Conditional Probability

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Conditional probability and its different appearances in contests.

Authors: Jaden Park, Alexander Ren

Introduction

Conditional probability is just probability of an event after another event has occured.

The conditional probability of an event $A$ given that event $B$ has occured is denoted by $P(A \mid B),$ and it is calculated using the formula:

P(A \mid B) = \frac{P(A \cap B)}{P(B)}

where:

$P(A \mid B)$ is the conditional probability of $A$ given $B,$
$P(A \cap B)$ is the probability that both $A$ and $B$ occur,
$P(B)$ is the probability that $B$ occurs (and $B > 0$ ).

Here are some examples.

Example

Imagine you have a standard deck of $52$ cards, and you draw one card. What is the probability that the card is an ace given that it is a spade?

Solution

First, we identify our events:

$A:$ The card is and ace.
$B:$ The card is a spade.

We also know that:

$P(A \cup B)$ is the probability of drawing the ace of spades, which is $\frac{1}{52}$ because there is only one ace of spades in a deck of $52$ cards.
$P(B)$ is the probability of drawing a spade, which is $\frac{13}{52}$ since there are $13$ spades in a deck of $52$ cards.

Using the conditional probability formula:

P(A \mid B) = \frac{P(A \cup B)}{P(B)} = \frac{\frac{1}{52}}{\frac{13}{52}} = \frac{1}{13}

Therefore, the probability of drawing an ace given that you've drawn a spade is $\frac{1}{13}.$

Example

Suppose you roll two six-sided dice. What is the probability that the sum of the dice if $8$ given that at least one of the dice shows a $3$ ?

Solution

First, we identify our events:

$A:$ The sum of the dice is $8.$
$B:$ At least one die shows a $3.$

We also know that:

$A \cup B:$ The combinations that result in a sum of $8$ and include at least one $3$ are $(3, 5)$ and $(5, 3).$ So, $P(A \cup B) = \frac{2}{36}$ since there are two favorable outcomes out of $36$ possible outcomes when rolling two dice.
$B:$ The combinations that include at least one $3$ are $(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),$ and the same combinations with the $3$ in the second position, minus the repeated $(3, 3),$ giving us $11$ combinations. Therefore, $P(B) = \frac{11}{36}.$

Using the conditional probability formula:

P(A \mid B) = \frac{P(A \cup B)}{P(B)} = \frac{\frac{2}{36}}{\frac{11}{36}} = \frac{2}{11}

Therefore, the probability of getting a sum of $8$ given that at least one die shows a $3$ is $\frac{2}{11}.$

Here are some harder practice problems.

Practice Problem

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?

Practice Problem

Final Notes and Tips

Not super common, if a problem can be solved using conditionals, it can be solved in another way (probably).
Consider different cases in a problem, especially extreme ones, and I would memorize the formula.

Recursion

Pascal's Triangle

Combinatorics Progress

Counting

Principles

Miscellaneous

Introduction

Example

Solution

Example

Solution

Practice Problem

Practice Problem

Final Notes and Tips

Introduction

Example

Solution

Example

Solution

Practice Problem

Practice Problem

Final Notes and Tips