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Conditional Probability & States

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

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Conditional Probability & StatesAMC 10 Step by Step

Name: Date: October 28, 2026

17 problems, easiest first

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

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

1.2023 AMC 10A #14
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 $11$ ?
1. A) $\frac{4}{100}$
2. B) $\frac{9}{200}$
3. C) $\frac{1}{20}$
4. D) $\frac{11}{200}$
5. E) $\frac{3}{50}$
2.2020 AMC 10B #18
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 urn contains six balls. What is the probability that the urn contains three balls of each color?
1. A) $\frac16$
2. B) $\frac15$
3. C) $\frac14$
4. D) $\frac13$
5. E) $\frac12$
3.2020 AMC 10A #13
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 $(0,0), (0,4), (4,4),$ and $(4,0)$ . What is the probability that the sequence of jumps ends on a vertical side of the square?
1. A) $\frac12$
2. B) $\frac 58$
3. C) $\frac 23$
4. D) $\frac34$
5. E) $\frac 78$
4.2017 AMC 10A #18
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 that Amelia wins is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $q-p$ ?
1. A) $1$
2. B) $2$
3. C) $3$
4. D) $4$
5. E) $5$
5.2012 AMC 10B #18
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 positive rate--in other words, for such people, $98\%$ of the time the test will turn out negative, but $2\%$ of the time the test will turn out positive and will incorrectly indicate that the person has the disease. Let $p$ be the probability that a person who is chosen at random from this population and gets a positive test result actually has the disease. Which of the following is closest to $p$ ?
1. A) $\frac{1}{98}$
2. B) $\frac{1}{9}$
3. C) $\frac{1}{11}$
4. D) $\frac{49}{99}$
5. E) $\frac{98}{99}$
6.2003 AMC 10B #21
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?
1. A) $\frac{1}{8}$
2. B) $\frac{5}{32}$
3. C) $\frac{9}{32}$
4. D) $\frac{3}{8}$
5. E) $\frac{7}{16}$
7.2025 AMC 10B #22
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 $61?$
1. A) $\frac{3}{14}$
2. B) $\frac{3}{11}$
3. C) $\frac{2}{7}$
4. D) $\frac{4}{11}$
5. E) $\frac{3}{7}$
8.2023 AMC 10B #21
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?
1. A) $\frac{2}{3}$
2. B) $\frac{3}{10}$
3. C) $\frac{1}{2}$
4. D) $\frac{1}{3}$
5. E) $\frac{1}{4}$
9.2021 AMC Fall 10B #20
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 the probability that Hugo's first roll was a $5,$ given that he won the game?
1. A) $\frac{61}{216}$
2. B) $\frac{367}{1296}$
3. C) $\frac{41}{144}$
4. D) $\frac{185}{648}$
5. E) $\frac{11}{36}$
10.2021 AMC 10A #23
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 edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?
1. A) $\frac{9}{16}$
2. B) $\frac{5}{8}$
3. C) $\frac{3}{4}$
4. D) $\frac{25}{32}$
5. E) $\frac{13}{16}$
11.2019 AMC 10B #22
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 bell has rung $2019$ times, each player will have $\$1$ ? (For example, Raashan and Ted may each decide to give $\$1$ to Sylvia, and Sylvia may decide to give her dollar to Ted, at which point Raashan will have $\$0$ , Sylvia will have $\$2$ , and Ted will have $\$1$ , and that is the end of the first round of play. In the second round Raashan has no money to give, but Sylvia and Ted might choose each other to give their $\$1$ to, and the holdings will be the same at the end of the second round.)
1. A) $\frac{1}{7}$
2. B) $\frac{1}{4}$
3. C) $\frac{1}{3}$
4. D) $\frac{1}{2}$
5. E) $\frac{2}{3}$
12.2019 AMC 10A #22
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 uniformly at random from the closed interval $[0,1]$ . Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $|x-y| > \tfrac{1}{2}$ ?
1. A) $\frac{1}{3}$
2. B) $\frac{7}{16}$
3. C) $\frac{1}{2}$
4. D) $\frac{9}{16}$
5. E) $\frac{2}{3}$
13.2019 AMC 10B #21
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 head?
1. A) $\frac{1}{36}$
2. B) $\frac{1}{24}$
3. C) $\frac{1}{18}$
4. D) $\frac{1}{12}$
5. E) $\frac{1}{6}$
14.2011 AMC 10A #21
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 the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?
1. A) $\frac{7}{11}$
2. B) $\frac{9}{13}$
3. C) $\frac{11}{15}$
4. D) $\frac{15}{19}$
5. E) $\frac{15}{16}$
15.2010 AMC 10A #23
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 $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles. What is the smallest value of $n$ for which $P(n) < \frac{1}{2010}$ ?
1. A) $45$
2. B) $63$
3. C) $64$
4. D) $201$
5. E) $1005$
16.2005 AMC 10A #18
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 first game?
1. A) $\frac{1}{5}$
2. B) $\frac{1}{4}$
3. C) $\frac{1}{3}$
4. D) $\frac{1}{2}$
5. E) $\frac{2}{3}$
17.2014 AMC 10B #25
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 frog reaches pad $0$ it will be eaten by a patiently waiting snake. If the frog reaches pad $10$ it will exit the pond, never to return. What is the probability that the frog will escape without being eaten by the snake?
1. A) $\frac{32}{79}$
2. B) $\frac{161}{384}$
3. C) $\frac{63}{146}$
4. D) $\frac{7}{16}$
5. E) $\frac{1}{2}$

Conditional Probability & States — answersAMC 10 Step by Step

Name: Date: October 28, 2026

Keep this page separate from the problems.

Answers

#	Answer	The idea it turns on
1	B	Only the 9 multiples of 11 up to 99 can work. For each, exactly half its divisors are multiples of 11, because each divisor d pairs with 11 times d.
2	B	Every order of two red and two blue draws has chance 1/30: denominators are always 2, 3, 4, 5 and numerators 1, 2, 1, 2. There are 6 orders.
3	B	The points (1,1), (1,3) and the center sit on diagonals of the square, so from them both kinds of side are equally likely. Look one jump ahead.
4	D	If both players miss in a round (chance 2/5), the game starts over. So Amelia's chance P satisfies P = 1/3 + (2/5)P.
5	C	Imagine 50,000 people: 100 are sick and test positive, but 2% of the 49,900 healthy people, 998 people, also test positive. So about 1 in 11 positives is real.
6	C	All beads end red exactly when two of the three draws are green. Remember the bag holds fewer greens after each green draw.
7	A	A digit sum of 61 is 2 short of seven 9s, leaving only 28 numbers. The test for 11 passes only when the whole shortfall is in even places.
8	E	Only odd or even matters for each bin. Track one question, 'do all three bins match?', ball by ball. The chance quickly settles at 1/4.
9	C	Hugo wins with chance 1/4 by fairness. After rolling 5 he wins with chance 41/96, splitting ties with other 5s evenly. Then divide.
10	D	Only three kinds of square matter: center, edge, corner. From an edge, half the hops reach a corner, a quarter return to center, a quarter wrap to an edge.
11	B	The money is always split 1, 1, 1 or 2, 1, 0 in some order. From either split, the next ring leaves everyone with 1 dollar with probability 1/4.
12	B	Each number is 0 (chance 1/4), 1 (chance 1/4) or a random point in between (chance 1/2). Split into cases by type and add: 1/8 + 1/4 + 1/16.
13	B	The first four flips are forced to be T, H, T, H. After that, list the ways to finish with two heads and add their probabilities.
14	D	The two pairs weigh the same only when both are all real coins or each holds exactly one fake. Count those two kinds of outcomes and compare.
15	A	Stopping at box n means white from every earlier box, then red. The chances 1/2 x 2/3 x ... cancel, leaving 1 / (n(n+1)).
16	A	List every way the series can go when B wins game 2 and A wins the series. Weight shorter series more, since each game is a coin flip.
17	C	Look at the gaps between escape chances on neighboring pads: each gap is the previous one times N/(10 - N), and all gaps add up to 1.

Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.