Probability: sample spaces, events and counting

Objectives, prerequisites and scope

Build a probability model before applying a formula; translate “and”, “or”, “not” and “at least” into events; use complements and inclusion–exclusion; and count finite equally likely outcomes without mixing ordered and unordered models. Prerequisites: sets, fractions, permutations and combinations. This lesson is the foundation for the conditional probability, Bayes theorem and random-variable topics in the GATE 2026 CS syllabus. It is not a complete probability course. Independence is introduced formally in the next lesson.

1. Outcomes are not automatically equally likely

A random experiment has a sample space Ω containing all possible outcomes under the chosen model. An event A is a subset of Ω. A single die roll can use Ω={1,2,3,4,5,6}; the event “even” is {2,4,6}. A probability assigns nonnegative weights to outcomes, totaling one in a finite model. An event receives the sum of the weights of its outcomes.

For a fair die, each outcome has weight 1/6. For a biased die, merely listing six outcomes does not establish those weights. The formula P(A)=|A|/|Ω| is valid only when Ω is finite and its elementary outcomes are equally likely. A sample space can be correct while a uniform probability assignment on it is wrong.

Two independent fair coin tosses have four equally likely ordered outcomes HH, HT, TH, TT. If we instead record only the number of heads, the possible values are 0,1,2, but their probabilities are 1/4,1/2,1/4. The three values are not equally likely. “Exactly one head” has two underlying outcomes, while “two heads” has one. Grouping outcomes changes their weights.

2. Probability rules from event sets

P(Ω)=1 and P(∅)=0. For any event, 0≤P(A)≤1. The complement Aᶜ consists of the outcomes outside A, so P(Aᶜ)=1−P(A). If A⊆B, then P(A)≤P(B). A negative probability or a probability exceeding one indicates an error or inconsistent input.

A∩B means both A and B occur. A∪B means at least one occurs, including both. “A or B” in mathematics is inclusive unless “but not both” is stated. For two events,

P(A∪B)=P(A)+P(B)−P(A∩B).

The intersection is subtracted because it was counted twice. If A and B are disjoint (mutually exclusive), A∩B=∅ and the formula reduces to a sum. Disjointness is not independence: two different faces of one die cannot occur together, which is strong information about their relationship.

Example: on a fair die, A={2,4,6} and B={4,5,6}. Then P(A)=P(B)=1/2, P(A∩B)=2/6=1/3, and P(A∪B)=1−1/3=2/3. The union is {2,4,5,6}. Adding 1/2+1/2 without the overlap would incorrectly give one.

“Exactly one of A,B” is (A\B)∪(B\A), with probability P(A)+P(B)−2P(A∩B). “Neither” is (A∪B)ᶜ; “not both” is (A∩B)ᶜ. These are different events.

3. Ordered models for repeated experiments

For two independent fair dice distinguished by color, use ordered pairs (i,j). All 36 pairs have probability 1/36. Sum seven has six outcomes: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1). Thus P(sum=7)=6/36=1/6. The possible sums 2,…,12 are not equally likely, so 1/11 is unjustified.

An unordered pair model requires care: {1,2} has two ordered preimages but {1,1} has one. Those unordered outcomes therefore are not equiprobable. A reliable shortcut is to keep the ordered model throughout and divide counts only after checking that every grouping has the same multiplicity.

With three independent fair coin tosses, there are 2³=8 equally likely strings. “At least one head” is the complement of TTT, so its probability is 1−1/8=7/8. “Exactly one head” has HTT,THT,TTH, giving 3/8. “At least” and “exactly” cannot be substituted.

4. Sampling without replacement

A box contains five distinguishable red balls and three distinguishable blue balls. Select two uniformly at random without replacement, with every two-ball subset equally likely. There are C(8,2)=28 subsets. Exactly one red and one blue can be chosen in C(5,1)C(3,1)=15 ways, so the probability is 15/28. The denominator and numerator both count unordered selections.

Check with ordered draws: there are 8×7=56 equally likely ordered pairs. Red then blue gives 5×3=15 pairs; blue then red gives 3×5=15. The probability is 30/56=15/28. Using 15/56 counts only one order while claiming to count either order. Using 15/64 would treat sampling as with replacement.

For at least one red, use the complement: 1−C(3,2)/C(8,2)=25/28. A single red draw has probability 5/8, but two-draw events require their own model. After a draw without replacement, the composition changes.

5. Three events and consistency checks

For three events, add individual probabilities, subtract the three pairwise intersections, and add the triple intersection:

P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C).

Pairwise intersection data include the triple intersection unless “only these two” is stated. Example: individual probabilities .4,.3,.2; pair intersections .1,.05,.04; triple .02. The union is .73 and neither of the three occurs with probability .27. Exactly two occur with probability .1+.05+.04−3(.02)=.13. The triple was counted once in each pair and must be removed from all three to get exactly two.

For two events with known marginal probabilities, max(0,P(A)+P(B)−1)≤P(A∩B)≤min(P(A),P(B)). The upper bound follows from containment; the lower bound ensures the union is at most one. If P(A)=.8 and P(B)=.7, an intersection of .2 is impossible because the union would be 1.3. Do not blindly calculate with inconsistent inputs.

Solving checklist and pitfalls

Describe one elementary outcome, establish its probability, and translate the target into sets. Use counting ratios only with equally likely outcomes. Keep replacement, ordering and distinguishability consistent in numerator and denominator. Use a complement when “at least one” has a simple failure event. Subtract overlaps for unions. Check the result against [0,1] and containment bounds. A count is an integer; a probability need not be. A probability of zero is equivalent to impossibility for positive-weight finite elementary outcomes, but that equivalence does not extend to all continuous models.

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

PF01. Two independent fair coins are tossed. What is the probability of exactly one head?

A. 1/3 B. 1/2 C. 1/4 D. 3/4

Answer: B

A: The head counts 0,1,2 are not equally likely.

B: HT and TH are two of four equally likely outcomes.

C: This counts only one of HT and TH.

D: This is the probability of at least one head.

PF02. P(A)=0.6, P(B)=0.5 and P(A∩B)=0.2. Find P(A∪B).

A. 1.1 B. 0.3 C. 0.9 D. 0.7

Answer: C

A: The overlap is counted twice and the result exceeds one.

B: Multiplication is not the union rule.

C: Use 0.6+0.5−0.2=0.9.

D: This subtracts the intersection twice and counts exactly one event.

PF03. Two independent fair dice are rolled. What is P(sum=7)?

A. 1/11 B. 1/12 C. 7/36 D. 1/6

Answer: D

A: The eleven possible sums are not equally likely.

B: Three unordered distinct-face pairs correspond to six ordered outcomes.

C: Seven is the target sum, not the favorable-outcome count.

D: Six of the 36 ordered pairs have sum seven.

PF04. Three independent fair coin tosses are made. What is P(at least one head)?

A. 7/8 B. 3/8 C. 1/8 D. 1/2

Answer: A

A: Only TTT fails, so 1−1/8=7/8.

B: This is exactly one head.

C: This is no heads, the complement of the target.

D: One-toss head probability does not answer a three-toss event.

PF05. Two balls are selected uniformly without replacement from five red and three blue distinguishable balls. What is the probability of one of each color?

A. 15/56 B. 15/28 C. 15/64 D. 25/28

Answer: B

A: This counts only one color order over ordered outcomes.

B: There are 5×3 favorable subsets among C(8,2)=28.

C: The denominator 64 corresponds to replacement and also requires consistent ordered counting.

D: This is at least one red, which includes two reds.

PF06. P(A)=0.8 and P(B)=0.7. Which value of P(A∩B) is impossible?

A. 0.5 B. 0.6 C. 0.2 D. 0.7

Answer: C

A: This yields a union of one and is feasible.

B: This yields a union of 0.9 and is feasible.

C: The intersection must be at least 0.8+0.7−1=0.5.

D: This is feasible when B is a subset of A.

Source and scope

Prerequisite foundations for Section 1 Probability and Statistics in the official GATE 2026 CS syllabus. Conditional probability, Bayes theorem and random variables receive separate lessons; this is not full-topic or full-syllabus coverage. The source establishes scope, not authorship of these original examples.

Official GATE 2026 CS syllabus (verified 2026-10-14).