Two events are considered **dependent** if the occurrence or outcome of the first event *changes the probability* of the next event occurring. For an example, let’s consider the following two events:

A = there is a blizzard in New York City

B = a flight bound for JFK Airport is delayed

These two events are **dependent**. This is because if a blizzard hits New York City, there is a much *greater* probability that any given inbound bound to JFK Airport will be delayed than there would be under non-blizzard conditions.

To clarify dependent events further, we should differentiate them from their opposite—independent events. As you might be able to conclude from the names, two events are independent if the occurrence of one event has **no impact** on the probability of the next event occurring. A classic example would be the tossing of a fair coin twice in a row. Each toss is **independent** of the last; the fact that the first toss lands heads does not change the fact that on the next toss — there remains a 50% chance of getting heads and a 50% chance of getting tails.

For more on distinguishing between dependent and independent events, check out this page from Yale University’s Statistics course. Otherwise, we will further explore the concept of dependent events using an example with playing cards.

### Example: Drawing cards without replacement

A classic example involves drawing cards without replacement (meaning, drawing a card and then not putting it back into the deck). Let’s consider the following two events:

C = drawing a club

D = drawing a diamond

What is the probability of drawing a club, P(C), from a full standard deck of fifty-two cards? Since there are four suits of thirteen cards each, P(C) = 13/52, or 1/4. The same is true for the probability of drawing a diamond; P(D) = 1/4.

Now let’s say we pick a card, and it’s a club. What is the probability that the **next card** is a club? Also, what is the probability that the next card is a diamond?

Since we’ve already *removed* one card from the deck, there are only a total of 51 cards remaining. The new probabilities are as follows:

P(C) = 12/51 (because there are only 12 clubs left in the deck)

P(D) = 13/51 (there are still all 13 diamonds left, but there are only 51 cards remaining to choose from)

Thus, by drawing a club from the deck, we changed the probabilities P(C) and P(D). Thus, the two are **dependent** events.

Need more practice with dependent events? Check out our statistics lessons and videos!

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