Independent and Dependent Events

Independent and Dependent Events — Moosa Academy

Two events are independent when the first has no effect on the second — like tossing a coin twice. To find the probability that both happen, you multiply their separate probabilities.

Independent The multiplication rule

If A and B are independent, the probability that both occur is the product of their probabilities.

P(A \cap B) = P(A) \times P(B)

Examples of independence: tossing a coin twice, rolling a die, or drawing from a box with replacement.

Multiplying one fraction by another always gives a smaller result — adding a second condition makes the combined event harder to achieve.

Example Two heads in a row

Each toss is heads with probability $\tfrac{1}{2}$ , and the tosses are independent:

P(H \cap H) = \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}

⟹ 25%.

Example Drawing with replacement

A box holds 3 balls. You draw one, put it back, and draw again. Replacing the ball keeps the two draws independent.

The probability of a yellow ball each time is

\tfrac{1}{3}

P = \tfrac{1}{3} \times \tfrac{1}{3} = \tfrac{1}{9}

⟹ ≈ 11.1%.

Note Dependent events

If you draw without replacing, the events become dependent — the first draw changes what is left, so the second probability changes too. You then multiply by the conditional probability $P(B \mid A)$ .

A bag has 5 balls, 2 of them red. Drawing two without replacement:

P(\text{red}, \text{red}) = \tfrac{2}{5} \times \tfrac{1}{4} = \tfrac{1}{10}

⟹ 10%.

Reference Independent events compared

Event	P(A)	P(B)	P(A ∩ B)
Coin — two heads	1/2	1/2	1/4 = 25%
Balls with replacement — yellow twice	1/3	1/3	1/9 ≈ 11.1%
Die — six twice	1/6	1/6	1/36 ≈ 2.8%

Summary

Independent events: the first does not affect the second — so you multiply.
Rule: $P(A \cap B) = P(A) \times P(B)$ ; the result is smaller than either probability.
Drawing with replacement keeps events independent; without replacement makes them dependent.
Every extra "and" condition lowers the overall probability.