Imagine reaching into a bag of sweets, eating the first one, then reaching in again. The second grab isn't the same as the first — there's one fewer sweet, and one fewer of whatever flavour you just ate. Sometimes one event changes the probability of the next; we call these dependent events, and the classic case is picking objects without putting them back ("without replacement").
With vs without replacement
The whole chapter hinges on one distinction — does the item go back?
- With replacement: you put the item back, so the totals and chances stay identical every time. The events are independent — the second pick has no memory of the first.
- Without replacement: you keep the item, so both the count and the total drop for the next pick. The events are dependent — what happened first genuinely changes what can happen next.
| With replacement | Without replacement | |
|---|---|---|
| Item goes back? | yes | no |
| Total each pick | stays the same | drops by 1 |
| Events are | independent | dependent |
• First red: 3 of 5 → 3/5.
• One red is now gone, so only 2 red remain out of 4 balls → second red = 2/4.
• Multiply along (the AND rule): 3/5 × 2/4 = 6/20 = 3/10.
Notice how both parts of the second fraction changed: the top went 3 → 2 (one red used up) and the bottom went 5 → 4 (one ball fewer altogether). If you'd done this with replacement instead, it would be 3/5 × 3/5 = 9/25 — a different answer, because the ball came back.
Tree diagrams still help
The tree-diagram method carries straight over, with one twist: the second set of branches changes depending on what happened first. Draw the first pick's branches, then draw a fresh second layer for each outcome using the updated counts. As always, multiply along each path and add the paths that count as a success.
In a nutshell: dependent events change the next probability. For "without replacement", reduce both the count and the total before the next pick, then multiply along the branches — and add across the branches when more than one path wins.