Counting & ProbabilityDifficulty 30-8588 tagged problems

Conditional Probability

Conditional Probability is the habit of updating a probability once partial information is known. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(A \mid B) = \frac{P(B \mid A)\,P(A)}{P(B)} \ (\text{Bayes})$$

The probability of $A$ given that $B$ happened is the chance both happen divided by the chance of $B$; Bayes' rule flips the conditioning when you know the reverse direction.

Why It Works

1
Once you learn $B$ occurred, the only relevant outcomes are those inside $B$, so $B$ becomes the new sample space.
2
Within $B$, the outcomes that also give $A$ are exactly $A \cap B$, and probabilities are rescaled to make $P(B \mid B) = 1$.
3
Dividing by $P(B)$ does that rescaling: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$.
4
Writing $P(A \cap B) = P(B \mid A)\,P(A)$ and expanding $P(B)$ over disjoint causes gives Bayes' rule, which converts an easy-to-know $P(B \mid A)$ into the wanted $P(A \mid B)$.

Worked Examples

Example 1

A disease affects $1\%$ of people. A test is positive for $99\%$ of those who have it and for $5\%$ of those who do not. Given a positive test, what is the probability the person has the disease?

1
Let $D$ be 'has disease' and $+$ be 'tests positive'. We know $P(D) = 0.01$, $P(+ \mid D) = 0.99$, $P(+ \mid D^c) = 0.05$.
2
Total positive rate (law of total probability): $P(+) = 0.99 \cdot 0.01 + 0.05 \cdot 0.99 = 0.0099 + 0.0495 = 0.0594$.
3
Bayes: $P(D \mid +) = \dfrac{P(+ \mid D)\,P(D)}{P(+)} = \dfrac{0.0099}{0.0594}$.
4
Simplify: $\dfrac{0.0099}{0.0594} = \dfrac{1}{6} \approx 0.167$.

Answer:

$\dfrac{1}{6} \approx 16.7\%$

Warm-up

A fair die is rolled and you are told the result is even. Given that, what is the probability the result is greater than $3$?

1
Condition on $B = \{2, 4, 6\}$ (even), the new sample space, with $P(B) = \tfrac{3}{6}$.
2
The event $A = \{\text{greater than } 3\} = \{4, 5, 6\}$; the overlap $A \cap B = \{4, 6\}$ has probability $\tfrac{2}{6}$.
3
Apply $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{2/6}{3/6} = \dfrac{2}{3}$.

Answer:

$\dfrac{2}{3}$

Contest level

Two cards are drawn without replacement from a standard $52$-card deck. Given that the first card is a heart, what is the probability that both cards are hearts?

1
We want $P(\text{both hearts} \mid \text{first is a heart})$. Given the first card is a heart, only the second draw is uncertain.
2
After removing one heart, $51$ cards remain, of which $12$ are hearts.
3
So the conditional probability the second is also a heart is $\dfrac{12}{51} = \dfrac{4}{17}$.

Answer:

$\dfrac{4}{17}$

Olympiad / Challenge

Box A has $2$ gold and $1$ silver coin; box B has $1$ gold and $3$ silver coins. A box is chosen at random (each with probability $\tfrac{1}{2}$) and one coin is drawn; it is gold. What is the probability the coin came from box A?

1
We want $P(A \mid G)$ where $G$ is 'drew gold'. We know $P(A) = P(B) = \tfrac{1}{2}$, $P(G \mid A) = \tfrac{2}{3}$, and $P(G \mid B) = \tfrac{1}{4}$.
2
Law of total probability: $P(G) = P(G \mid A)P(A) + P(G \mid B)P(B) = \tfrac{2}{3}\cdot\tfrac{1}{2} + \tfrac{1}{4}\cdot\tfrac{1}{2} = \tfrac{1}{3} + \tfrac{1}{8} = \tfrac{11}{24}$.
3
Bayes: $P(A \mid G) = \dfrac{P(G \mid A)P(A)}{P(G)} = \dfrac{1/3}{11/24}$.
4
Simplify: $\dfrac{1/3}{11/24} = \dfrac{1}{3}\cdot\dfrac{24}{11} = \dfrac{8}{11}$.

Answer:

$\dfrac{8}{11}$

Going Deeper

Bayes' rule generalizes via the law of total probability: if $B_1, \ldots, B_k$ partition the sample space, then $P(B_i \mid A) = \dfrac{P(A \mid B_i)P(B_i)}{\sum_j P(A \mid B_j)P(B_j)}$. The chain rule $P(A_1 \cap \cdots \cap A_n) = P(A_1)P(A_2 \mid A_1)\cdots P(A_n \mid A_1 \cap \cdots \cap A_{n-1})$ is the multiplicative analog, central to without-replacement and sequential problems.
Conditioning appears throughout AMC/AIME probability: 'given that,' updating beliefs after evidence (medical-test, two-children, Monty Hall), and computing draws without replacement step by step. The one-step conditioning trick (condition on the first move, then recurse) also drives expected-value and absorbing-state problems.
Pitfall: $P(A \mid B) \ne P(B \mid A)$ in general (the 'prosecutor's fallacy') — the disease example shows a $99\%$-accurate test still yields only $\tfrac{1}{6}$ posterior because the prior $P(D)$ is tiny. Always weight by the prior, and never forget the base rate when inverting a conditional.

Spot the Signal

Use it when an event's probability changes given that another event occurred.
You can describe the hard part as updating a probability once partial information is known, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by writing $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ and computing each piece.
Name the relation that makes Conditional Probability legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Confusing $P(A \mid B)$ with $P(B \mid A)$, or forgetting to restrict to the conditioning event.
Applying Conditional Probability because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Condition by shrinking the world to $B$, then measure $A$ inside it.

Try it on:

Compute a conditional probability, or apply Bayes' rule, in a contest problem.

Back to techniques

Counting & ProbabilityDifficulty 30-8588 tagged problems

Conditional Probability

Practice this technique Match my level

The Key Result

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(A \mid B) = \frac{P(B \mid A)\,P(A)}{P(B)} \ (\text{Bayes})$$

The probability of $A$ given that $B$ happened is the chance both happen divided by the chance of $B$; Bayes' rule flips the conditioning when you know the reverse direction.

Why It Works

1
Once you learn $B$ occurred, the only relevant outcomes are those inside $B$, so $B$ becomes the new sample space.
2
Within $B$, the outcomes that also give $A$ are exactly $A \cap B$, and probabilities are rescaled to make $P(B \mid B) = 1$.
3
Dividing by $P(B)$ does that rescaling: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$.
4
Writing $P(A \cap B) = P(B \mid A)\,P(A)$ and expanding $P(B)$ over disjoint causes gives Bayes' rule, which converts an easy-to-know $P(B \mid A)$ into the wanted $P(A \mid B)$.

Worked Examples

Example 1

A disease affects $1\%$ of people. A test is positive for $99\%$ of those who have it and for $5\%$ of those who do not. Given a positive test, what is the probability the person has the disease?

1
Let $D$ be 'has disease' and $+$ be 'tests positive'. We know $P(D) = 0.01$, $P(+ \mid D) = 0.99$, $P(+ \mid D^c) = 0.05$.
2
Total positive rate (law of total probability): $P(+) = 0.99 \cdot 0.01 + 0.05 \cdot 0.99 = 0.0099 + 0.0495 = 0.0594$.
3
Bayes: $P(D \mid +) = \dfrac{P(+ \mid D)\,P(D)}{P(+)} = \dfrac{0.0099}{0.0594}$.
4
Simplify: $\dfrac{0.0099}{0.0594} = \dfrac{1}{6} \approx 0.167$.

Answer:

$\dfrac{1}{6} \approx 16.7\%$

Warm-up

A fair die is rolled and you are told the result is even. Given that, what is the probability the result is greater than $3$?

1
Condition on $B = \{2, 4, 6\}$ (even), the new sample space, with $P(B) = \tfrac{3}{6}$.
2
The event $A = \{\text{greater than } 3\} = \{4, 5, 6\}$; the overlap $A \cap B = \{4, 6\}$ has probability $\tfrac{2}{6}$.
3
Apply $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{2/6}{3/6} = \dfrac{2}{3}$.

Answer:

$\dfrac{2}{3}$

Contest level

Two cards are drawn without replacement from a standard $52$-card deck. Given that the first card is a heart, what is the probability that both cards are hearts?

1
We want $P(\text{both hearts} \mid \text{first is a heart})$. Given the first card is a heart, only the second draw is uncertain.
2
After removing one heart, $51$ cards remain, of which $12$ are hearts.
3
So the conditional probability the second is also a heart is $\dfrac{12}{51} = \dfrac{4}{17}$.

Answer:

$\dfrac{4}{17}$

Olympiad / Challenge

1
We want $P(A \mid G)$ where $G$ is 'drew gold'. We know $P(A) = P(B) = \tfrac{1}{2}$, $P(G \mid A) = \tfrac{2}{3}$, and $P(G \mid B) = \tfrac{1}{4}$.
2
Law of total probability: $P(G) = P(G \mid A)P(A) + P(G \mid B)P(B) = \tfrac{2}{3}\cdot\tfrac{1}{2} + \tfrac{1}{4}\cdot\tfrac{1}{2} = \tfrac{1}{3} + \tfrac{1}{8} = \tfrac{11}{24}$.
3
Bayes: $P(A \mid G) = \dfrac{P(G \mid A)P(A)}{P(G)} = \dfrac{1/3}{11/24}$.
4
Simplify: $\dfrac{1/3}{11/24} = \dfrac{1}{3}\cdot\dfrac{24}{11} = \dfrac{8}{11}$.

Answer:

$\dfrac{8}{11}$

Going Deeper

Bayes' rule generalizes via the law of total probability: if $B_1, \ldots, B_k$ partition the sample space, then $P(B_i \mid A) = \dfrac{P(A \mid B_i)P(B_i)}{\sum_j P(A \mid B_j)P(B_j)}$. The chain rule $P(A_1 \cap \cdots \cap A_n) = P(A_1)P(A_2 \mid A_1)\cdots P(A_n \mid A_1 \cap \cdots \cap A_{n-1})$ is the multiplicative analog, central to without-replacement and sequential problems.
Conditioning appears throughout AMC/AIME probability: 'given that,' updating beliefs after evidence (medical-test, two-children, Monty Hall), and computing draws without replacement step by step. The one-step conditioning trick (condition on the first move, then recurse) also drives expected-value and absorbing-state problems.
Pitfall: $P(A \mid B) \ne P(B \mid A)$ in general (the 'prosecutor's fallacy') — the disease example shows a $99\%$-accurate test still yields only $\tfrac{1}{6}$ posterior because the prior $P(D)$ is tiny. Always weight by the prior, and never forget the base rate when inverting a conditional.

Spot the Signal

Use it when an event's probability changes given that another event occurred.
You can describe the hard part as updating a probability once partial information is known, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by writing $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ and computing each piece.
Name the relation that makes Conditional Probability legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Confusing $P(A \mid B)$ with $P(B \mid A)$, or forgetting to restrict to the conditioning event.
Applying Conditional Probability because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Condition by shrinking the world to $B$, then measure $A$ inside it.

Try it on:

Compute a conditional probability, or apply Bayes' rule, in a contest problem.