Probability measures how likely an event is to occur — from 0 (impossible) to 1 (certain). In aptitude tests it often involves coins, dice, cards, and balls.
P(E) = (Number of Favorable Outcomes) / (Total Number of Possible Outcomes)
P(not E) = 1 − P(E).
| Coins | Sample Space | Total |
|---|---|---|
| 1 coin | {H, T} | 2 |
| 2 coins | {HH, HT, TH, TT} | 4 |
| 3 coins | {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} | 8 |
General rule: for n coins, total outcomes = 2n.
| Dice | Sample Space | Total |
|---|---|---|
| 1 die | {1, 2, 3, 4, 5, 6} | 6 |
| 2 dice | (1,1), (1,2) … (6,6) | 36 |
General rule: for n dice, total outcomes = 6n.
Events that do not affect each other (e.g., rolling a die and tossing a coin).
P(A and B) = P(A) × P(B)
Events that cannot happen at the same time (e.g., drawing a red card OR a black card).
P(A or B) = P(A) + P(B)
Events that can overlap (e.g., drawing a heart OR a face card — some cards are both).
P(A or B) = P(A) + P(B) − P(A and B)
When an item is picked and not put back, the total decreases for the next pick.
Example: probability of drawing 2 kings from a pack.
1st draw: 4/52.
2nd draw (only 3 kings left, 51 cards total):
3/51.
Total P = (4/52) × (3/51).
Used when there are exactly two outcomes (success / failure) repeated n times.
P(r successes in n trials) = nCr × pr × q(n−r)
n: total trialsr: number of successes requiredp: probability of success in one trialq: probability of failure (1 − p)Calculating "at least one" directly is often hard. It is easier to calculate "none" and subtract from 1.
P(At least one) = 1 − P(None)
| Symbol | Meaning |
|---|---|
S |
Sample space |
E |
Event |
n(E) |
Number of favorable outcomes |
n(S) |
Total number of outcomes |
P(A) |
Probability of event A |
nCr
|
Combination (selection) |
For n coins, total outcomes are?
For n dice, total outcomes are?
Independent events A and B: P(A and B) = ?
Mutually exclusive A or B (cannot co-occur): P(A or B) = ?
Non-mutually exclusive (can overlap): P(A or B) = ?
Probability of drawing 2 kings without replacement?
Binomial: P(r successes in n trials) = ?
P(At least one) is most easily computed as?
P(E) ⇒ favorable / total ; range 0→1
S ⇒ sample space (all outcomes) ; E ⇒ event ; P(not E) = 1 − P(E)
Coins ⇒ n coins → 2ⁿ (1→2, 2→4, 3→8) ; Dice ⇒ n dice → 6ⁿ (1→6, 2→36)
Cards ⇒ 52 total ; 4 suits ×13 ; face 12 (J,Q,K×4) ; aces 4
a) Independent (AND) ⇒ do not affect each other → P(A and B) = P(A)×P(B)
b) Mutually exclusive (OR, no overlap) ⇒ P(A or B) = P(A)+P(B)
c) Non-mutually exclusive (overlap) ⇒ P(A or B) = P(A)+P(B) − P(A and B)
Without replacement ⇒ totals shrink (eg 2 kings: 4/52 × 3/51)
Binomial ⇒ ⁿCr × pʳ × qⁿ⁻ʳ ; p = success, q = 1−p
At least one ⇒ 1 − P(None) (complement easier)
S ⇒ sample space ; n(S) ⇒ total outcomes ; n(E) ⇒ favorable
P(A) ⇒ prob of A ; ⁿCr ⇒ combinations
Primary source: none pinned yet — drop your preferred video/book resource into RESOURCES.md and it will be linked here. Ask me anything that's unclear.
Questions? Ask your agent — you can follow up on any concept, quiz answer, or get extra practice problems tuned to this module.