A straightforward calculation is to find the probability that a card drawn from a standard deck of cards is a king. There is a total of four kings out of 52 cards, and so the probability is simply 4/52. Related to this calculation is the following question: "What is the probability that we draw a king given that we have already drawn a card from the deck and it is an ace?" Here we consider the contents of the deck of cards. There are still four kings, but now there are only 51 cards in the deck. The probability of drawing a king given that an ace has already been drawn is 4/51.

This calculation is an example of conditional probability. Conditional probability is defined to be the probability of an event given that another event has occurred. If we name these events *A* and *B*, then we can talk about the probability of *A* given *B*. We could also refer to the probability of *A* dependent upon *B*.

### Notation

The notation for conditional probability varies from textbook to textbook. In all of the notations, the indication is that the probability we are referring to is dependent upon another event. One of the most common notations for the probability of *A* given *B* is *P( A | B )*. Another notation that is used is *P _{B}( A )*.

### Formula

There is a formula for conditional probability that connects this to the probability of *A* and *B*:

*P( A | B ) = P( A ∩ B ) / P( B )*

Essentially what this formula is saying is that to calculate the conditional probability of the event *A* given the event *B*, we change our sample space to consist of only the set *B*. In doing this, we don’t consider all of the even *A*, but only the part of *A* that is also contained in *B*. The set that we just described can be identified in more familiar terms as the intersection of *A* and *B*.

We can use algebra to express the above formula in a different way:

* P( A ∩ B ) = P( A | B ) P( B )*

### Example

We will revisit the example we started with in light of this information. We want to know the probability of drawing a king given that an ace has already been drawn. Thus the event *A* is that we draw a king. The event *B* is that we draw an ace.

The probability that both events happen and we draw an ace and then a king corresponds to P( A ∩ B ). The value of this probability is 12/2652. The probability of the event *B*, that we draw an ace is 4/52. Thus we use the conditional probability formula and see that the probability of drawing a king given than an ace has been drawn is (16/2652) / (4/52) = 4/51.

### Another Example

For another example, we will look at the probability experiment where we roll two dice. A question that we could ask is, “What is the probability that we have rolled a three, given that we have rolled a sum of less than six?”

Here the event *A* is that we have rolled a three, and the event *B* is that we have rolled a sum less than six. There are a total of 36 ways to roll two dice. Out of these 36 ways, we can roll a sum less than six in ten ways:

- 1 + 1 = 2
- 1 + 2 = 3
- 1 + 3 = 4
- 1 + 4 = 5
- 2 + 1 = 3
- 2 + 2 = 4
- 2 + 3 = 5
- 3 + 1 = 4
- 3 + 2 = 5
- 4 + 1 = 5

### Independent Events

There are some instances in which the conditional probability of *A* given the event *B* is equal to the probability of *A*. In this situation we say that the events *A* and *B* are independent of one another. The above formula becomes:

*P( A | B ) = P( A ) = P( A ∩ B ) / P( B ),*

and we recover the formula that for independent events the probability of both *A* and *B* is found by multiplying the probabilities of each of these events:

*P( A ∩ B ) = P( B ) P( A )*

When two events are independent, this means that one event has no effect on the other. Flipping one coin and then another is an example of independent events. One coin flip has no effect on the other.

### Cautions

Be very careful to identify which event depends upon the other. In general *P( A | B) * is not equal to *P( B | A)*. That is the probability of *A* given the event *B* is not the same as the probability of *B* given the event *A*.

In an example above we saw that in rolling two dice, the probability of rolling a three, given that we have rolled a sum of less than six was 4/10. On the other hand, what is the probability of rolling a sum less than six given that we have rolled a three? The probability of rolling a three and a sum less than six is 4/36. The probability of rolling at least one three is 11/36. So the conditional probability in this case is (4/36) / (11/36) = 4/11.