Graphs, data and probability
Probability outcomes: lists, trees and fair comparisons
List possible outcomes, calculate probabilities, and compare replacement with no replacement. Six questions include complete explanations.
Grades 6–7 practice · 6 questions · Free to print · Answers included
Make sense of the method
An outcome is a possible result. For equally likely elementary outcomes, probability is the number of favourable outcomes divided by the total number of possible outcomes.
Different colour names are not automatically equally likely. For two-stage experiments, an organised list or probability tree helps track what can happen and whether the second stage changes.
- State the experiment and what makes the elementary outcomes equally likely, if that rule applies.
- List outcomes systematically, or draw a tree labelled with branch probabilities.
- Multiply probabilities along a path and add probabilities of separate favourable paths.
- Check whether an object is replaced: without replacement, both the total and some colour counts change.
A worked example
A fair spinner has four equal sectors labelled A, B, C and D. What is the probability of landing on A or D?
- The four equal sectors are equally likely elementary outcomes.
- Two sectors, A and D, are favourable.
- Probability = 2/4 = 1/2.
1/2.
Try the worksheet
Start with the first two questions, then build toward the final challenge. Show your method with calculations, drawings or a short explanation. These are practice questions with published answers, not a full-topic readiness check.
Build the foundation
1.A fair standard die has faces 1 to 6. List the outcomes greater than 4 and find their probability.
Show your work on paper.Build the foundation
2.A bag contains 3 orange, 2 purple and 1 white counter, identical except for colour. Mix well and choose one without looking. Find the probability of purple. Are the three colours equally likely? Explain.
Show your work on paper.Strengthen your method
3.Toss a fair coin and independently spin a fair three-sector spinner labelled 1, 2, 3. List all ordered outcomes and find the probability of heads with an odd spinner number.
Show your work on paper.Strengthen your method
4.A bag has 4 green and 2 yellow identical counters. Draw one at random, replace it, mix, then draw again. Draw a probability tree and find the probability of two green counters.
Show your work on paper.Apply and explain
5.Repeat the experiment with 4 green and 2 yellow counters, now WITHOUT replacement. Draw the green-first branches, find the probability of two greens, and compare it with 4/9 for replacement.
Show your work on paper.Apply and explain
6.Draw two counters without replacement from the same bag of 4 green and 2 yellow counters. Find the probability of one of each colour in either order. A class records one of each in 31 of 60 repeated trials, resetting the bag each trial. Compare the experimental proportion with the theoretical probability. Does a difference prove the model is wrong?
Show your work on paper.
Worked answers
Try the questions first. Then compare the reasoning, not just the final result.
Show all six worked solutions
1. Outcomes 5 and 6; probability 1/3.
- There are six equally likely faces.
- Two faces are greater than 4: 5 and 6.
- Probability = 2/6 = 1/3.
2. P(purple) = 1/3. The colours are not equally likely.
- There are 3 + 2 + 1 = 6 counters, each assumed equally likely to be selected.
- Purple has 2 favourable counters, so 2/6 = 1/3.
- Orange has probability 3/6 and white 1/6. Counting three colour names as three equal outcomes would be incorrect.
3. H1, H2, H3, T1, T2, T3; probability 1/3.
- Pair each coin result with all three spinner results to list six equally likely outcomes.
- H1 and H3 are favourable: heads and an odd number.
- Probability = 2/6 = 1/3, also (1/2) × (2/3).
4. 4/9.
- At each draw, green has probability 4/6 = 2/3 and yellow 2/6 = 1/3.
- Replacement restores the original counts, so each first branch has the same second-stage probabilities.
- The green–green path has probability (2/3) × (2/3) = 4/9.
5. 2/5, which is 2/45 less than 4/9.
- First green: 4/6. After a green is removed, 3 green and 2 yellow remain.
- On the green-first branch, second green is 3/5 and second yellow is 2/5.
- P(two greens) = (4/6) × (3/5) = 12/30 = 2/5.
- 4/9 − 2/5 = 20/45 − 18/45 = 2/45. Removing a green lowers the next green probability.
6. Theoretical probability 8/15; experimental proportion 31/60, which is 1/60 lower. The difference alone does not prove the model is wrong.
- Green then yellow: (4/6) × (2/5) = 4/15.
- Yellow then green: (2/6) × (4/5) = 4/15.
- The two orders are separate outcomes, so add: 4/15 + 4/15 = 8/15 = 32/60.
- Experimental proportion is 31/60, one trial out of 60 below 32/60.
- Random variation can produce a difference. Check the procedure and collect more trials; even a larger sample need not match theory exactly.
Mistakes worth catching
Assuming named categories are equally likely.
Count the equally likely objects or sectors behind the names.
Keeping the original probabilities after removing a counter.
Update both the available counts and the total on each branch.
Counting only one order when either order works.
List both paths and add their probabilities.
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Skills covered
Graphs, data and probability
- Organize two-stage outcomes using lists or trees.
- Compare dependent and independent events and theoretical and experimental results.
Choose this practice by the skills you need. Grade placement and strand names vary between school systems; one worksheet covers selected skills rather than every expectation in a topic.
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