Probability questions test whether you can identify possible outcomes, account for overlap and choose the shortest calculation method. This guide covers the essential formulas, common question patterns, 25 SNAP-style probability questions with solutions and a focused preparation plan.
Reviewed by Kaushik Kapoor
Co-founder and Director of Erudite, a premium MBA entrance coaching institute headquartered in Kolkata. An alumnus of XLRI Jamshedpur and St. Xavier’s College, Kolkata, he has over 25 years of experience in mentoring students, MBA entrance preparation, test-prep strategy and academic training.
| Concept | Formula | When to Use It |
|---|---|---|
| Basic probability | (P(E) = (Favourable outcomes)/(Total equally likely outcomes)) | Coins, dice, cards and number-selection questions |
| Complement | (P(E') = 1-P(E)) | “At least one,” “none,” “not” or “neither” |
| Addition rule | (P(A ∪ B) = P(A)+P(B)-P(A ∩ B)) | When the question uses “A or B” |
| Independent events | (P(A ∩ B) = P(A) × P(B)) | When one event does not affect the other |
| Conditional probability | (P(A | B) = (P(A ∩ B))/(P(B))) | Dependent events or selection without replacement |
| Combinations | (nCr = (n!)/(r!(n-r)!)) | Selecting objects when order does not matter |
| Permutations with repetition | ((n!)/(p!q!…)) | Arranging letters when some letters repeat |
Look for these trigger phrases:
Three fair coins are tossed. What is the probability of obtaining at least two heads?
Answer: C. (1/2)
Solution: There are (2^3=8) possible outcomes. The favourable outcomes are HHT, HTH, THH and HHH.
P = 4/8 = 1/2
Two fair dice are rolled. What is the probability that their sum is a prime number?
Answer: B. (5/12)
Solution: The possible prime sums are 2, 3, 5, 7 and 11.
Their respective numbers of outcomes are:
1+2+4+6+2 = 15
Since two dice have 36 possible outcomes:
P = 15/36 = 5/12
Two fair dice are rolled. What is the probability that the sum is at least 10?
Answer: B. (1/6)
Solution: The required sums are 10, 11 and 12.
Therefore:
P = 3+2+1/36 = 6/36 = 1/6
A fair coin is tossed twice. What is the probability of obtaining at most one tail?
Answer: C. (3/4)
Solution: The sample space is HH, HT, TH and TT. The outcomes with zero or one tail are HH, HT and TH.
P = 3/4
Three fair dice are rolled. What is the probability that their total is 18?
Answer: C. (1/216)
Solution: Three dice produce (6^3=216) outcomes. A total of 18 is possible only through ((6,6,6)).
P = 1/216
Two fair dice are rolled. What is the probability that the product of the two numbers is odd?
Answer: B. (1/4)
Solution: A product is odd only when both numbers are odd. A die has three odd faces out of six.
P = 3/6 × 3/6 = 1/2 × 1/2 = 1/4
A fair die is rolled once. What is the probability of obtaining a multiple of 3 or an even number?
Answer: C. (2/3)
Solution: Multiples of 3 are ({3,6}), while even numbers are ({2,4,6}). Their union is ({2,3,4,6}).
P = 4/6 = 2/3
Four fair coins are tossed. What is the probability of obtaining exactly two tails?
Answer: B. (3/8)
Solution: The two tail positions can be selected in:
4C2 = 6
There are (2^4=16) total outcomes.
P = 6/16 = 3/8
One card is drawn from a standard deck of 52 cards. What is the probability of drawing a king or a red card?
Answer: B. (7/13)
Solution: There are 4 kings and 26 red cards. The two red kings are counted in both groups, so subtract the overlap.
P = 4+26-2/52 = 28/52 = 7/13
A bag contains 5 red, 4 blue and 3 green balls. Two balls are drawn without replacement. What is the probability that both are blue?
Answer: B. (1/11)
Solution:
P = (4C2)/(12C2) = 6/66 = 1/11
Two cards are drawn without replacement from a standard deck. What is the probability that one is a spade and the other is a heart?
Answer: B. (13/102)
Solution: Select one card from the 13 spades and one from the 13 hearts.
P = (13 × 13)/(52C2) = 169/1326 = 13/102
A bag contains 6 black and 8 white balls. One ball is drawn. What is the probability that it is white?
Answer: B. (4/7)
Solution:
P = 8/14 = 4/7
Two cards are drawn without replacement from a standard deck. What is the probability that both are aces?
Answer: B. (1/221)
Solution:
P = 4/52 × 3/51 = 12/2652 = 1/221
A bag contains 4 red, 5 yellow and 6 pink balls. Three balls are drawn without replacement. What is the probability of obtaining at least one pink ball?
Answer: C. (371/455)
Solution: Use the complement.
There are 9 non-pink balls. Therefore:
P(no pink) = (9C3)/(15C3) = 84/455
P(at least one pink) = 1-84/455 = 371/455
A card is drawn from a standard deck. What is the probability that it is a face card?
Answer: B. (3/13)
Solution: A standard deck has 12 face cards: four jacks, four queens and four kings.
P = 12/52 = 3/13
A box contains 10 bulbs, of which 3 are defective. Two bulbs are selected without replacement. What is the probability that neither is defective?
Answer: B. (7/15)
Solution: There are 7 non-defective bulbs.
P = (7C2)/(10C2) = 21/45 = 7/15
The probability that A solves a problem is (1/3), while the probability that B solves it is (2/5). If they work independently, what is the probability that the problem is solved by at least one of them?
Answer: C. (3/5)
Solution: Calculate the probability that both fail.
P(both fail) = 2/3 × 3/5 = 2/5
P(problem solved) = 1-2/5 = 3/5
A husband and wife apply for two vacancies. Their respective probabilities of selection are (1/7) and (1/5). Assuming that their selection outcomes are independent, what is the probability that exactly one is selected?
Answer: B. (2/7)
Solution:
P(husband only) = 1/7 × 4/5 = 4/35
P(wife only) = 6/7 × 1/5 = 6/35
P(exactly one) = 4/35+6/35 = 10/35 = 2/7
A speaks the truth in 75% of cases and B in 80% of cases. Assuming their responses are independent, in what percentage of cases will they contradict each other?
Answer: C. 35%
Solution: They contradict each other when A tells the truth and B lies, or when A lies and B tells the truth.
(0.75 × 0.20)+(0.25 × 0.80) = 0.15+0.20 = 0.35
Therefore, they contradict each other in 35% of cases.
Three shooters hit a target with probabilities (1/2), (1/3) and (1/4). Assuming independent attempts, what is the probability that the target is hit by at least one shooter?
Answer: D. (3/4)
Solution: The respective miss probabilities are (1/2), (2/3) and (3/4).
P(all miss) = 1/2 × 2/3 × 3/4 = 1/4
P(at least one hit) = 1-1/4 = 3/4
A student guesses the answers to two independent multiple-choice questions, each containing four options and one correct answer. What is the probability of answering both correctly?
Answer: C. (1/16)
Solution:
P = 1/4 × 1/4 = 1/16
Two independent events have probabilities 0.3 and 0.6. What is the probability that neither event occurs?
Answer: C. 0.28
Solution: The probabilities of non-occurrence are 0.7 and 0.4.
P(neither) = 0.7 × 0.4 = 0.28
A bag contains 3 red and 7 black keys. Two keys are drawn with replacement. What is the probability that both are red?
Answer: C. 0.09
Solution: Replacement keeps the probability of selecting a red key at (3/10) for both draws.
P = 3/10 × 3/10 = 9/100 = 0.09
One integer is selected at random from 1 to 50, inclusive. What is the probability that it is a multiple of 4 or 6?
Answer: C. (8/25)
Solution:
Using inclusion–exclusion:
12+8-4 = 16
P = 16/50 = 8/25
The letters of LEADER are arranged randomly. What is the probability that all three vowels are together?
Answer: C. (1/5)
Solution: The word contains six letters, with E repeated twice.
The total number of distinct arrangements is:
(6!)/(2!) = 360
Treat the vowels E, A and E as one block. The block and the consonants L, D and R form four objects, which can be arranged in (4!) ways.
The vowels can be arranged internally in:
(3!)/(2!) = 3
Therefore, the favourable arrangements are:
4! × (3!)/(2!) = 24 × 3 = 72
P = 72/360 = 1/5
| Question | Answer |
|---|---|
| 1 | C. (1/2) |
| 2 | B. (5/12) |
| 3 | B. (1/6) |
| 4 | C. (3/4) |
| 5 | C. (1/216) |
| 6 | B. (1/4) |
| 7 | C. (2/3) |
| 8 | B. (3/8) |
| 9 | B. (7/13) |
| 10 | B. (1/11) |
| 11 | B. (13/102) |
| 12 | B. (4/7) |
| 13 | B. (1/221) |
| 14 | C. (371/455) |
| 15 | B. (3/13) |
| 16 | B. (7/15) |
| 17 | C. (3/5) |
| 18 | B. (2/7) |
| 19 | C. 35% |
| 20 | D. (3/4) |
| 21 | C. (1/16) |
| 22 | C. 0.28 |
| 23 | C. 0.09 |
| 24 | C. (8/25) |
| 25 | C. (1/5) |
When two events overlap, adding their probabilities counts the overlap twice. Use:
P(A ∪ B) = P(A)+P(B)-P(A ∩ B)
With replacement, the probabilities remain unchanged. Without replacement, both the numerator and denominator may change after the first selection.
For “at least one” questions, calculating every acceptable case may be unnecessarily slow. It is often faster to calculate the probability of zero occurrences and subtract it from 1.
Probabilities can be multiplied directly only when the events are independent or when conditional probabilities are used correctly.
For words such as LEADER, repeated letters reduce the total number of distinct arrangements.
Use permutations when order matters and combinations when only the selected group matters.
| Day | Focus | Practice Target |
|---|---|---|
| Day 1 | Basic probability and sample spaces | 15 coin and dice questions |
| Day 2 | Addition rule and overlapping events | 15 “A or B” questions |
| Day 3 | Independent and dependent events | 15 mixed questions |
| Day 4 | Complements and “at least one” problems | 15 timed questions |
| Day 5 | Cards, balls and selections | 20 questions |
| Day 6 | Permutations and combinations | 15 arrangement questions |
| Day 7 | Mixed timed test and error review | 25 questions in 25 minutes |
During the first few sessions, prioritise method selection and accuracy. Once your process is reliable, reduce your average solving time. Because the latest published SNAP structure includes negative marking, blind guessing can reduce the value of otherwise accurate work.
No fixed number should be assumed. The official test structure publishes the overall Quantitative, Data Interpretation and Data Sufficiency allocation rather than a topic-wise probability count. Prepare probability as part of the broader quantitative aptitude syllabus, but do not depend on a guaranteed number of questions.
Focus on basic probability, sample spaces, complements, the addition rule, independent and dependent events, cards, coins, dice, selection without replacement, combinations and simple arrangements.
Yes. Combination methods are particularly useful when objects are selected without regard to order. Permutations are required for arrangement questions, especially when letters must remain together or when some letters repeat.
Use the complement method:
P(at least one) = 1-P(none)
This usually replaces several separate calculations with one multiplication and one subtraction.
Practise mental arithmetic and written simplification rather than depending on a calculator. The latest published official test instructions prohibit calculators and electronic calculating devices in the test hall. Confirm the applicable rule again in the SNAP 2026 test-day guide.
A practical initial target is 75–100 varied questions. The goal is not merely volume. Maintain an error log identifying whether each mistake came from counting, formula selection, arithmetic, replacement or an unstated assumption.
Probability becomes faster when you classify the problem before calculating. Train yourself to recognise the wording, choose the appropriate rule and write only the steps needed to avoid an error.
After completing the 25 questions above, repeat the set under timed conditions. Review every incorrect or slow solution by method rather than by answer alone. This will help you improve both speed and accuracy before SNAP 2026.