Probability
CSAT · Quantitative Aptitude
Introduction
CSAT probability uses coins, dice, cards and balls. Every question is the ratio favourable ÷ total, provided the outcomes are equally likely – so careful counting is everything.
Key concepts
Definition
P(event) = number of favourable outcomes ÷ total number of equally likely outcomes. 0 ≤ P ≤ 1; P(not E) = 1 − P(E).
Sample spaces
| Experiment | Outcomes |
|---|---|
| One coin | 2 |
| n coins | 2ⁿ |
| One die | 6 |
| Two dice | 36 |
| Pack of cards | 52 (4 suits × 13; 26 red, 26 black; 12 face cards: J, Q, K) |
Addition rule
P(A or B) = P(A) + P(B) − P(A and B). A heart or a king: 13/52 + 4/52 − 1/52 = 16/52 = 4/13.
Multiplication rule
Independent events: P(A and B) = P(A) × P(B). Without replacement, the second probability changes: two reds from 5 red and 3 blue = 5/8 × 4/7 = 5/14.
'At least one'
P(at least one) = 1 − P(none). At least one head in 3 tosses = 1 − 1/8 = 7/8.
Odds
Odds in favour = favourable : unfavourable. Odds of 3 : 5 mean P = 3/8.
Sums of two dice
Number of ways to get a sum s (2–12): 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1. Seven is the most likely sum (6/36).
- P = favourable/total
- P(not E) = 1 − P(E)
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
- Independent: P(A ∩ B) = P(A)P(B)
- At least one = 1 − P(none)
Solved examples
Example 1
One card is drawn at random from a well-shuffled pack of 52. What is the probability that it is an ace or a king?
(a) 9/52 (b) 2/13 (c) 1/6 (d) 7/52
- Favourable cards = 8.
- P = 8/52 = 2/13.
Answer: (b) 2/13
Example 2
4 fair coins are tossed. What is the probability of getting at least 1 head?
(a) 5/6 (b) 15/16 (c) 1/4 (d) 1/16
- Total outcomes = 2^4 = 16.
- Favourable = 15.
- P = 15/16.
Answer: (b) 15/16
Example 3
A bag contains 4 red and 6 blue balls. Two balls are drawn at random without replacement. What is the probability that both are red?
(a) 2/15 (b) 13/15 (c) 4/25 (d) 6/49
- Ways to choose 2 from 10 = C(10, 2) = 45.
- Favourable ways = 6.
- P = 6/45 = 2/15.
Answer: (a) 2/15
Example 4
A bag contains 3 red and 3 blue balls. Two balls are drawn at random without replacement. What is the probability that both are red?
(a) 1/5 (b) 4/5 (c) 1/6 (d) 1/4
- Ways to choose 2 from 6 = C(6, 2) = 15.
- Favourable ways = 3.
- P = 3/15 = 1/5.
Answer: (a) 1/5
Example 5
A bag contains 3 red and 5 blue balls. Two balls are drawn at random without replacement. What is the probability that both are red?
(a) 3/31 (b) 25/28 (c) 9/64 (d) 3/28
- Ways to choose 2 from 8 = C(8, 2) = 28.
- Favourable ways = 3.
- P = 3/28.
Answer: (d) 3/28
Example 6
3 fair coins are tossed. What is the probability of getting exactly 1 head?
(a) 3/8 (b) 5/8 (c) 3/10 (d) 1/3
- Total outcomes = 2^3 = 8.
- Favourable = 3 (C(3,1)).
- P = 3/8.
Answer: (a) 3/8
Example 7
Two fair dice are thrown. What is the probability that the sum is at least 10?
(a) 1/2 (b) 5/6 (c) 1/6 (d) 7/36
- There are 6 × 6 = 36 equally likely outcomes.
- Favourable outcomes: 6.
- P = 6/36 = 1/6.
Answer: (c) 1/6
Example 8
One card is drawn at random from a well-shuffled pack of 52. What is the probability that it is a face card?
(a) 1/4 (b) 55/182 (c) 11/52 (d) 3/13
- Favourable cards = 12.
- P = 12/52 = 3/13.
Answer: (d) 3/13
Example 9
A bag contains 4 red and 4 blue balls. Two balls are drawn at random without replacement. What is the probability that one is red and one is blue?
(a) 1/4 (b) 4/7 (c) 3/7 (d) 1/2
- Ways to choose 2 from 8 = C(8, 2) = 28.
- Favourable ways = 16.
- P = 16/28 = 4/7.
Answer: (b) 4/7
Example 10
A bag contains 7 red and 6 blue balls. Two balls are drawn at random without replacement. What is the probability that at least one is blue?
(a) 57/85 (b) 7/26 (c) 19/26 (d) 49/169
- Ways to choose 2 from 13 = C(13, 2) = 78.
- Favourable ways = 57.
- P = 57/78 = 19/26.
Answer: (c) 19/26
- List small sample spaces (two dice, three coins) – it is faster than formulas.
- Use the complement for 'at least' questions.
- Use combinations for draws without replacement: C(red, 2)/C(total, 2).
- Double-counting cards that belong to two categories (the king of hearts).
- Treating 'without replacement' as independent.
- Forgetting that face cards are only J, Q, K (12 cards).
Practice set
Attempt all 12 questions (about 90 seconds each), then check the worked solutions below.
Q1. 4 fair coins are tossed. What is the probability of getting at least 3 heads?
(a) 5/18 (b) 5/16 (c) 3/4 (d) 11/16
Q2. Two fair dice are thrown. What is the probability that the sum is exactly 11?
(a) 17/18 (b) 1/6 (c) 1/18 (d) 1/12
Q3. One card is drawn at random from a well-shuffled pack of 52. What is the probability that it is a red card?
(a) 27/52 (b) 13/24 (c) 1/2 (d) 25/52
Q4. A bag contains 7 red and 2 blue balls. Two balls are drawn at random without replacement. What is the probability that at least one is blue?
(a) 15/43 (b) 49/81 (c) 7/12 (d) 5/12
Q5. One card is drawn at random from a well-shuffled pack of 52. What is the probability that it is a red king?
(a) 1/26 (b) 3/52 (c) 1/52 (d) 1/24
Q6. One card is drawn at random from a well-shuffled pack of 52. What is the probability that it is a heart?
(a) 1/4 (b) 13/48 (c) 3/13 (d) 7/26
Q7. A bag contains 4 red and 2 blue balls. Two balls are drawn at random without replacement. What is the probability that both are red?
(a) 6/19 (b) 4/9 (c) 3/5 (d) 2/5
Q8. A bag contains 6 red and 2 blue balls. Two balls are drawn at random without replacement. What is the probability that one is red and one is blue?
(a) 9/16 (b) 6/17 (c) 3/7 (d) 4/7
Q9. 3 fair coins are tossed. What is the probability of getting at least 3 heads?
(a) 1/8 (b) 1 (c) 7/8 (d) 1/10
Q10. Two fair dice are thrown. What is the probability that the sum is exactly 4?
(a) 11/12 (b) 1/9 (c) 1/4 (d) 1/12
Q11. 4 fair coins are tossed. What is the probability of getting exactly 2 heads?
(a) 1/3 (b) 3/8 (c) 5/8 (d) 1/2
Q12. A bag contains 4 red and 3 blue balls. Two balls are drawn at random without replacement. What is the probability that at least one is blue?
(a) 16/49 (b) 5/7 (c) 3/5 (d) 2/7
Q1 – (b) 5/16. Total outcomes = 2^4 = 16. Favourable = 5. P = 5/16.
Q2 – (c) 1/18. There are 6 × 6 = 36 equally likely outcomes. Favourable outcomes: 2. P = 2/36 = 1/18.
Q3 – (c) 1/2. Favourable cards = 26. P = 26/52 = 1/2.
Q4 – (d) 5/12. Ways to choose 2 from 9 = C(9, 2) = 36. Favourable ways = 15. P = 15/36 = 5/12.
Q5 – (a) 1/26. Favourable cards = 2. P = 2/52 = 1/26.
Q6 – (a) 1/4. Favourable cards = 13. P = 13/52 = 1/4.
Q7 – (d) 2/5. Ways to choose 2 from 6 = C(6, 2) = 15. Favourable ways = 6. P = 6/15 = 2/5.
Q8 – (c) 3/7. Ways to choose 2 from 8 = C(8, 2) = 28. Favourable ways = 12. P = 12/28 = 3/7.
Q9 – (a) 1/8. Total outcomes = 2^3 = 8. Favourable = 1. P = 1/8.
Q10 – (d) 1/12. There are 6 × 6 = 36 equally likely outcomes. Favourable outcomes: 3. P = 3/36 = 1/12.
Q11 – (b) 3/8. Total outcomes = 2^4 = 16. Favourable = 6 (C(4,2)). P = 6/16 = 3/8.
Q12 – (b) 5/7. Ways to choose 2 from 7 = C(7, 2) = 21. Favourable ways = 15. P = 15/21 = 5/7.
- P = fav/total
- At least one = 1 − none
- Cards: 52, 26 red, 12 face
- Two dice: 36; sum 7 most likely
