📌 Where it comes in CGL: the SSC CGL Tier 2 Paper I syllabus includes "simple probability", and Paper II (JSO) has a full unit. All questions here are practice questions prepared by SF Prep (not PYQs); JSO (Paper II) PYQs are not yet available on SF Prep.
📑 Topics on this page (Jump to Topic):
📌 1. Basic Ideas and Rules
- The set of all possible outcomes of an experiment = sample space (S); any part of it = an event (E).
- Impossible event: P = 0; sure event: P = 1; for every event 0 ≤ P(E) ≤ 1.
- Mutually exclusive: cannot happen together (2 and 5 on one die). Independent: one result does not affect the other (two separate coins).
📘 Formulas:
- P(E) = favourable outcomes ÷ total equally likely outcomes
- P(not E) = 1 − P(E)
- Odds in favour a : b → P(E) = a/(a + b); odds against a : b → P(E) = b/(a + b)
- ⁿCᵣ = n!/[r!(n − r)!]
✏️ Worked Example: The odds in favour of an event are 3 : 5. Its probability?
→ 3 parts favourable, 5 parts unfavourable → 3 + 5 = 8 parts in all
→ P = 3/8; probability of it not happening = 5/8
→ 3 parts favourable, 5 parts unfavourable → 3 + 5 = 8 parts in all
→ P = 3/8; probability of it not happening = 5/8
📌 2. Coins and Dice
📘 Formulas:
- n coins → 2ⁿ outcomes; exactly r heads → ⁿCᵣ/2ⁿ; at least one head → 1 − 1/2ⁿ
- n dice → 6ⁿ outcomes; doublets on two dice: 6
| Sum of two dice | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Ways (out of 36) | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
✏️ Worked Example: 3 coins are tossed. Probability of exactly 2 heads?
→ Total outcomes 2³ = 8: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
→ Exactly 2 heads: HHT, HTH, THH = 3 (or ³C₂ = 3)
→ P = 3/8
→ Total outcomes 2³ = 8: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
→ Exactly 2 heads: HHT, HTH, THH = 3 (or ³C₂ = 3)
→ P = 3/8
✏️ Worked Example: Two dice are thrown. Probability that the sum is 8?
→ (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) = 5 ways
→ P = 5/36
→ (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) = 5 ways
→ P = 5/36
✏️ Worked Example: Two dice are thrown. Probability that the product of the numbers is even?
→ The product is odd only when both are odd: 3 × 3 = 9 ways
→ P(even) = 1 − 9/36 = 3/4
→ The product is odd only when both are odd: 3 × 3 = 9 ways
→ P(even) = 1 − 9/36 = 3/4
⚠️ Exam Traps:
- For "at least one" use 1 − P(none) — direct counting invites mistakes.
- With two dice, (2, 6) and (6, 2) are different outcomes.
📌 3. Cards and Balls
| Cards (52) | Number |
|---|---|
| 4 suits: spades ♠, clubs ♣ (black); hearts ♥, diamonds ♦ (red) | 13 in each suit; 26 red, 26 black |
| Face cards (king, queen, jack) | 12 |
| Honour cards (ace, king, queen, jack) | 16 |
| Aces / kings / queens / jacks | 4 each |
📘 Formulas:
- Choosing r items together: P = (favourable selections)/(ᴺCᵣ)
- One after another without replacement: P = P(first) × P(second | first)
- For "or" with cards subtract the overlap: P(A or B) = P(A) + P(B) − P(both)
✏️ Worked Example: One card is drawn. Probability that it is a face card or a heart?
→ Face cards 12, hearts 13, both (J, Q, K of hearts) 3
→ P = (12 + 13 − 3)/52 = 22/52 = 11/26
→ Face cards 12, hearts 13, both (J, Q, K of hearts) 3
→ P = (12 + 13 − 3)/52 = 22/52 = 11/26
✏️ Worked Example: A bag has 5 red and 3 blue balls. Two balls are drawn. Probability of one red and one blue?
→ Total selections ⁸C₂ = 28; favourable ⁵C₁ × ³C₁ = 15
→ P = 15/28
→ Total selections ⁸C₂ = 28; favourable ⁵C₁ × ³C₁ = 15
→ P = 15/28
✏️ Worked Example: 3 balls are drawn from 4 red and 5 green balls. Probability that all three are green?
→ ⁵C₃ = 10; ⁹C₃ = 84
→ P = 10/84 = 5/42
→ ⁵C₃ = 10; ⁹C₃ = 84
→ P = 10/84 = 5/42
📌 4. Addition and Multiplication Rules, Independent Events
📘 Formulas:
- P(A or B) = P(A) + P(B) − P(A and B); if mutually exclusive, P(A) + P(B)
- Independent: P(A and B) = P(A) × P(B)
- At least one = 1 − (1 − p₁)(1 − p₂)…; exactly one (two events) = p₁(1 − p₂) + (1 − p₁)p₂
- Conditional: P(A | B) = P(A and B)/P(B) — knowing B has happened shrinks the sample space
✏️ Worked Example: A hits a target with probability 1/3 and B with probability 1/4. Each fires once. Probability that the target is hit?
→ Both miss: (2/3)(3/4) = 1/2
→ P(target hit) = 1 − 1/2 = 1/2
→ Both miss: (2/3)(3/4) = 1/2
→ P(target hit) = 1 − 1/2 = 1/2
✏️ Worked Example: In the same situation, probability that exactly one of them hits?
→ A hits, B misses: (1/3)(3/4) = 3/12; A misses, B hits: (2/3)(1/4) = 2/12
→ P = 3/12 + 2/12 = 5/12
→ A hits, B misses: (1/3)(3/4) = 3/12; A misses, B hits: (2/3)(1/4) = 2/12
→ P = 3/12 + 2/12 = 5/12
✏️ Worked Example: A number from 1 to 30 is chosen. Probability that it is divisible by 3 or 5?
→ Multiples of 3: 10, multiples of 5: 6, both (multiples of 15): 2
→ Favourable = 10 + 6 − 2 = 14 → P = 14/30 = 7/15
→ Multiples of 3: 10, multiples of 5: 6, both (multiples of 15): 2
→ Favourable = 10 + 6 − 2 = 14 → P = 14/30 = 7/15
⚠️ Exam Traps:
- Add P(A) + P(B) for mutually exclusive events; multiply P(A) × P(B) for independent events — do not mix them up.
📌 5. Miscellaneous: Numbers, Letters, Calendar, Conditional
- Letters: a random letter from a word → frequency of that letter ÷ total letters (count repeated letters separately).
- Calendar: an ordinary year = 52 weeks + 1 day; a leap year = 52 weeks + 2 days (7 possible pairs).
- In conditional questions, form the new (smaller) sample space from the information given.
✏️ Worked Example: Probability that a leap year has 53 Sundays?
→ 366 = 52 weeks + 2 extra days; pairs: (Sun, Mon), (Mon, Tue), …, (Sat, Sun) = 7
→ Pairs containing Sunday: 2 → P = 2/7
→ 366 = 52 weeks + 2 extra days; pairs: (Sun, Mon), (Mon, Tue), …, (Sat, Sun) = 7
→ Pairs containing Sunday: 2 → P = 2/7
✏️ Worked Example: A letter is chosen at random from the word EXAMINATION. Probability that it is a vowel?
→ 11 letters in all; vowels E, A, I, A, I, O = 6
→ P = 6/11
→ 11 letters in all; vowels E, A, I, A, I, O = 6
→ P = 6/11
✏️ Worked Example: A die is thrown and you are told the number is even. Probability that it is 6?
→ New sample space {2, 4, 6} — 3 outcomes
→ P = 1/3 (it would be 1/6 without the information)
→ New sample space {2, 4, 6} — 3 outcomes
→ P = 1/3 (it would be 1/6 without the information)
Next: Bayes’ theorem, random variables, and the binomial, Poisson and normal distributions — Statistics (Advanced) — Unit 6 and Unit 7.
📌 6. Practice MCQs
📝 Practice MCQs (prepared by SF Prep — not PYQs)
Q1. 4 coins are tossed. The probability of at least 3 heads is: Practice
Correct answer: (C)
Solution: Exactly 3 heads: ⁴C₃ = 4; exactly 4 heads: 1 → 5 favourable; total 2⁴ = 16 → P = 5/16.
Solution: Exactly 3 heads: ⁴C₃ = 4; exactly 4 heads: 1 → 5 favourable; total 2⁴ = 16 → P = 5/16.
Q2. Two dice are thrown. The probability that the sum is at most 4 is: Practice
Correct answer: (A)
Solution: Sum 2: 1 way, sum 3: 2 ways, sum 4: 3 ways → 6 ways; P = 6/36 = 1/6.
Solution: Sum 2: 1 way, sum 3: 2 ways, sum 4: 3 ways → 6 ways; P = 6/36 = 1/6.
Q3. Two dice are thrown. The probability of a doublet or a sum of 10 is: Practice
Correct answer: (D)
Solution: Doublets 6; sum 10: (4, 6), (5, 5), (6, 4) = 3; (5, 5) is in both → 6 + 3 − 1 = 8; P = 8/36 = 2/9.
Solution: Doublets 6; sum 10: (4, 6), (5, 5), (6, 4) = 3; (5, 5) is in both → 6 + 3 − 1 = 8; P = 8/36 = 2/9.
Q4. Two dice are thrown. The probability that the two numbers differ by exactly 1 is: Practice
Correct answer: (B)
Solution: (1, 2), (2, 3), (3, 4), (4, 5), (5, 6) and their reverses → 10 ways; P = 10/36 = 5/18.
Solution: (1, 2), (2, 3), (3, 4), (4, 5), (5, 6) and their reverses → 10 ways; P = 10/36 = 5/18.
Q5. A card is drawn from a pack of 52. The probability that it is a red king or a black queen is: Practice
Correct answer: (B)
Solution: Red kings 2 + black queens 2 = 4 (no overlap); P = 4/52 = 1/13.
Solution: Red kings 2 + black queens 2 = 4 (no overlap); P = 4/52 = 1/13.
Q6. A card is drawn from a pack of 52. The probability that it is neither a heart nor a king is: Practice
Correct answer: (D)
Solution: Heart or king = 13 + 4 − 1 = 16; the rest = 52 − 16 = 36; P = 36/52 = 9/13.
Solution: Heart or king = 13 + 4 − 1 = 16; the rest = 52 − 16 = 36; P = 36/52 = 9/13.
Q7. Two cards are drawn without replacement from a pack of 52. The probability that one is red and one is black is: Practice
Correct answer: (A)
Solution: Favourable = 26 × 26 = 676; total ⁵²C₂ = 1326; P = 676/1326 = 26/51.
Solution: Favourable = 26 × 26 = 676; total ⁵²C₂ = 1326; P = 676/1326 = 26/51.
Q8. A bag has 4 red and 6 blue balls. Two balls are drawn. The probability that at least one is red is: Practice
Correct answer: (C)
Solution: P(both blue) = ⁶C₂/¹⁰C₂ = 15/45 = 1/3; P(at least one red) = 1 − 1/3 = 2/3.
Solution: P(both blue) = ⁶C₂/¹⁰C₂ = 15/45 = 1/3; P(at least one red) = 1 − 1/3 = 2/3.
Q9. A bag has 3 red, 4 white and 5 black balls. The probability that a ball drawn is not black is: Practice
Correct answer: (A)
Solution: Not black = 3 + 4 = 7; total 12; P = 7/12.
Solution: Not black = 3 + 4 = 7; total 12; P = 7/12.
Q10. From 5 red and 5 blue balls, two are drawn one after another without replacement. The probability that the first is red and the second is blue is: Practice
Correct answer: (D)
Solution: P = 5/10 × 5/9 = 25/90 = 5/18.
Solution: P = 5/10 × 5/9 = 25/90 = 5/18.
Q11. The odds against an event are 7 : 4. The probability that the event happens is: Practice
Correct answer: (C)
Solution: Odds against 7 : 4 → 7 parts unfavourable, 4 parts favourable; P = 4/(7 + 4) = 4/11.
Solution: Odds against 7 : 4 → 7 parts unfavourable, 4 parts favourable; P = 4/(7 + 4) = 4/11.
Q12. A number from 1 to 50 is chosen. The probability that it is prime is: Practice
Correct answer: (B)
Solution: 15 primes in 1–50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47; P = 15/50 = 3/10.
Solution: 15 primes in 1–50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47; P = 15/50 = 3/10.
Q13. A number from 1 to 50 is chosen. The probability that it is divisible by 4 or 6 is: Practice
Correct answer: (D)
Solution: Multiples of 4: 12, of 6: 8, of both (12, 24, 36, 48): 4 → 12 + 8 − 4 = 16; P = 16/50 = 8/25.
Solution: Multiples of 4: 12, of 6: 8, of both (12, 24, 36, 48): 4 → 12 + 8 − 4 = 16; P = 16/50 = 8/25.
Q14. A and B are independent with P(A) = 0.3 and P(B) = 0.6. The probability that A happens but B does not is: Practice
Correct answer: (B)
Solution: P(A and B′) = P(A) × [1 − P(B)] = 0.3 × 0.4 = 0.12.
Solution: P(A and B′) = P(A) × [1 − P(B)] = 0.3 × 0.4 = 0.12.
Q15. Three shooters hit a target with probabilities 0.5, 0.6 and 0.8 (independently). The probability that the target is hit at least once is: Practice
Correct answer: (A)
Solution: All miss: 0.5 × 0.4 × 0.2 = 0.04; P(at least one hit) = 1 − 0.04 = 0.96.
Solution: All miss: 0.5 × 0.4 × 0.2 = 0.04; P(at least one hit) = 1 − 0.04 = 0.96.
Q16. A solves a problem with probability 1/2 and B with probability 1/3. The probability that exactly one of them solves it is: Practice
Correct answer: (C)
Solution: A solves, B does not: (1/2)(2/3) = 1/3; A does not, B solves: (1/2)(1/3) = 1/6; total = 1/3 + 1/6 = 1/2.
Solution: A solves, B does not: (1/2)(2/3) = 1/3; A does not, B solves: (1/2)(1/3) = 1/6; total = 1/3 + 1/6 = 1/2.
Q17. A die is thrown and the number is known to be even. The probability that it is prime is: Practice
Correct answer: (C)
Solution: New sample space {2, 4, 6}; the only prime among them is 2 → P = 1/3.
Solution: New sample space {2, 4, 6}; the only prime among them is 2 → P = 1/3.
Q18. A card is drawn and it is red. The probability that it is a king is: Practice
Correct answer: (B)
Solution: 26 red cards, 2 of them kings → P = 2/26 = 1/13.
Solution: 26 red cards, 2 of them kings → P = 2/26 = 1/13.
Q19. The probability that a leap year has 53 Fridays is: Practice
Correct answer: (A)
Solution: A leap year = 52 weeks + 2 days; of the 7 possible pairs, (Thu, Fri) and (Fri, Sat) contain a Friday → P = 2/7.
Solution: A leap year = 52 weeks + 2 days; of the 7 possible pairs, (Thu, Fri) and (Fri, Sat) contain a Friday → P = 2/7.
Q20. A letter is chosen at random from the word STATISTICS. The probability that it is T is: Practice
Correct answer: (D)
Solution: 10 letters: S 3, T 3, A 1, I 2, C 1 → P(T) = 3/10.
Solution: 10 letters: S 3, T 3, A 1, I 2, C 1 → P(T) = 3/10.
Q21. The letters C, A, T are written in a random order. The probability that they spell "CAT" is: Practice
Correct answer: (B)
Solution: 3 different letters have 3! = 6 orders; only one is "CAT" → P = 1/6.
Solution: 3 different letters have 3! = 6 orders; only one is "CAT" → P = 1/6.
Q22. The probability that two people were born on the same day of the week is: Practice
Correct answer: (C)
Solution: The first person can be any day; the chance the second matches it is 1/7 (or 7/49).
Solution: The first person can be any day; the chance the second matches it is 1/7 (or 7/49).
Q23. A die is thrown twice. The probability of getting a 6 at least once is: Practice
Correct answer: (D)
Solution: P(no 6 at all) = (5/6)² = 25/36; P(at least one 6) = 1 − 25/36 = 11/36.
Solution: P(no 6 at all) = (5/6)² = 25/36; P(at least one 6) = 1 − 25/36 = 11/36.
Q24. 2 people are chosen at random from 3 men and 2 women. The probability that both are women is: Practice
Correct answer: (A)
Solution: Total ⁵C₂ = 10; both women ²C₂ = 1 → P = 1/10.
Solution: Total ⁵C₂ = 10; both women ²C₂ = 1 → P = 1/10.
Q25. Bag I has 2 red and 3 black balls; bag II has 4 red and 1 black. A bag is chosen at random and the ball drawn is red. The probability that it came from bag II is: Practice
Correct answer: (B)
Solution: P(red ∩ I) = ½ × 2/5 = 1/5; P(red ∩ II) = ½ × 4/5 = 2/5; P(II | red) = (2/5)/(1/5 + 2/5) = 2/3.
Solution: P(red ∩ I) = ½ × 2/5 = 1/5; P(red ∩ II) = ½ × 4/5 = 2/5; P(II | red) = (2/5)/(1/5 + 2/5) = 2/3.