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PROBABILITY LAB
Learn it by playing it
Four short missions on live bags, a spinner and a probability tree. Weigh the counters, expect the spinner, then chase two draws with nothing going back. When a mission glows green, read why it worked.
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Mission 1 of 4
Weigh the bag
The bag holds 3 red and 5 blue counters - 8 in total. One counter comes out at random. Dial in the probability it is red: the top dial is the numerator (ways to win), the bottom dial the denominator (all the ways).
dialled 0/1 · as a decimal 0
Before you touch the dials: what is P(red)?
Currently dialled 0/1 - count the red counters for the top, ALL the counters for the bottom.
Study notes
The probability scale and single events C8.1 / E8.1
Every probability lives on a scale from 0 to 1: 0 is impossible, 1 is certain, and 1/2 is an even chance. 'Very likely' means close to 1, never more than 1.
For equally likely outcomes, P(event) = favourable outcomes / total outcomes. A bag with 3 red and 5 blue counters gives P(red) = 3/8.
Give the answer as a fraction, decimal or percentage - all three earn the mark: 3/8 = 0.375 = 37.5%.
The complement rule: P(not A) = 1 - P(A). If P(blue) = 0.8, then P(not blue) = 0.2. Extended writes this as P(A') = 1 - P(A).
Questions can start from tables, graphs or Venn diagrams (two sets at most): count the matching region for the numerator and the whole total for the denominator.
A probability outside the 0 to 1 range, or a negative one, is a red flag that the counting went wrong - go back and recount.
Relative and expected frequencies C8.2 / E8.2
Relative frequency = times the outcome happened / total trials. It is an estimate of probability built from experiment, and it usually settles closer to the true value as trials pile up.
A spinner that lands on red 34 times in 100 spins gives an estimated P(red) = 34/100 = 0.34 - use that estimate for further predictions.
Expected frequency = probability x number of trials. With P(win) = 1/4 over 200 games, expect about 50 wins.
Expectation is an average, not a promise: getting 47 or 53 wins is normal wobble. Only a large, persistent gap suggests something is off.
Fair means every outcome has its theoretical chance; biased means it does not; random means every counter (or number) has an equal chance of being picked.
To test bias, compare a large number of trials against the theoretical expectation - a handful of spins proves nothing.
Combined events: sample space diagrams C8.3 / E8.3
A sample space diagram lists every paired outcome in a grid - two dice rolled together give a 6 by 6 table of 36 equally likely pairs.
Count from the grid: P(total of 7 with two dice) = 6/36 = 1/6, because 6 of the 36 cells sum to 7.
P(at least one six) = 11/36: count the top row and the last column once each - the (6, 6) corner belongs to both, so count it only once.
Every cell is equally likely only because the two dice are independent: what the first shows never changes what the second can show.
Venn diagrams work the same way for overlapping events: put the overlap in the middle, and Extended may ask with the notation P(A intersect B) for the overlap and P(A union B) for everything in either circle.
Tree diagrams C8.3 / E8.3
A tree diagram maps repeated events: each pair of branches from a point lists the outcomes with their probabilities, and each pair adds to 1.
Multiply ALONG the branches to get one path: P(red then blue) = P(red) x P(blue after red).
Add BETWEEN the paths that all count as success: P(exactly one red) = P(red then blue) + P(blue then red).
With replacement the bag resets, so both draws use the same fractions: P(two reds) = 3/5 x 3/5 = 9/25.
Without replacement the bag shrinks: after keeping a red, 4 counters and 2 reds remain, so P(two reds) = 3/5 x 2/4 = 6/20 = 3/10. Core questions stay with replacement; without replacement is Extended territory.
On the exam tree, outcomes are written at the ends of the branches and the probabilities sit beside the branches.
Conditional probability (Extended) E8.4
Conditional probability is a probability taken AFTER something has already happened: 'given that the first counter was red' shrinks the world to the cases where that is true.
On a tree, the second row of branches is already conditional - 2/4 after a red leaves is P(second red given first red).
From a two-way table, shrink the denominator: if 12 of 30 students study French and 5 of those 12 also study German, then P(German given French) = 5/12, not 5/30.
From a Venn diagram, work only inside the given set: the part of B inside A over the whole of A.
The notation P(A|B) and conditional formulas are NOT required - just read the restricted group directly from the tree, table or diagram.
Without-replacement questions are conditional in disguise: each kept-out counter changes every later branch.
EduTribe study notes - original material aligned to the Cambridge IGCSE Mathematics 0580 syllabus (2025-2027). Automated checks completed; not yet reviewed by an educator.