Conditional probability

1. The definition. Given that event $B$ has occurred, the probability of event $A$ is written $P(A|B)$ and defined as the fraction of $B$'s probability that also belongs to $A$:

where: $P(A\cap B)$ is the probability that both $A$ and $B$ occur, and $P(B)$ is the probability of the conditioning event, which becomes the new "whole" once you know $B$ has happened.

2. Reading it as restricted counting. When all outcomes of a sample space $S$ are equally likely, the formula above is equivalent to simply counting inside $B$:

where: $n(A\cap B)$ is the number of outcomes common to both events and $n(B)$ is the number of outcomes in $B$ alone — you never divide by $n(S)$ once you have conditioned on $B$, because $B$ itself is now the sample space.

3. The multiplication theorem falls straight out. Multiply both sides of the definition by $P(B)$ and you get the probability that two events both occur:

This is the single most-used tool in the chapter: it lets you build up the probability of a compound event ("first this, then that") from a chain of simpler, conditional steps — exactly the situation in drawing cards or balls without replacement.

4. Independence is a special case. Two events $A$ and $B$ are independent when knowing one occurred does not change the probability of the other:

If this equality fails, the events are dependent and you must use the full conditional form, not the simple product.

5. The addition theorem handles "or." For any two events, whether or not they are independent:

which drops the last term when $A$ and $B$ are mutually exclusive, since then $P(A\cap B) = 0$.

6. The chain extends to three events. The same logic used in Step 3 of the multiplication theorem below applies repeatedly, so for three events $A$, $B$, $C$:

where: each successive factor conditions on everything assumed to have already happened — first $A$ alone, then $B$ given $A$, then $C$ given both $A$ and $B$ together. This is the pattern behind problems that draw three cards or three balls in sequence without replacement: multiply a chain of shrinking conditional probabilities, one for each draw.

A bag contains 5 red and 3 black balls. Two balls are drawn one after another without replacement. Find the probability that both balls drawn are red.

Step 1. Define the events. Let $A$ = "the first ball is red" and $B$ = "the second ball is red." The question asks for $P(A\cap B)$, the probability both happen.

Step 2. Find $P(A)$. There are 5 red balls among 8 total, so $P(A) = \dfrac{5}{8}$.

Step 3. Find $P(B|A)$. Because the first ball is not replaced, once one red ball is removed only 4 red and 3 black balls remain, i.e. 7 balls total, so $P(B|A) = \dfrac{4}{7}$.

Step 4. Apply the multiplication theorem from the derivation above.

The probability that both balls drawn are red is $\dfrac{5}{14}$. Because the draws are without replacement, $A$ and $B$ are dependent — using $P(B|A)=4/7$ instead of the plain $P(B)=5/8$ (which would be correct only for sampling with replacement) is exactly what makes this a conditional-probability problem rather than a simple product of two independent chances.

Competitive stretch

Same bag, but now find the probability that the second ball is black, given the first ball drawn was red. This is asking directly for $P(B'|A)$ where $B'$ = "second ball is black." After removing one red ball, 4 red and 3 black balls remain out of 7, so $P(B'|A) = \dfrac{3}{7}$ — read straight off the reduced sample space, no formula manipulation needed, because the conditioning event has already fixed what remains in the bag.

• Confusing $P(A|B)$ with $P(B|A)$ — these restrict to different sample spaces and are generally not equal.
• Assuming independence without checking $P(A\cap B) = P(A)P(B)$; drawing without replacement almost always makes successive draws dependent.
• Using $P(A\cup B) = P(A)+P(B)$ when the events are not mutually exclusive, forgetting to subtract $P(A\cap B)$ and double-counting the overlap.
• Writing the conditional formula with $n(A)$ instead of $n(A\cap B)$ in the numerator — only outcomes common to both events count once you condition on $B$.

Ask the tutor: if a word problem with dice, cards, or bags of balls has you stuck on whether events are independent or dependent, ask the AI tutor to walk through identifying $A$, $B$, and the right conditional probability step by step.