Probability of Dependent Events

Dependent events occur when the outcome of one event changes the probabilities available for the next. This lesson explains how removing an item without replacing it shifts the counts a later draw depends on, and shows how to chain several dependent draws together by multiplying updated probabilities stage by stage.

What you will learn

• Define conditional probability informally
• Compute probabilities for events without replacement
• Chain dependent probabilities across multiple draws

Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.

When one outcome changes the next

Two events are dependent when knowing that one of them happened changes what is possible, or how likely it is, for the other. The clearest example is drawing items from a shared pool without putting anything back: once one item is removed, both the total number of items and the number of items in its category shrink for the next draw.

Contrast this with the independent situations from the previous lesson, where an item is replaced before the next draw, restoring the original counts. Whenever a problem describes drawing several objects from the same fixed group and never mentions returning them, treat every draw after the first as depending on what has already been removed.

Computing probabilities without replacement

To find the probability of a second draw given the first draw's outcome, update the counts before computing a new fraction: subtract 1 from the total number of items, and subtract 1 from the count of whichever category the first item belonged to. If the first item drawn shares a category with what the second draw is looking for, that category's count also drops by 1; otherwise only the total drops.

For example, from a bag of 5 red and 3 blue marbles, drawing a red marble first leaves 4 red and 3 blue among 7 remaining marbles, so the next draw's probability of red becomes $\dfrac{4}{7}$ instead of the original $\dfrac{5}{8}$.

Chaining multiple dependent draws

When a problem asks for the probability of a specific sequence of several dependent draws, multiply the stage-by-stage probabilities in order, updating the counts after each stage before computing the next fraction. This mirrors the multiplication principle used for counting arrangements, except each factor is itself a probability computed from the pool as it exists at that stage.

A useful way to check a chained answer is to imagine every individual object as distinguishable and count favorable ordered sequences directly, then divide by the total number of ordered sequences; this direct count should match the product of the stage-by-stage probabilities.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Reusing the original total or original category count for a later draw instead of updating both after each item is removed.
• Treating draws without replacement as independent and multiplying the first-stage probability by itself.
• Forgetting that removing an item of the same category as the next draw changes both the numerator and denominator for that stage, while removing an item of a different category only changes the denominator.

Try with support

Guided practice

Choose a strategy, take a first step, and use the hints when you need direction.

Put it into practice

Independent practice

Work on your own first. If you reveal a solution, work through it and mark it reviewed.

Bring it together

• Dependent events occur when the outcome of one draw changes the probabilities available for the next.
• Sampling without replacement updates both the total count and the relevant category count at each stage.
• Chain dependent probabilities by multiplying each stage's updated probability in sequence, and check the result against a direct count of ordered outcomes.