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Two things happening together, counted with a list, a table or a tree.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you find the probability of two or more things happening together. You write down every outcome in an organized list, a table or a tree diagram, then count the ones you want. You will also see why, for independent events, you can simply multiply the probabilities, and how a simulation estimates a chance that is too hard to count.
You know that a probability is a number from 0 to 1. When all outcomes are equally likely, you find it by counting: favorable outcomes over all outcomes. A fair number cube gives $P(4) = \frac{1}{6}$. You can also multiply fractions, such as $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$. In this lesson you put those two skills together to find the chance of two or more things happening at once.
| Term | What it means |
|---|---|
| Compound event | An event made of two or more simpler events, such as flipping heads and rolling a 6. |
| Sample space | The list of every possible outcome of an experiment. |
| Outcome table | A grid with one row for each result of the first event and one column for each result of the second. Each cell is one outcome. |
| Tree diagram | A drawing that branches once for each stage of an experiment. Each path from start to end is one outcome. |
| Organized list | A list written in a fixed order so that no outcome is missed or written twice. |
| Counting principle | If one choice can be made in $a$ ways and a second in $b$ ways, the pair can be made in $a \times b$ ways. |
| Independent events | Events where the result of one does not change the probability of the other, such as two separate coin flips. |
| Simulation | An experiment with a simple tool, such as random numbers, that acts out a real situation many times. |
A compound event is two or more things happening together: heads on a coin and a 5 on a number cube, or two free throws both going in.
The key idea is simple. Write down every possible outcome first. Once you have the whole sample space, and the outcomes are equally likely, the probability is just counting:
$$P(\text{event}) = \frac{\text{number of favorable outcomes}}{\text{number of all outcomes}}$$
There are three good ways to write down the sample space:
All three give the same count. A coin (2 outcomes) and a number cube (6 outcomes) give $2 \times 6 = 12$ outcomes, however you draw them. This is the counting principle: multiply the number of outcomes at each stage.
Another way: picture
Picture a table with the coin's two sides down the left, H and T, and the numbers 1 to 6 across the top. There are 2 rows of 6 cells, so 12 cells in all. The event heads and an even number is the three cells H2, H4 and H6, so its probability is $\frac{3}{12} = \frac{1}{4}$.
Another way: shortcut
When the stages do not affect each other, you can skip the table and multiply probabilities: $P(\text{H}) \times P(\text{even}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. The table shows why this works: half of the rows, and half of the cells in each of those rows.
An organized list works well when there are only a few outcomes. The trick is to hold the first choice still while you run through every second choice, then move to the next first choice.
Suppose a sandwich shop offers turkey, ham or veggie on white or wheat bread. Hold turkey still: turkey-white, turkey-wheat. Then ham: ham-white, ham-wheat. Then veggie: veggie-white, veggie-wheat. That is 6 sandwiches, and $3 \times 2 = 6$ agrees.
If you write outcomes in a random order, it is easy to miss one or to write one twice. A fixed order is what makes the list trustworthy.
When there are exactly two stages, a table is often the clearest. Here are the sums when two number cubes are rolled, first cube down the side and second across the top.
| + | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
There are 36 cells, and each one is equally likely. A sum of 12 happens in only one cell, so $P(12) = \frac{1}{36}$. A sum of 7 happens in six cells, along the diagonal, so $P(7) = \frac{6}{36} = \frac{1}{6}$. That is why 7 is the most common sum in board games that use two cubes.
Notice that a 3 on the first cube and a 5 on the second is a different cell from a 5 on the first and a 3 on the second. Both count.
A table has only two directions, so for three or more stages a tree diagram is better. Start at a point. Draw one branch for each outcome of the first stage. From the end of every branch, draw one branch for each outcome of the second stage, and so on.
For three coin flips, the tree splits into 2 branches, then 4, then 8. Reading along each path gives the 8 outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. The number of paths is $2 \times 2 \times 2 = 8$, the counting principle again.
Trees are also useful when the branches have different probabilities. Write the probability on each branch. To get the probability of a whole path, multiply the numbers along it. The probabilities of all the paths add to 1.
For independent events, the probability that both happen is the product of their probabilities:
$$P(A \text{ and } B) = P(A) \times P(B)$$
Rolling a 6 and flipping heads: $\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$. Drawing a red marble, putting it back, and drawing red again from a jar that is 40% red: $0.4 \times 0.4 = 0.16$.
The shortcut needs independence. If the first marble is not put back, the jar changes, and the second probability is different. Then you have to think about what is left in the jar, or go back to listing outcomes.
Also notice that the product is always smaller than either probability, because multiplying by a number less than 1 shrinks a number. Two things together are harder to get than either one alone.
Some compound events are too messy to count. Suppose a cereal company puts one of 4 different toys in each box, each equally likely. What is the chance of getting all 4 toys in the first 6 boxes? Listing $4^6 = 4{,}096$ outcomes by hand is not practical.
Instead you can simulate. Let the digits 1, 2, 3 and 4 stand for the toys. Use a random number generator to make a string of 6 digits from 1 to 4, which is one trial, and check whether all four digits appear. Repeat 50 or 100 times. If all four toys appear in 38 out of 100 trials, the estimate is about 0.38. More trials give a better estimate, just as in the last lesson.
The simulation must match the real situation: each digit must be equally likely because each toy is, and each trial must use 6 digits because there are 6 boxes.
Every problem in this lesson follows the same plan.
Each step has a reason. Counting the whole sample space first means nothing is missed. Marking outcomes one by one keeps you honest when the event is something like a sum of 8, where many different pairs work.
How to check. The total number of outcomes should equal the product of the stage counts. The probability must be between 0 and 1. The probability of A and B should be smaller than the probability of A alone. If you can, work the problem both ways, by counting and by multiplying: the two answers must agree.
A small luggage lock has 3 dials, each with the digits 0 to 9. Each dial is a stage with 10 outcomes, so the lock has $10 \times 10 \times 10 = 1{,}000$ possible codes, from 000 to 999. A stranger who guesses one code has a probability of $\frac{1}{1{,}000} = 0.001$ of opening it. A bike lock with 4 dials has $10^4 = 10{,}000$ codes, so one guess has a chance of only 0.0001. Adding one dial makes the lock ten times harder to guess, because every old code now splits into ten new ones. The same counting principle explains why a 4-digit phone passcode has 10,000 possibilities and a 6-digit one has 1,000,000.
In the 1850s and 1860s, Gregor Mendel grew pea plants. Each plant carries two copies of the gene for flower color, and passes one of them, chosen at random, to each seed. Call the purple copy P and the white copy p. A plant with one of each, Pp, has purple flowers, because P hides p. When two Pp plants are crossed, the outcome table has 2 rows and 2 columns: PP, Pp, pP and pp. Only pp gives a white flower, so $P(\text{white}) = \frac{1}{4}$ and $P(\text{purple}) = \frac{3}{4}$. Mendel counted 705 purple and 224 white plants in one such experiment. That is $224 \div 929 \approx 0.24$ white, very close to the predicted 0.25. Doctors use the same table, called a Punnett square, to explain the chance that a child inherits a family trait.
The most common mistake is adding instead of multiplying to count outcomes. A coin and a number cube do not give $2 + 6 = 8$ outcomes. Each of the 2 sides goes with each of the 6 numbers, so there are 12.
The second is treating different orders as the same outcome. Rolling a 1 and a 2 can happen two ways, (1, 2) and (2, 1). If you count it once, sums like 3 look less likely than they really are.
The third is adding probabilities for 'and'. The chance of red on two spins of a spinner that is half red is not $\frac{1}{2} + \frac{1}{2} = 1$. Red twice is far from certain. It is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
The last is using the multiplication shortcut when the events are not independent, for example when a marble is not put back. Then the second probability must be worked out from what is left.
A coin is flipped and a spinner with equal red, blue and yellow sections is spun. Name the stages and count each one.
$\text{coin}: 2, \quad \text{spinner}: 3$
Each stage has equally likely outcomes, so counting works.
Write the organized list, holding heads still first.
$\text{HR, HB, HY, TR, TB, TY}$
Running through every color for each side misses nothing.
Check the count with the counting principle.
$2 \times 3 = 6$
The list has 6 outcomes, which matches the product.
Mark the outcomes for heads and blue.
$\text{HB}: 1 \text{ outcome}$
Only one outcome has both heads and blue.
Divide favorable by total.
$P(\text{H and blue}) = \tfrac{1}{6}$
One of six equally likely outcomes; also $\tfrac{1}{2} \times \tfrac{1}{3} = \tfrac{1}{6}$.
Two number cubes are rolled. Count all the outcomes.
$6 \times 6 = 36$
The table has 6 rows and 6 columns.
List the pairs whose sum is 8, first cube first.
$(2,6), (3,5), (4,4), (5,3), (6,2)$
The first cube must be at least 2, or the second would need more than 6.
Check that the pairs really are different outcomes.
$(3,5) \ne (5,3)$
A 3 then a 5 is a different cell of the table from a 5 then a 3.
Count the favorable outcomes.
$5 \text{ outcomes}$
Five cells of the table show a sum of 8.
Divide favorable by total.
$P(\text{sum} = 8) = \tfrac{5}{36}$
All 36 cells are equally likely.
Check it against a nearby sum.
$P(\text{sum} = 7) = \tfrac{6}{36} > \tfrac{5}{36}$
Seven has one more pair than 8, so it should be a little more likely, and it is.
Leo guesses on three true-or-false questions. Name the stages.
$\text{Q1}, \; \text{Q2}, \; \text{Q3}: \text{right (R) or wrong (W)}$
Each question is a stage with two equally likely results when he guesses.
Draw the first level of the tree.
$\text{R}, \; \text{W}: 2 \text{ branches}$
One branch for each result of question 1.
Draw the second level from each branch.
$\text{RR, RW, WR, WW}: 4 \text{ paths}$
Every result of question 1 is followed by both results of question 2.
Draw the third level.
$2 \times 2 \times 2 = 8 \text{ paths}$
Every path splits in two again.
List all eight outcomes.
$\text{RRR, RRW, RWR, RWW, WRR, WRW, WWR, WWW}$
Reading each path from start to end gives one outcome.
Mark the outcomes with exactly two right.
$\text{RRW, RWR, WRR}$
The one wrong answer can be on question 3, 2 or 1.
Divide favorable by total.
$P(\text{exactly 2 right}) = \tfrac{3}{8}$
Three of the eight equally likely paths.
Check that the whole model adds to 1.
$\tfrac{1}{8} + \tfrac{3}{8} + \tfrac{3}{8} + \tfrac{1}{8} = 1$
The chances of 0, 1, 2 and 3 right cover every path exactly once.
Two players each choose rock, paper or scissors at random. Count the outcomes.
$3 \times 3 = 9$
Each of the first player's 3 choices meets each of the second player's 3.
Count the ties.
Divide favorable by total.
A number cube is rolled and a spinner with $7$ equal sections is spun. How many outcomes should a complete outcome table show?
A coin is flipped and a spinner with $11$ equal sections, numbered 1 to $11$, is spun. Complete the worked solution for the probability of heads and a number greater than $3$.
Count the coin's outcomes.
$\text{heads, tails} \to 2$
A coin has two equally likely sides.
Multiply by the spinner's outcomes to count every pair.
$2 \times 11 =$ t
The outcome table has a row for each side and a column for each section.
In the heads row, count the numbers greater than $3$.
$4 \text{ up to } 11 \to$ f
Only heads cells can count, and only those with a big enough number.
Divide the favorable count by the total.
$P(\text{heads and more than } 3) =$ p
The pairs are equally likely, so the probability is favorable over total.
A spinner lands on red with probability $0.7$. It is spun twice. Which statement about the probability of red on both spins is true?
A food truck's lunch deal lets you pick one of its $5$ hot dog styles and one of its $9$ sides. How many different lunch deals are possible?
Answer:
A spinner has $7$ equal sections numbered 1 to $7$. A bag holds $8$ tiles, each with a different letter, and one of them is Q. You spin once and draw one tile. How many (number, letter) pairs are possible, and what is the probability of spinning a 1 and drawing Q? Give the probability as a fraction.
Pairs: n. $P(1 \text{ and Q}) =$ p
A jar holds $4$ green marbles and $6$ yellow marbles. You pick a marble without looking, put it back, shake the jar and pick again. How many marbles are in the jar, and what is the probability that both marbles are green? Give the probability as a fraction.
Marbles: p. Both green: b
A crosswalk signal shows WALK for $10$ seconds out of every $40$-second cycle. Jada reaches the crosswalk at a random moment on Monday and again on Tuesday. What is the probability of WALK on one arrival, and what is the probability that the signal shows WALK both times she arrives? Give fractions.
One arrival: p. Both days: b
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Mateo enters two separate contests. He has a $30\%$ chance of winning the art contest and a $35\%$ chance of winning the essay contest, and the results do not affect each other. What is the probability that he loses each contest, and what is the probability that he wins neither? Give decimals.
Loses art: p. Loses essay: q. Wins neither: n
You can find the probability of a compound event. Without looking: how many outcomes are there when you flip a coin and roll a number cube, and what is the probability of heads and a number less than 3?
17. Your turn: a tie in rock, paper, scissors, step 2
$\text{RR, PP, SS}: 3$
A tie happens when both players choose the same thing.
17. Your turn: a tie in rock, paper, scissors, step 3
$P(\text{tie}) = \tfrac{3}{9} = \tfrac{1}{3}$
Three of the nine equally likely outcomes are ties.