Let \(A\) and \(B\) be events. The conditional probability is defined as

\[ \Pr(A | B) = \frac{ \Pr(A \cap B) }{ \Pr(B) } \]as long as \(Pr(B) \ne 0 \).

This conditional probability can be interpreted as the probability that A happens *assuming that we know that B is true*. In other words, this conditional probability is simply the probability of A given some extra information about B.

We normally refer \(\Pr(A | B)\) as *the probability of A given B*. This means, assuming that B is true, we need to compute the probability of A.

__Example:__ A study shows that if we choose a person randomly, the probability that the person will go out to a mall during the weekend is 0.74, the probability that the person will go to get some ice cream is 0.45, and the probability that the person will do both is 0.34. Find the probability that the person will get some ice cream *given* that she will go to mall.

** Answer:** Let's define the following events

This means that

\[\Pr(A | B) = \frac{\Pr(A \cap B)}{\Pr(B)} = \frac{\Pr(\text{The person goes to a mall and goes to eat ice cream})}{\Pr(\text{The person goes to a mall})}\] \[ = \frac{0.34}{0.74} = 0.459\]≫__Another way to use conditional probabilities__

The conditional probability formula can be written in the following very useful way:

\[ \Pr(A \cap B)= \Pr(A | B) Pr(B)\]This formula makes some calculations really simple, as shown in the example below:

**Application Example:** An urn contains 8 black balls and 4 white balls. Two balls are taken from the urn without replacement. Compute the probability that both balls are white.

** Answer:** This problem can be tricky without the proper preliminaries. First, we define the following events:

We need to compute the probability that both balls are white, which means that the need to compute \(\Pr (A \cap B) \). Using the last formula for the conditional probability:

\[\Pr(A \cap B)= \Pr(A | B) Pr(B) = \frac{3}{11}\times \frac{4}{12} = \frac{1}{11} = 0.0909\](Notice that if the first ball is white, then there are only 11 balls left: 3 white balls and 8 black balls)

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