Let's say you roll a fair six-sided die. * Event $A$: Rolling a 2. * Event $B$: Rolling an even number.
Without any conditions, what is the probability of rolling a 2? * $P(A) = \frac{1}{6}$.
With a condition: Now, suppose I look at the die, cover it with my hand, and tell you, "The number rolled is even." What is the new probability that the number is a 2? * You are finding $P(A|B)$. * Because you know the number is even, your sample space is no longer ${1, 2, 3, 4, 5, 6}$. It has shrunk to just the even numbers: ${2, 4, 6}$. * Out of those 3 possibilities, only one is a '2'. * So, $P(A|B) = \frac{1}{3}$.
Using the formula: * $A \cap B$ (Rolling a 2 AND an even number) = Rolling a 2. So, $P(A \cap B) = \frac{1}{6}$. * $P(B)$ (Rolling an even number) = $\frac{3}{6}$. * $P(A|B) = \frac{1/6}{3/6} = \frac{1}{3}$.
The new information (that the roll was even) doubled the probability of the event from $\frac{1}{6}$ to $\frac{1}{3}$.
By rearranging the conditional probability formula, we derive one of the most useful rules in probability: The General Multiplication Rule.
If we multiply both sides of the conditional probability formula by $P(B)$, we get a formula to find the probability of both events happening (the intersection):
$$ P(A \cap B) = P(B) \times P(A|B) $$ or $$ P(A \cap B) = P(A) \times P(B|A) $$
This rule is extremely useful for calculating the probability of a sequence of events. * In English: The probability of both $A$ and $B$ happening is the probability of the first event happening, multiplied by the probability of the second event happening given that the first event already happened.
You draw two cards from a standard 52-card deck without replacing the first card. What is the probability of drawing two Aces? * Event $A$: Drawing an Ace on the first draw. $P(A) = \frac{4}{52}$. * Event $B$: Drawing an Ace on the second draw. * To find $P(A \cap B)$, we need $P(B|A)$. If $A$ happened, there are 3 Aces left in a deck of 51 cards. So, $P(B|A) = \frac{3}{51}$. * Using the Multiplication Rule: $P(\text{Two Aces}) = P(A) \times P(B|A) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$.
Conditional probability gives us a formal mathematical definition of independent events.
Two events $A$ and $B$ are independent if the occurrence of $B$ gives absolutely no information about the occurrence of $A$. In other words, knowing that $B$ happened does not change the probability of $A$.
If we substitute $P(A|B) = P(A)$ into the General Multiplication Rule ($P(A \cap B) = P(B) \times P(A|B)$), we get the familiar formula for independent events: $$ P(A \cap B) = P(A) \times P(B) $$
If you draw an Ace, put it back, shuffle, and draw again, the deck resets. * $P(B|A)$ (Drawing a second Ace given you drew a first Ace) remains $\frac{4}{52}$. * Since $P(B|A) = P(B)$, the events are independent.