Conditional Probability & States

What you need to know

• Conditional probability. $P(A\mid B)$, read "the probability of $A$ given $B$," is the chance of $A$ when you know $B$ happened. Keep only the outcomes in $B$, and ask what fraction of them are in $A$: $P(A\mid B) = \dfrac{P(A\text{ and }B)}{P(B)}$. Example: if a die roll is even, it is $2$, $4$, or $6$, so it is a $6$ with probability $\frac13$ (the formula gives $\frac{1/6}{1/2}$).
• Multiplication rule. $P(A\text{ and }B) = P(B)\cdot P(A\mid B)$, used for draws in a row and for experiments in stages. Example: the top two cards of a deck are both aces with probability $\frac{4}{52}\cdot\frac{3}{51} = \frac{1}{221}$.
• Splitting into cases, and Bayes' Theorem. If the cases $B_1, B_2, \dots$ do not overlap and cover every possibility, then $P(A) = P(B_1)P(A\mid B_1)+P(B_2)P(A\mid B_2)+\cdots$, and $P(B_1\mid A) = \dfrac{P(B_1)\,P(A\mid B_1)}{P(A)}$. Example: box 1 has $3$ red balls and $1$ blue, box 2 has $1$ red and $3$ blue, and a coin flip picks the box: $P(\text{red}) = \frac12\cdot\frac34+\frac12\cdot\frac14 = \frac12$, and $P(\text{box 1}\mid\text{red}) = \frac{3/8}{1/2} = \frac34$.
• States. For a process moving among a few situations (states), let $p_s$ be the chance of reaching the goal from state $s$. Write one equation per state based on the next step, with $p = 1$ at the goal and $p = 0$ at a failure, and solve.
• Restarts. A process that can return to its start gives an equation like $p = a+b\,p$: you win now with probability $a$, or start over with probability $b$. Example: Amy and Ben take turns flipping a coin, Amy first; the first head wins. For Amy's chance $p$, $a = \frac12$ (heads at once) and $b = \frac14$ (two tails), so $p = \frac12+\frac14p$ and $p = \frac23$.

How AMC 10 tests it

• Problems 12–20: two-stage experiments (choose a box, then a ball; roll a die, then flip that many coins), asking for an outcome's probability or the reverse, "given the outcome, which box was it?"
• Problems 15–22: "given that at least one is a girl" or "given that the sum is even," where the trap is which outcomes the condition has thrown away.
• Problems 18–25: a frog or token jumps at random until it reaches an end; find the probability of a particular end, using states and a small system of equations.
• Draws without replacement where you must keep track of what is left in the bag.

Standard approaches

1. Draw the tree. Multiply along each branch, then add the branches that end in the event.
2. For "given that," find $P(A\text{ and }B)\div P(B)$, or count only among the outcomes where $B$ happened.
3. For repeated processes, give each state an unknown $p_s$, merge states that are alike by symmetry, and solve.
4. Look at the first step when the future depends only on where you are, not how you got there.
5. Check that the probabilities of all possible endings add up to $1$.

Worked example

A token starts on square $1$ of a row of squares numbered $0$ through $4$. Each second it moves to an adjacent square, right with probability $\frac{2}{3}$ and left with probability $\frac{1}{3}$. The process stops when the token reaches square $0$ or square $4$. What is the probability that it stops at square $4$?

(A) $\frac{4}{15}$ (B) $\frac{8}{15}$ (C) $\frac{2}{5}$ (D) $\frac{8}{27}$ (E) $\frac{16}{27}$

Solution.

Step 1: unknowns. Let $p_k$ be the probability of stopping at $4$ from square $k$. At the ends, $p_0 = 0$ and $p_4 = 1$.

Step 2: equations. From square $k$, the first move goes right (chance $\frac23$) or left (chance $\frac13$), so $p_k = \frac23p_{k+1}+\frac13p_{k-1}$. Using $p_0 = 0$ and $p_4 = 1$:

$$p_1 = \tfrac{2}{3}p_2, \qquad p_2 = \tfrac{2}{3}p_3 + \tfrac{1}{3}p_1, \qquad p_3 = \tfrac{2}{3} + \tfrac{1}{3}p_2.$$

Step 3: solve for $p_2$. Substitute the first and third into the middle: $p_2 = \frac23\left(\frac23+\frac13p_2\right)+\frac13\cdot\frac23p_2 = \frac49+\frac29p_2+\frac29p_2 = \frac49+\frac49p_2$. Subtract $\frac49p_2$ from both sides: $\frac59p_2 = \frac49$, so $p_2 = \frac45$.

Step 4: answer. $p_1 = \frac23\cdot\frac45 = \frac{8}{15}$.

The answer is $\boxed{\textbf{(B)}\ \frac{8}{15}}$.

Pitfalls

• Conditioning on the wrong event: "at least one child is a boy" is not "the first child is a boy." With two children, two boys has probability $\frac13$ given that at least one is a boy, and $\frac12$ given that the first child is a boy.
• Giving $P(A\text{ and }B)$ as the answer and forgetting to divide by $P(B)$.
• Writing the state equations but leaving out the values at the ends ($p = 1$ at the goal, $p = 0$ at a failure).
• Treating draws without replacement as independent: after an ace is drawn, the next card is an ace with probability $\frac{3}{51}$, not $\frac{4}{52}$.

Traps that recur

• Solving the ten linear equations by brute substitution and making a fraction error; the difference recurrence turns it into one clean sum. (2014 AMC 10B #25)
• Noticing only one of the two ways to fall 2 short of 63; keeping just the numbers with a single 7 gives 3 out of 7, which is (E). (2025 AMC 10B #22)
• Treating the three parities as independent coin flips or guessing 1/3, instead of noticing the odd total leaves four (nearly) equally likely parity patterns. (2023 AMC 10B #21)
• Ignoring the wraparound (an edge square's outward hop lands on the opposite edge, not off the board) or forgetting that Frieda stops once she reaches a corner. (2021 AMC 10A #23)