A tray has 10 eggs, of which 3 are cracked. Three eggs are taken one by one without replacement. Find the probability that (i) none of them is cracked, (ii) only the third egg is cracked.

1. 1 · Write what is given and what is to be found. Given: 10 eggs, 3 cracked and 7 good, taken without replacement. To find: two probabilities.
2. 2 · Write the theorem. Multiplication theorem: P(E ∩ F ∩ G) = P(E) · P(F|E) · P(G|(E ∩ F)).
3. 3 · Work out (i).
   • P(1st good) = 7/10, P(2nd good | 1st good) = 6/9, P(3rd good | first two good) = 5/8
   • P = 7/10 × 6/9 × 5/8 = 210/720 = 7/24
4. 4 · Work out (ii).
   • Good, good, then cracked: 7/10 × 6/9 × 3/8
   • = 126/720 = 7/40
5. 5 · Say what it means. A shopkeeper who picks 3 eggs blindly from such a tray gets all good eggs only about 7 times in 24.
6. 6 · Check and box the answer. The denominators fall 10, 9, 8 because nothing is put back. (i) 7/24 (ii) 7/40.

Scoring tip: Write each factor with its meaning, such as 'P(2nd good | 1st good) = 6/9'. That shows the method.