When do I multiply probabilities and when do I add them?

Multiply when you need two things to both happen, like heads on the first flip and heads on the second — that is the "and" situation, and multiplying makes the answer smaller. Add when you are combining separate favorable outcomes from your sample space, like the HT path plus the TH path for "exactly one head." A tree diagram makes this clear: multiply along a path, add across different paths.

How do I know whether a problem is with or without replacement?

Look for words about what happens to the first item. "Puts it back," "replaces it," "returns it," or spinning or flipping again all mean with replacement, so the fractions stay the same. "Keeps it," "eats it," "does not replace it," "gives it away," or "picks two at once" all mean without replacement, so the second draw has one fewer item in the total and one fewer of whatever type you removed.

Do I have to list every outcome if I can just multiply?

For the total count, multiplying is enough. But if the question asks about a specific condition — a sum, "at least one," "different colors" — you usually need the organized list, table, or tree to count the favorable outcomes correctly. A good habit is to multiply first to know how many outcomes to expect, then list them and check that your list has exactly that many.

Why do HT and TH count as two different outcomes?

Because the two flips happen at different times, so the results are in a specific order. "Heads then tails" and "tails then heads" are genuinely different ways the experiment can turn out, and each is equally likely. Merging them into one outcome shrinks your sample space from 4 to 3 and makes every probability wrong. The same idea applies to two dice: (2,5) and (5,2) are separate cells in the table.

M7MATH-10.4

Compound Events & Counting Outcomes

Learn to list sample spaces with organized lists, tables, and tree diagrams, multiply probabilities for independent events, and tell with-replacement from without-replacement draws.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Compound Events & Counting Outcomes, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Compound Events & Counting Outcomes free

What this lesson covers

So far you have found the probability of one thing happening: one coin flip, one spin, one card. But real situations usually pile events together. You flip a coin AND roll a die. You pick a shirt AND a pair of shoes. You draw two marbles from a bag. Events like these are called compound events, and the big challenge is simply keeping track of every possible result without missing any or counting one twice.

This lesson gives you three organizing tools — the organized list, the table, and the tree diagram — plus a shortcut for counting outcomes and a rule for multiplying probabilities. You will also learn the one distinction that trips up the most students: whether the first item goes back into the bag before the second draw. That single detail changes the second fraction, and therefore the whole answer.

Compound Events and the Sample Space

A compound event is an event made of two or more simple events happening together or one after another. Flipping a coin twice, rolling two dice, and choosing a sandwich and a drink are all compound events.

The sample space is the complete set of possible outcomes. For one coin flip the sample space is H, T — two outcomes. For two coin flips it is HH, HT, TH, TT — four outcomes. Notice that HT and TH are different outcomes: the first flip was heads and the second was tails, or the reverse. Students who write the sample space for two flips as HH, HT, TT are treating HT and TH as the same thing, and they end up with three outcomes instead of four. Order matters when you are listing outcomes of a compound event, even if the final question does not care about order.

Once you have the sample space, probability works exactly as it did for simple events:

P(\text{event}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}

So for two coin flips,

P(\text{exactly one head}) = \frac{2}{4} = \frac{1}{2}

, because HT and TH are the two favorable outcomes out of four.

The reason we bother with organized lists, tables, and tree diagrams is that they guarantee completeness. A random scribble of outcomes almost always misses one. An organized method walks through the possibilities in a fixed pattern, so nothing escapes.

Three Ways to Organize Outcomes

All three tools produce the same sample space. Pick the one that fits the situation.

An organized list works when you can hold one thing steady and vary the other. Rolling a die and flipping a coin: list all the die values with H first (1H, 2H, 3H, 4H, 5H, 6H), then all with T (1T, 2T, 3T, 4T, 5T, 6T). Twelve outcomes, and the pattern proves none are missing.

A table (sometimes called a grid) is ideal for exactly two events. Put one event's outcomes across the top and the other's down the side, then fill in each cell. Two dice make a 6-by-6 table with 36 cells, and it makes questions about sums easy to answer by counting cells along a diagonal.

A tree diagram handles two, three, or more stages. Each stage adds a new set of branches, and each complete path from left to right is one outcome.

Tool	Best for	Total outcomes shown as
Organized list	Small sample spaces, any number of stages	Items in the list
Table	Exactly two events	Cells in the grid
Tree diagram	Two or more stages, or changing probabilities	Complete paths

The tree diagram has one advantage the others lack: you can write a probability on every branch. That makes it the tool of choice when the second draw depends on the first, which is exactly what happens in without-replacement problems later in this lesson.

The Counting Principle and Multiplying Probabilities

Listing every outcome is reliable but slow. The Fundamental Counting Principle gives you the total instantly: if the first event has

m

outcomes and the second has

n

outcomes, the compound event has

m \times n

outcomes. A menu with 4 sandwiches and 3 drinks gives

4 \times 3 = 12

meals. Add a choice of 2 desserts and it becomes

4 \times 3 \times 2 = 24

.

Two events are independent when the outcome of the first does not change the probability of the second. Coins, dice, and spinners are independent — a coin has no memory of the last flip. For independent events,

P(A \text{ and } B) = P(A) \times P(B)

Example: rolling a 5 on a die and flipping heads.

P = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}

. Check it against the 12-outcome list: exactly one outcome is 5H, so

\frac{1}{12}

matches.

A very common mistake is adding the probabilities instead of multiplying. Ask yourself whether the answer should be bigger or smaller than the pieces. Requiring two things to both happen is harder than requiring just one, so the answer must be smaller than either individual probability. Since

\frac{1}{12}

is less than both

\frac{1}{6}

and

\frac{1}{2}

, multiplying was right. Adding would have given

\frac{2}{3}

, which is larger than both — an immediate signal that something went wrong.

When a question asks for two or more favorable outcomes ("at least one head"), multiply within each path and then add the paths that qualify.

With Replacement Versus Without Replacement

A bag holds 4 green and 6 yellow tiles, 10 tiles total. You draw two.

With replacement means you put the first tile back and mix before drawing again. The bag is identical both times, so the draws are independent:

P(\text{green, green}) = \frac{4}{10} \times \frac{4}{10} = \frac{16}{100} = \frac{4}{25}

Without replacement means you keep the first tile. Now the second draw happens in a changed bag — one fewer green and one fewer tile overall. The draws are dependent:

P(\text{green, green}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}

Both numerator and denominator drop by one, and only for the color you already removed. If the first tile were yellow, the second draw would be

\frac{4}{9}

green, because the green count did not change but the total did.

	Total for 2nd draw	Favorable for 2nd draw	Independent?
With replacement	10	4	Yes
Without replacement	9	3 (if 1st was green)	No

Watch the wording. Phrases like "puts it back," "replaces it," and "spins again" signal replacement. Phrases like "keeps it," "eats it," "does not replace," "picks two at the same time," and "hands one to a friend" all mean without replacement. Grabbing two tiles simultaneously counts as without replacement, because you cannot get the same physical tile twice.

Where Students Actually Go Wrong

Four errors account for most wrong answers on this topic.

First, an incomplete sample space. If you list outcomes in random order, you will drop one. Always sweep systematically: fix the first stage, run through all of the second stage, then move on.

Second, miscounting favorable outcomes. "At least one head" in two flips means HH, HT, and TH — three outcomes, not one. "Exactly one head" means only HT and TH — two outcomes. Read the words "at least," "exactly," and "both" very carefully, and highlight the qualifying outcomes in your list before you write the fraction.

Third, forgetting to change the denominator on a without-replacement second draw. Writing

\frac{4}{10} \times \frac{3}{10}

mixes the two ideas together and is not a valid answer for either situation.

Fourth, reducing too early. If a problem asks for the probability out of the total number of outcomes, keep the unreduced form long enough to check it against your list or table, then simplify at the end.

One more habit worth building: sanity-check the size of your answer. Any probability must land between 0 and 1. If you multiply two proper fractions and get something bigger than either one, you made an arithmetic slip. And if a question asks about a compound event with many stages, the probability of one specific full path gets small fast —

\frac{1}{2}

for one flip becomes

\frac{1}{32}

for five specific flips.

Key terms

Compound event.: An event consisting of two or more simple events, such as flipping a coin and rolling a die, or drawing two marbles.
Sample space.: The complete set of all possible outcomes of an experiment. For two coin flips it is HH, HT, TH, TT.
Tree diagram.: A branching diagram in which each stage of an experiment adds a level of branches; each complete path from start to end is one outcome, and probabilities can be written on the branches.
Fundamental Counting Principle.: If one event has $m$ outcomes and a second has $n$ outcomes, the compound event has $m \times n$ total outcomes.
Independent events.: Events for which the outcome of one does not change the probability of the other, so $P(A \text{ and } B) = P(A) \times P(B)$ .
Dependent events.: Events for which the first outcome changes the probability of the second, as in drawing without replacement.
With replacement.: The selected item is returned before the next selection, so the total and the favorable counts stay the same on every draw.
Without replacement.: The selected item is kept out, so the total drops by one and the count of that item's type drops by one for the next draw.

Worked example

A bag contains 5 red counters and 3 blue counters. Maya draws one counter, does not put it back, then draws a second counter. (a) Draw a tree diagram and label each branch with its probability. (b) Find the probability that both counters are red. (c) Find the probability that the two counters are different colors. (d) How would the answer to part (b) change if Maya replaced the first counter?

Start with the totals:

5 + 3 = 8

counters.

(a) The first stage has two branches.

P(\text{red}) = \frac{5}{8}

and

P(\text{blue}) = \frac{3}{8}

. Each of those splits into two second-stage branches, and because the first counter is kept out, only 7 counters remain. Down the red branch, 4 red and 3 blue are left, so the second-stage branches are

\frac{4}{7}

red and

\frac{3}{7}

blue. Down the blue branch, 5 red and 2 blue are left, so the branches are

\frac{5}{7}

red and

\frac{2}{7}

blue. Four complete paths: RR, RB, BR, BB.

(b) Multiply along the RR path:

P(\text{RR}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}

P(\text{RB}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}, \qquad P(\text{BR}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}

P(\text{different}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}

Quick check: the fourth path is

P(\text{BB}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}

. All four paths add to

\frac{20 + 15 + 15 + 6}{56} = \frac{56}{56} = 1

, so the diagram is complete.

(d) With replacement, the bag is back to 8 counters with 5 red for the second draw, so the events are independent:

P(\text{RR}) = \frac{5}{8} \times \frac{5}{8} = \frac{25}{64}

That is larger than

\frac{5}{14}

, which makes sense: replacing a red counter leaves more red available for the second draw.

Practice questions

A spinner has 4 equal sections labeled 1, 2, 3, 4. You spin it once and flip a fair coin. How many outcomes are in the sample space, and what is the probability of spinning a 3 and getting tails?

6 outcomes, probability $\frac{1}{6}$
8 outcomes, probability $\frac{1}{8}$
8 outcomes, probability $\frac{3}{8}$
6 outcomes, probability $\frac{5}{8}$

Answer: 8 outcomes, probability $\frac{1}{8}$

By the Counting Principle,

4 \times 2 = 8

outcomes: 1H, 2H, 3H, 4H, 1T, 2T, 3T, 4T. Adding 4 and 2 to get 6 is the most common error — you multiply the numbers of outcomes, you do not add them. Only one of the 8 outcomes is 3T, so the probability is

\frac{1}{8}

, which also equals

\frac{1}{4} \times \frac{1}{2}

from the independent-events rule.

A drawer holds 6 black socks and 4 white socks. Jordan reaches in and pulls out two socks at once, without looking. Find the probability that both socks are black. Then explain why this is a without-replacement problem even though the socks are grabbed at the same time.

Answer: $\frac{1}{3}$ , because grabbing two at once means the same sock cannot be chosen twice.

Treat the simultaneous grab as two draws in a row with no replacement. There are 10 socks, 6 black, so the first sock is black with probability

\frac{6}{10}

. After one black sock is out, 5 black remain among 9 socks, so the second is black with probability

\frac{5}{9}

. Multiply:

\frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}

. The reason it is a without-replacement situation is physical: a single sock cannot appear in both positions of the pair, so the second selection comes from a smaller pool. Using

\frac{6}{10} \times \frac{6}{10} = \frac{9}{25}

would wrongly allow the same sock twice.

Two fair six-sided dice are rolled. Using a table of the 36 outcomes, find the probability that the sum is 9 or greater.

Answer: $\frac{10}{36} = \frac{5}{18}$

Build the 6-by-6 table and count cells whose sum is at least 9. Sum 9: (3,6), (4,5), (5,4), (6,3) — 4 cells. Sum 10: (4,6), (5,5), (6,4) — 3 cells. Sum 11: (5,6), (6,5) — 2 cells. Sum 12: (6,6) — 1 cell. That is

4 + 3 + 2 + 1 = 10

favorable outcomes out of 36, or

\frac{5}{18}

. Notice that (3,6) and (6,3) count separately because the dice are distinguishable; treating them as one outcome is why some students get

\frac{6}{36}

instead.

FAQ

When do I multiply probabilities and when do I add them?: Multiply when you need two things to both happen, like heads on the first flip and heads on the second — that is the "and" situation, and multiplying makes the answer smaller. Add when you are combining separate favorable outcomes from your sample space, like the HT path plus the TH path for "exactly one head." A tree diagram makes this clear: multiply along a path, add across different paths.
How do I know whether a problem is with or without replacement?: Look for words about what happens to the first item. "Puts it back," "replaces it," "returns it," or spinning or flipping again all mean with replacement, so the fractions stay the same. "Keeps it," "eats it," "does not replace it," "gives it away," or "picks two at once" all mean without replacement, so the second draw has one fewer item in the total and one fewer of whatever type you removed.
Do I have to list every outcome if I can just multiply?: For the total count, multiplying is enough. But if the question asks about a specific condition — a sum, "at least one," "different colors" — you usually need the organized list, table, or tree to count the favorable outcomes correctly. A good habit is to multiply first to know how many outcomes to expect, then list them and check that your list has exactly that many.
Why do HT and TH count as two different outcomes?: Because the two flips happen at different times, so the results are in a specific order. "Heads then tails" and "tails then heads" are genuinely different ways the experiment can turn out, and each is equally likely. Merging them into one outcome shrinks your sample space from 4 to 3 and makes every probability wrong. The same idea applies to two dice: (2,5) and (5,2) are separate cells in the table.

Learn this with a teacher, not a page

The Crimsora tutor teaches Compound Events & Counting Outcomes live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

Try a live session free