Independent vs. Dependent Events

April 13, 2026

Problem

A coin flip and a die roll happen together. Show P(heads AND 6) = P(heads) × P(6) = 1/12. Then compare to drawing two cards without replacement to see how dependence changes the rule.

Explanation

What does "independent" mean?

Two events are independent when the outcome of one has no effect on the other. If I flip a coin and roll a die at the same time, the coin has no idea what the die did — they are independent.

Two events are dependent when the first outcome changes the probability of the second. Drawing a card from a deck and not putting it back changes what is left for the second draw — that is dependence.

The rule for each case

  • Independent events: P(A and B)=P(A)P(B)P(A \text{ and } B) = P(A) \cdot P(B)
  • Dependent events: P(A and B)=P(A)P(BA)P(A \text{ and } B) = P(A) \cdot P(B \mid A), where P(BA)P(B \mid A) is the probability of BB given that AA has already happened.

The dependent rule reduces to the independent rule when P(BA)=P(B)P(B \mid A) = P(B), i.e. when AA did not affect BB.

Worked example — independent

Event A: flip a coin, get heads. P(A)=12P(A) = \dfrac{1}{2}. Event B: roll a die, get a 6. P(B)=16P(B) = \dfrac{1}{6}.

Step 1 — Confirm independence: Flipping the coin cannot influence the die. ✓

Step 2 — Multiply: P(A and B)=1216=112P(A \text{ and } B) = \tfrac{1}{2} \cdot \tfrac{1}{6} = \boxed{\tfrac{1}{12}}

Step 3 — Sanity check by counting: There are 2×6=122 \times 6 = 12 equally likely combined outcomes. Only one of them is (H, 6). So P=1/12P = 1/12. ✓

Worked example — dependent

Draw two cards from a standard 52-card deck without replacement. What is P(both aces)P(\text{both aces})?

Step 1 — First ace: 4 aces out of 52 cards. P(A)=452=113P(A) = \dfrac{4}{52} = \dfrac{1}{13}.

Step 2 — Second ace given first was an ace: Only 3 aces remain in 51 cards. P(BA)=351=117P(B \mid A) = \dfrac{3}{51} = \dfrac{1}{17}.

Step 3 — Multiply: P(both aces)=113117=12210.0045P(\text{both aces}) = \tfrac{1}{13} \cdot \tfrac{1}{17} = \tfrac{1}{221} \approx 0.0045

Common mistakes

  • Assuming events are independent without checking. Always ask: "Does the first event change what is possible or change the odds for the second?"
  • Using P(A)P(B)P(A) \cdot P(B) when there is no replacement. If items are removed, the denominator shrinks and P(BA)P(B)P(B \mid A) \ne P(B).
  • Confusing "mutually exclusive" with "independent." They are different: mutually exclusive means both can't happen at once (P(A and B)=0P(A \text{ and } B) = 0); independent means they don't influence each other.

Try it in the visualization

Toggle between the two scenarios and watch the probability tree update live. You can adjust how many aces remain or how many faces the die has, and the product rule is recomputed.

Interactive Visualization

Parameters

Independent: coin + die
0.50
6.00
6.00
52.00
4.00
Your turn

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