Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — probability by counting
A committee of 3 is chosen at random from 6 men and 4 women. Find P(exactly two women).
1Total selections: (310)=120Order does not matter in a committee.
2Choose 2 women from 4: (24)=6
3Choose 1 man from 6: (16)=6
4The choices are made together, so multiply: 6×6=36'And' multiplies.
5P=12036=103Combinations top and bottom — consistent throughout.
Example 2 — conditional probability from a table
Of 110 candidates, 45 of the 60 women passed and 30 of the 50 men passed. Find P(passed∣female) and test for independence.
1'Given female' restricts attention to the 60 women.The condition becomes the new denominator.
2P(passed∣female)=6045=0.75
3Overall, P(passed)=11075=0.682
40.75=0.682, so the events are not independent.Equivalently P(A∩B)=0.409 but P(A)P(B)=0.372.
5Women passed at a higher rate than the group as a whole.Note P(female∣passed)=7545=0.6 — a different number entirely.
Example 3 — total probability and Bayes
Machines A, B and C make 50%, 30% and 20% of output with defect rates 2%, 3% and 5%. A defective item is found — what is the probability it came from A?
1P(A∩D)=0.5×0.02=0.010Multiply along the branch.
2P(B∩D)=0.3×0.03=0.009
3P(C∩D)=0.2×0.05=0.010
4P(D)=0.010+0.009+0.010=0.029Law of total probability — add between branches.
5P(A∣D)=0.0290.010=0.345The branch you want, over all branches ending in a defect.
6A is the most likely source despite having the lowest defect rate.Because it makes half of everything.