Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — filling in a Venn diagram
Given P(A)=0.5, P(B)=0.4 and P(A∩B)=0.2, find P(A∪B) and the probability of neither event.
1Fill in the intersection first: 0.2.Always start with the overlap.
2Only A: 0.5−0.2=0.3P(A) is the total for A, including the overlap.
3Only B: 0.4−0.2=0.2
4P(A∪B)=0.3+0.2+0.2=0.7Or by the addition rule: 0.5+0.4−0.2.
5Neither: 1−0.7=0.3The four regions total 1 ✓
Example 2 — a tree diagram without replacement
A bag holds 4 red and 6 blue balls. Two are drawn without replacement. Find P(both red) and P(one of each).
1First draw: P(R)=104, P(B)=106.
2After a red is removed, 9 balls remain with 3 red.Both numerator and denominator change.
3P(RR)=104×93=9012=152Multiply along the branch.
4P(RB)=104×96=9024 and P(BR)=106×94=9024
5P(one of each)=9048=158Add between branches.
6Check: 9012+24+24+30=1 ✓All four outcomes must total 1.
Example 3 — Bayes' theorem
A disease affects 1% of a population. A test detects it in 95% of sufferers and gives a false positive for 4% of non-sufferers. Someone tests positive — what is the probability they have the disease?
1P(D)=0.01, P(+∣D)=0.95, P(+∣D′)=0.04Write down what is given before reaching for the formula.
2True positives: (0.95)(0.01)=0.009595% of the 1% who have it.
3False positives: (0.04)(0.99)=0.03964% of the 99% who do not — far more people.
4P(+)=0.0095+0.0396=0.0491Law of total probability.
5P(D∣+)=0.04910.0095=0.193
6Fewer than one positive in five is genuine.The disease's rarity outweighs the test's accuracy.