D. Data · Data Literacy and ProbabilityTwo Independent Events: Theoretical Probability
D2.2 D2.1About 50 minutes
Learning goals- I can explain what makes two events independent.
- I can list all the outcomes of two independent events using a tree diagram or a table.
- I can find the theoretical probability of two events happening, as a fraction, decimal and percent.
Words to know
independent events
Two events where the result of one does not change the chances of the other, like flipping a coin and rolling a die.
theoretical probability
The probability you work out by thinking about the equally likely outcomes, without doing an experiment.
tree diagram
A diagram with branches that shows every possible outcome, step by step.
sample space
The list of all possible outcomes.
Remember
In Grade 5 you found the theoretical probability of one event, like spinning blue. In Grade 3 you sorted things with tree diagrams. Today you'll use tree diagrams and tables to find the probability of two events happening together.
Two things at once
Flip a coin and spin this spinner. What could happen?
The coin doesn't know what the spinner did, and the spinner doesn't know what the coin did. Getting heads doesn't make a 3 more or less likely. These are independent events.
To find all the outcomes, use a tree diagram. First draw a branch for each coin result (H, T). Then, from each of those, draw a branch for each spinner result (1, 2, 3). Follow each path to read one outcome.
There are 2 × 3 = 6 outcomes: H1, H2, H3, T1, T2, T3. They are all equally likely.
So P(heads and 3) = 1/6, because only 1 of the 6 outcomes (H3) is heads and 3.
Counting tip: when you do one thing with 2 outcomes and another with 3 outcomes, there are 2 × 3 = 6 combined outcomes. A coin and a die give 2 × 6 = 12.
Worked exampleExample 1: A coin and a die (using a table)
Flip a coin and roll a die. Find P(heads and an even number).
- Make a table: coin results down the side, die results across the top. Each box is one outcome. (See the table below.)
- There are 2 × 6 = 12 equally likely outcomes.
- Heads and even: H2, H4, H6. That's 3 outcomes.
- P = 3/12 = 1/4 = 0.25 = 25%.
P(heads and even) = 3/12 = 1/4 = 25%.
| 1 | 2 | 3 | 4 | 5 | 6 |
|---|
| H | H1 | H2 | H3 | H4 | H5 | H6 |
| T | T1 | T2 | T3 | T4 | T5 | T6 |
Worked exampleExample 2: Two dice
Roll two dice and add the numbers. What is P(a sum of 7)?
- The first die has 6 outcomes and the second has 6, so there are 6 × 6 = 36 equally likely outcomes.
- Make a table with each box showing the sum (see the table below).
- Find the 7s: 1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1. That's 6 boxes.
- P(sum of 7) = 6/36 = 1/6 ≈ 0.17 ≈ 17%.
P(sum of 7) = 6/36 = 1/6, about 17%.
| First die → / Second die ↓ | 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 |
Careful: count every outcome
Flip two coins. Many people think there are 3 outcomes: two heads, two tails, or one of each. But a tree diagram shows 4: HH, HT, TH, TT. "One of each" happens two ways (HT and TH), so P(one of each) = 2/4 = 50%, not 1/3.
Just for funCoins have no memory. After 5 heads in a row, a coin is not thinking "I'd better land on tails now." It isn't thinking at all. It's a coin.
Not independent
Some events are not independent. A drawer has 2 red socks and 2 blue socks. You pull out a red sock and don't put it back. Now there is only 1 red sock out of 3, so the chance of red on the second pull has changed. When the first result changes the chances for the second, the events are not independent. In this lesson we only work with independent events, like coins, dice and spinners, or picking something and putting it back.
Hands-on activityMystery lunch generator
You need: a coin, a die and paper.
- Make a menu: heads = sandwich, tails = wrap; die 1 = apple, 2 = orange, 3 = banana, 4 = grapes, 5 = carrots, 6 = cucumber.
- Draw a tree diagram or a table that shows every possible lunch. How many are there? (12)
- Find the probability of: a sandwich with a fruit (1 to 4); a wrap with a vegetable (5 or 6); a sandwich with an apple.
- Flip and roll to choose a pretend lunch. Ask a grown-up if you can actually make it together (a grown-up should do or supervise any cutting).
Your turn
1A spinner has 4 equal sections: A, B, C, D. You spin it and flip a coin. How many outcomes are there? List them.
8 outcomes: AH, AT, BH, BT, CH, CT, DH, DT. (4 × 2 = 8)
2Same spinner and coin: find P(A and heads).
3Same spinner and coin: find P(tails and a letter that is A or B).
2/8 = 1/4 = 25%. The outcomes are AT and BT.
4Coin and die table: find P(tails and a number greater than 4).
2/12 = 1/6, about 17%. The outcomes are T5 and T6.
5Two dice: find P(a sum of 12).
1/36, about 3%. Only 6 + 6.
6Two dice: find P(a sum of 2 or 3).
3/36 = 1/12, about 8%. Sum of 2: (1,1). Sum of 3: (1,2) and (2,1).
7Are "rolling a die" and "spinning a spinner" independent events? Explain.
Yes. What the die shows doesn't change the chances on the spinner.
More practice: warm-up
1Flip a coin and spin the 1-2-3 spinner. How many outcomes are there?
2Flip a coin and roll a die. How many outcomes?
3Flip two coins. List all the outcomes.
4Two coins: what is P(two heads)?
5What does independent mean?
The result of one event doesn't change the chances of the other.
More practice: on your own
1Coin and 1-2-3 spinner: find P(tails and an odd number).
2/6 = 1/3, about 33%. The outcomes are T1 and T3.
2Two dice: which sum is most likely? What is its probability?
7, with a probability of 6/36 = 1/6. It has the most boxes in the table.
3Two dice: find P(both dice show the same number).
6/36 = 1/6, about 17%. (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
4A drawer has 3 red socks and 3 blue socks. You take one and keep it, then take another. Are the two picks independent? Explain.
No. After the first sock is taken, there are only 5 left, and the number of that colour is smaller. The chances for the second pick change.
5You have 3 shirts and 4 pairs of pants. You pick one of each at random. How many different outfits are possible? What is P(one particular outfit)?
12 outfits. 3 × 4 = 12. P(one particular outfit) = 1/12, about 8%.
Challenge
1Flip three coins. How many outcomes are there? What is P(exactly two heads)?
8 outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT (2 × 2 × 2 = 8). Exactly two heads: HHT, HTH, THH. P = 3/8 = 0.375 = 37.5%.
2Spinner X has 2 equal sections (1, 2). Spinner Y has 5 equal sections (1 to 5). You spin both and multiply. What is P(the product is even)?
There are 2 × 5 = 10 outcomes. Products: 1×1=1, 1×2=2, 1×3=3, 1×4=4, 1×5=5, 2×1=2, 2×2=4, 2×3=6, 2×4=8, 2×5=10. Even products: 2, 4, 2, 4, 6, 8, 10 → 7. P = 7/10 = 70%.
3Notice: P(heads) = 1/2 and P(rolling a 6) = 1/6, and P(heads and 6) = 1/12. Check with the table, then explain why it makes sense that 1/12 is smaller than both 1/2 and 1/6.
From the table, only H6 is heads and 6: 1 of 12 outcomes. ✓ It makes sense because both things must happen. Only half of the times you roll a 6 will the coin also be heads, so the chance is half of 1/6, which is 1/12.
Quiz
1. Which pair of events is independent?
2. You flip a coin and roll a die. How many outcomes are there?
3. Two coins are flipped. What is P(one head and one tail)?
4. Coin and 1-2-3 spinner: what is P(H and 2)?
5. Two dice: what is P(sum of 7)?
6. A spinner with 4 equal sections and a coin: what is P(a particular section and tails), as a percent?
What this lesson teaches: finding the theoretical probability of two independent events happening, using tree diagrams and tables to list all the outcomes, and writing the probability as a fraction, decimal or percent (D2.2, D2.1). Two independent events are new in Grade 6.
How to help at home:
- Use real coins, dice and spinners. Ask your child to list every outcome before calculating.
- Board games with two dice are a great place to talk about why 7 comes up so often.
- Ask "Does the first result change the chances for the second?" to check independence.
Common difficulties:
- Missing outcomes (treating HT and TH as one outcome).
- Adding instead of multiplying to count outcomes (2 + 6 instead of 2 × 6).
- Forgetting that each outcome in the table must be equally likely.