Topics / Counting & Probability

Conditional Probability & States

Conditional probability, states/Markov-style processes, sequential random events

Conditional Probability & StatesAMC 10 Step by Step

Name: Date: September 13, 2026

Counting & Probability · 17 problems have this as their main idea

primary-topic problems (1.3% of all)

more as a secondary topic

Where it appears

P1-10

P11-15

P16-20

P21-25

## What you need to know
- Conditional probability: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ ; "given that $B$ " means restrict the sample space to $B$ and renormalize.
- Multiplication rule: $P(A \cap B) = P(B)\,P(A \mid B)$ , the engine behind sequential draws and multi-stage experiments.
- Total probability and Bayes: $P(A) = \sum_i P(B_i)\,P(A \mid B_i)$ over a partition, and $P(B_i \mid A) = \dfrac{P(B_i)\,P(A \mid B_i)}{P(A)}$ .
- States: for a process moving among finitely many situations, let $p_s$ be the probability of the goal from state $s$ ; write one linear equation per state and solve, with $p_{\text{goal}} = 1$ and $p_{\text{fail}} = 0$ .
- A process that restarts gives equations like $p = a + b\,p$ .

## 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?"
- Problems 15–22: "given that at least one is a girl / the sum is even," where the trap is which outcomes have been conditioned away.
- Problems 18–25: a frog or token jumps randomly until it reaches an end; find the probability of a particular end, intended for states and a small linear system.
- Dependent draws without replacement that require tracking the remaining bag.

## Standard approaches
1. Draw the tree; multiply down each branch, add the branches that end in the event.
2. For "given that," compute $P(A \cap B) / P(B)$ from counts inside the reduced sample space.
3. For repeated processes, set unknowns $p_s$ per state, merge symmetric states, and solve.
4. Condition on the first step when the process forgets its past.
5. Check that the probabilities of all terminal outcomes sum 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. Let $p_k$ be the probability of finishing at $4$ from square $k$ , with $p_0 = 0$ and $p_4 = 1$ . Then
$p_1 = \tfrac{2}{3}p_2, \qquad p_2 = \tfrac{2}{3}p_3 + \tfrac{1}{3}p_1, \qquad p_3 = \tfrac{2}{3} + \tfrac{1}{3}p_2.$
Substituting the first and third into the second: $p_2 = \frac{4}{9} + \frac{2}{9}p_2 + \frac{2}{9}p_2$ , so $\frac{5}{9}p_2 = \frac{4}{9}$ and $p_2 = \frac{4}{5}$ . Hence $p_1 = \frac{8}{15}$ . Answer $\boxed{\textbf{(B)}\ \frac{8}{15}}$ .

## Pitfalls
- Conditioning on the wrong event: "at least one boy" is not "the first child is a boy."
- Reporting $P(A \cap B)$ as the conditional probability, forgetting to divide by $P(B)$ .
- Writing the state equations but omitting the boundary conditions.
- Assuming independence for draws without replacement.

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)

Problems, easiest first

2023 AMC 10A · #14Conditional Probability & States

A number is chosen at random from among the first $100$ positive integers, and a positive integer divisor of that number is then chosen at random. What is the probability that the chosen divisor is divisible by $11$ ?

2020 AMC 10B · #18Conditional Probability & States

An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the …

2020 AMC 10A · #13Conditional Probability & States

A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1$ , and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices …

2017 AMC 10A · #18Conditional Probability & States

Amelia has a coin that lands heads with probability $\tfrac{1}{3}$ , and Blaine has a coin that lands on heads with probability $\tfrac{2}{5}$ . Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability …

2012 AMC 10B · #18Conditional Probability & States

Suppose that one of every 500 people in a certain population has a particular disease, which displays no symptoms. A blood test is available for screening for this disease. For a person who has this disease, the test always turns out positive. For a person who does not have the disease, however, there is a $2\%$ false …

2003 AMC 10B · #21Conditional Probability & States

A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?

2025 AMC 10B · #22Conditional Probability & States

A seven-digit positive integer is chosen at random. What is the probability that the number is divisible by $11$ , given that the sum of its digits is $61?$

2023 AMC 10B · #21Conditional Probability & States

Each of $2023$ balls is randomly placed into one of $3$ bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?

2021 AMC 10A · #23Conditional Probability & States

Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite …

2021 AMC Fall 10B · #20Conditional Probability & States

In a particular game, each of $4$ players rolls a standard $6{ }$ -sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is …

2019 AMC 10B · #21Conditional Probability & States

Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second …

2019 AMC 10A · #22Conditional Probability & States

Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen …

2019 AMC 10B · #22Conditional Probability & States

Raashan, Sylvia, and Ted play the following game. Each starts with $\$1$ . A bell rings every $15$ seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives $\$1$ to that player. What is the probability that after the …

2011 AMC 10A · #21Conditional Probability & States

Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from …

2010 AMC 10A · #23Conditional Probability & States

Each of 2010 boxes in a line contains a single red marble, and for $1 \le k \le 2010$ , the box in the $k\text{th}$ position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n)$ …

2005 AMC 10A · #18Conditional Probability & States

Team $A$ and team $B$ play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team $B$ wins the second game and team $A$ wins the series, what is the probability that team $B$ wins the …

2014 AMC 10B · #25Conditional Probability & States

In a small pond there are eleven lily pads in a row labeled $0$ through $10$ . A frog is sitting on pad $1$ . When the frog is on pad $N$ , $0<N<10$ , it will jump to pad $N-1$ with probability $\frac{N}{10}$ and to pad $N+1$ with probability $1-\frac{N}{10}$ . Each jump is independent of the previous jumps. If the …