NUMERICAL — TIER 2
Seven topics worth real time, but after Tier 1 · Days 6–8
425 questions across these seven. Probability and P&C alone are 175 — bigger than any single Tier 1 topic except Number System. Do not treat "Tier 2" as "optional".
§1 · PROBABILITY — 93 questions
| Sub-type | Share | Priority |
|---|---|---|
| Cards | 27% | MUST |
| Dice | 25% | MUST |
| Balls / marbles from a bag | 18% | MUST |
| At least / at most (complement) | 5% | IMPORTANT — saves the most time |
| Conditional probability | — | WORTH KNOWING |
The only formula you need
P(event) = favourable outcomes / total outcomesEverything else is counting correctly. Probability questions are really combinatorics questions.
Cards — memorise the deck cold
52 cards · 4 suits × 13 · 26 red / 26 black
Suits: hearts + diamonds (red), spades + clubs (black)
Face cards: J, Q, K → 12 total (3 per suit)
Aces: 4 Number cards 2–10: 36Most card questions are one C(n,r) division once the counts are right.
P(a king or a heart) = 4/52 + 13/52 − 1/52 = 16/52 — subtract the overlap (king of hearts).
Dice
Two dice → 36 outcomes. Know the sum distribution:
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Ways | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
P(sum = 7) = 6/36 = 1/6 — the most likely sum.
Balls from a bag — the one fork that decides the answer
WITH replacement → denominator stays fixed
WITHOUT replacement → denominator drops by 1 each drawRead the question twice for that phrase. It is the difference between right and wrong, not between fast and slow.
3 red, 5 blue. Two drawn without replacement, both red: (3/8) × (2/7) = 6/56 = 3/28.
"At least one" → use the complement
P(at least one) = 1 − P(none)P(at least one 6 in two dice throws) = 1 − (25/36) = 11/36. Doing this directly costs three times as long and invites mistakes.
Worked example — the full method on one question
A bag has 4 red and 6 blue balls. Three are drawn at random. What is the probability that at least one is red?
Step 1 — spot the "at least one". Go straight to the complement. Step 2 — P(none red) = all three blue.
Total ways to draw 3 from 10 = C(10,3) = 120
Ways to draw 3 blue from 6 = C(6,3) = 20
P(no red) = 20/120 = 1/6Step 3 — complement. P(at least one red) = 1 − 1/6 = **5/6**
Why this is the whole topic in miniature: the probability was one division. The work was counting — and the "at least one" phrasing told you to count the opposite case, which had one term instead of three.
Conditional probability
P(A | B) = P(A ∩ B) / P(B)"Given that…" is the signal.