Conditional Probability and Tree Diagrams

1. 1 A bag has 5 red and 3 blue counters. P(red on first draw) = 5/8. Now draw without replacement. If first was red, bag now has 4 red and 3 blue. P(red second | red first) = 4/7. 'The probability changed because of what happened first — this is conditional probability.'
2. 2 P(A|B) = 'probability of A given B has occurred' = P(A∩B)/P(B). Example: in a class, 60% study maths, 40% study physics, 25% study both. P(physics|maths) = P(both)/P(maths) = 0.25/0.60 = 5/12. Students apply to two examples.
3. 3 Flip a coin twice. Both outcomes independent. P(H,H) = 1/2 × 1/2 = 1/4. Full tree: four branches, each with probability 1/4. 'For independent events, multiply along branches, add for alternatives.' Students draw the tree and verify all probabilities sum to 1.
4. 4 Draw two cards without replacement from a standard deck (52 cards). P(both aces). P(1st ace) = 4/52. P(2nd ace | 1st ace) = 3/51. P(both) = 4/52 × 3/51 = 12/2652 = 1/221. Students complete the full tree for (ace/not ace) × 2 draws.
5. 5 Draw a Venn diagram from a frequency table: 200 students, 120 play sport, 80 study music, 40 do both. Find P(music|sport) = 40/120 = 1/3. Compared to P(music) = 80/200 = 2/5. 'Are sport and music independent?' (No — P(music|sport) ≠ P(music).)
6. 6 Events A and B are independent iff P(A|B) = P(A) iff P(A∩B) = P(A)×P(B). Students test three pairs of events from given data to determine independence.
7. 7 A disease affects 1% of the population. A test is 95% accurate. If you test positive, what is the probability you have the disease? Students set up a tree diagram with realistic numbers (use 10,000 people as a reference population). Discuss why the answer surprises most people — and what it means for medical decision-making.

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