Conditional Probability
A conditional probability is a probability that a certain event will occur given some knowledge about the outcome or some other event.
\(P(A\mid B)\) is a conditional probability. It is read as "The probability of \(A\) given \(B\)."
If events \(A\) and \(B\) are in a uniform sample space, then:
\(P(A\mid B)=\dfrac{|A\cap B|}{|B|}\)
If events \(A\) and \(B\) are not in a uniform sample space, then (more generally):
\(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\)
The concept of conditional probability is closely tied to the concepts of independent and dependent events. Probability problems that provide knowledge about the outcome can often lead to surprising results. A good example of this is the Monty Hall Problem. Often, one's initial intuition is incorrect on these kind of problems. However, an understanding of conditional probability can help one obtain the correct result.
Uniform Conditional Probability
A fair 12-sided die is rolled. What is the probability that the roll is a 3 given that the roll is odd?
Let \(A\) be the event that a 3 is rolled. Let \(B\) be the event that an odd number is rolled. Using the definitions above, this problem can be restated as \(P(A\mid B)\).
The entire sample space of the 12-sided die is as follows:
\(S=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\).
Because the die is fair, the sample space will be uniform.
The event, B, is the following subset of S:
\(B=\{1,3,5,7,9,11\}\).
\(A\cap B\) is the event that the roll is 3 and is odd. A is a subset of B and \(A\cap B=\{3\}\).
\(P(A\mid B)=\dfrac{|A\cap B|}{|B|}=\boxed{\dfrac{1}{6}}\).
A deck of 10 cards numbered 1 through 10 is shuffled, and a card is drawn from the deck. Let \(A\) be the event that an odd number is drawn and \(B\) be the event that a prime number is drawn. What is \(P(B\mid A)\)?
A player is dealt 5 cards from a shuffled standard playing card deck. What is the probability that the player will obtain a 3-of-a-kind given that two of the cards are the same rank?
In poker, a 3-of-a-kind is a hand in which 3 cards are of the same rank, and the two other cards are different ranks.
Let \(A\) be the event that a 3-of-a-kind is obtained.
Let \(B\) be the event that two of the cards are the same rank.
This problem can be restated as finding \(P(A\mid B)\).
\(A\cap B=A\), because two of the cards will always be the same rank when the player obtains a 3-of-a-kind.
Using the above formulas, this gives \(P(A\mid B)=\dfrac{|A\cap B|}{|B|}=\dfrac{|A|}{|B|}\)
When calculating \(|A|\), it is important to keep in mind that in a 3-of-a-kind hand of 5 cards, the other 2 cards are different ranks.
\(|A|=\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}\binom{4}{1}=54912\)
\(|B|\) can be calculated with the complement of the event that all cards are different.
\(|B|=\binom{52}{5}-\binom{13}{5}\binom{4}{1}^5=1281072\)
\(P(A\mid B)=\dfrac{54912}{1281072}\approx 0.042864\)