SPM Mathematics: How to Master Probability of Combined Events

When two events must both happen (and), multiply their probabilities; when either one will do (or), add them. The catch is that adding only works cleanly when the events are mutually exclusive; if they can both happen you must subtract the overlap.

Deciding and versus or from the wording is the first move in almost every question.

Independent events do not affect each other (a coin does not remember its last toss), while dependent events do: drawing a second ball without replacement changes the second probability. A tree diagram lays the branches out so you multiply along a path and add between paths without guessing.

It is slower to draw but far safer under exam pressure.

Many combined-events questions involve drawing objects, balls from a bag, cards from a deck, and whether the first object is replaced before the second draw changes every probability that follows. With replacement, the total number of objects stays the same for each draw, so the probabilities on later branches of a tree diagram are identical to the first.

Without replacement, the object taken out is gone, so the total decreases by one and the count of whatever was removed also decreases by one, meaning the branches for the second draw look different depending on what happened on the first draw. This is the single most common source of lost marks in this topic: students copy the first draw's fractions onto the second draw without adjusting them.

Before drawing a tree diagram, decide and write down whether the situation is with or without replacement, then recalculate the denominator and relevant numerator for every branch after the first. For example, a bag containing 5 red and 3 blue balls has 8 balls in total.

Drawing two balls without replacement, the probability the first is red is 5/8; if the first ball was red, only 4 red and 3 blue balls remain out of 7, so the probability the second is also red becomes 4/7, not 5/8 again. This small numerical check, writing out how many of each colour are left and how many balls remain in total after each draw, is the fastest way to confirm your tree diagram is correct before doing any further calculation.

Questions that ask for the probability of "at least one" successful outcome across several trials are often easier to solve backwards. Calculating "at least one" directly usually means adding up several separate cases, exactly one, exactly two, and so on, which is slow and easy to get wrong.

Instead, find the probability that the event does not happen at all across every trial, then subtract that from 1. This works because "at least one" and "none at all" cover every possible outcome between them, so their probabilities must add to 1.

This shortcut is especially useful in combined-events questions involving three or more independent trials, where listing every successful combination by hand would take far too long for the time available. Recognising the phrase "at least one" as a signal to try the complement first, before attempting a direct calculation, saves time on exactly the questions where time is tightest.