Dependent Events and Conditional Probability

Dependent Events and Conditional Probability — Moosa Academy

Two events are dependent when the first changes the second. Drawing a ball and not replacing it leaves fewer balls, so the second probability shifts. We handle this with the conditional probability $P(B \mid A)$ — the probability of $B$ given that $A$ has happened.

Dependent With or without replacement

With replacement → independent: the count stays the same, so the first draw does not affect the second.

Without replacement → dependent: the count drops, so the first draw changes the second.

Quick test: does the first draw change how many items are left for the second? If yes, the events are dependent.

Formula Conditional probability

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

P(B \mid A)

is the probability of

B

given that

A

has already happened.

Dependent:

P(B \mid A)

changes after the first event.

Independent: it stays

P(B)

, unchanged.

Example Yellow then green, no replacement

Step 1 — 3 balls, so

P(\text{yellow}) = \tfrac{1}{3}

Step 2 — one ball gone, 2 left, so

P(\text{green} \mid \text{yellow}) = \tfrac{1}{2}

Step 3 —

\tfrac{1}{3} \times \tfrac{1}{2} = \tfrac{1}{6}

⟹ ≈ 16.7%.

Example Two more

The same draw, but with replacement

Replacing the ball keeps 3 balls, so the draws are independent.

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

— here, smaller than the

\tfrac{1}{6}

without replacement.

⟹ ≈ 11.1%.

Green then green again (no replacement)

There is only one green ball, so after drawing it none is left.

P(\text{green} \mid \text{green}) = 0

, so

\tfrac{1}{3} \times 0 = 0

⟹ impossible event, P = 0.

Reference With vs without replacement

Feature	With replacement	Without replacement
Balls for draw 2	stays the same	decreases by one
Effect of first draw	none	changes the second
Law	P(A) × P(B)	P(A) × P(B \| A)
Yellow then green	1/3 × 1/3 = 1/9	1/3 × 1/2 = 1/6

Summary

Dependent events: the first changes the probability of the second — use $P(B \mid A)$ , not $P(B)$ .
Law: $P(A \cap B) = P(A) \times P(B \mid A)$ .
Without replacement, the count drops after each draw, so the probabilities change at every step.
If an item is used up, $P(B \mid A) = 0$ and the whole result is 0 — an impossible event.