UPSC CSE · CSAT · Numeracy and data interpretation · TOPIC LESSON
Counting and elementary probability
Count the right outcomes and state the random model before turning counts into probabilities.
UPSC Civil Services45 min study guide8 indexed sections
BY THE END, YOU CAN
Distinguish ordered choices, unordered selection and repeated objects.
Use conditional probability, complements and independence correctly.
Read a complete no-replacement tree and check its total.
BEFORE YOU START
Multiply fractions and interpret ratios.
Original educational lesson. Work through the examples before revealing the practice answers. This original common-core lesson supports its declared paper scope. Optional subjects, qualifying language papers and the interview are outside this resource.
LEARN THROUGH ORIGINAL PRACTICE
Original UPSC-style worked applications and practice
All teaching examples, cases, prompts and model answers are original. Official exam questions and answer keys are not reproduced. This common-core lesson does not imply complete historical paper coverage or an official score.
01 / UNDERSTAND & APPLY
Add alternatives, multiply stages
The addition principle counts disjoint alternatives: if an object can be chosen from one of two nonoverlapping groups, add their sizes. If groups overlap, simple addition double-counts the intersection and must be corrected.
The multiplication principle counts successive choices when the number available at each stage is known. Later choices may depend on earlier ones; use branches or cases when their counts differ. Multiplication does not require probabilistic independence as a counting rule, but it does require correctly identifying each stage’s available choices.
Label distinct options so that enumeration can verify the product. A restriction may remove a particular combination without eliminating either item from all other combinations.
ORIGINAL WORKED EXAMPLE
Outfits with one forbidden pairing
Three distinct shirts S 1,S 2,S 3 can be paired with two distinct trousers T 1,T 2. Initially all pairings are permitted. Then the single pairing S 2 with T 1 is forbidden. Count outfits before and after the restriction.
For each of 3 shirts there are 2 trousers, giving 3×2=6 pairings.
List them: S 1 T 1,S 1 T 2,S 2 T 1,S 2 T 2,S 3 T 1,S 3 T 2.
Remove only S 2 T 1; S 2 T 2 and the other T 1 pairings remain permitted.
Answer Six unrestricted outfits and five after the stated restriction.
ORIGINAL PRACTICE · 1
There are four outward routes and three return routes, with every outward route compatible with every return route. How many ordered round-trip route choices exist?
Reveal answer & reasoning
12.
Each of four first-stage choices has three second-stage choices, so multiply 4×3. Compatibility is explicitly supplied.
02 / UNDERSTAND & APPLY
Roles make order matter
A permutation distinguishes order or assigned roles. Selecting a captain and deputy from five people gives five captain choices and four remaining deputy choices. An unordered two-person committee treats the same pair as one selection regardless of who was named first.
For n distinct objects, choosing r in order without repetition gives n(n−1)…(n−r+1). If order is irrelevant, divide by the r! arrangements of each selected set. Use the formula only when the distinctness and no-repetition conditions fit.
Repeated identical objects create fewer distinct arrangements. For AAB, swapping the two A copies produces no new visible word. Dividing 3! by 2! gives 3, which can be checked by listing AAB,ABA,BAA.
ORIGINAL WORKED EXAMPLE
Two roles or one committee
Five distinct people are eligible. Count assignments of captain and deputy, with no person holding both roles. Then count unordered two-person committees.
Captain has 5 choices, and deputy has 4 after captain is selected:20 assignments.
Each unordered pair appears twice in those assignments, once in each role order.
Divide 20 by 2 to obtain 10 committees.
Answer 20 role assignments and 10 unordered committees. The difference comes from whether exchanging the two selected people creates a new outcome.
ORIGINAL PRACTICE · 1
How many distinct arrangements can be made from the letters A,A,B?
Reveal answer & reasoning
Three: AAB,ABA,BAA.
The two A letters are indistinguishable. Treating them as separate labelled objects would count each displayed word twice.
ORIGINAL PRACTICE · 2
How many unordered committees of three can be chosen from four distinct people?
Reveal answer & reasoning
Four.
Each committee is determined by the one person left out. Equivalently,4 choose 3=4. Committee order is irrelevant.
03 / UNDERSTAND & APPLY
Count the easier event
The complement of an event contains all outcomes where it does not occur. If the full sample space is known, count or calculate the complement and subtract from the total or from 1. At least one success often has the simpler complement no successes.
Exactly one differs from at least one and at most one. For two trials, at least one includes one or two successes; exactly one excludes the double-success case. Adding two single-event probabilities without correcting their overlap can overcount.
ORIGINAL WORKED EXAMPLE
At least one six
Two independent fair six-sided dice are rolled. Find the probability of at least one 6.
The complement is no 6 on either die. Each die has probability 5/6 of not showing 6.
Adding 1/6+1/6 gives 12/36 but counts the double-six outcome twice; subtract its 1/36 to get 11/36.
Answer 11/36.
04 / UNDERSTAND & APPLY
Equal likelihood must be justified
For a finite model with equally likely elementary outcomes, probability is favourable outcome count divided by total outcome count. A list of outcome names is not enough to establish equal likelihood. On two fair coins, the four ordered outcomes HH,HT,TH,TT are equally likely, but the three head-count labels 0,1,2 are not: one head has two underlying outcomes.
An event’s probability lies between 0 and 1. Exhaustive disjoint elementary outcomes sum to 1. These checks can expose a missing branch or double-counted event. Probability describes uncertainty in the model, not a promise that a short experiment realizes exact proportions.
If selection is described as uniform from numbered objects, each object is equally likely. If physical objects differ in ways affecting selection, uniformity must be stipulated rather than inferred from the word random alone.
ORIGINAL PRACTICE · 1
A fair die is rolled. Find the probability of an even result and the probability of a result that is both even and greater than 4.
Reveal answer & reasoning
1/2 and 1/6 respectively.
Even outcomes are 2,4,6: three of six. Only 6 meets both conditions, so the intersection has one of six outcomes.
05 / UNDERSTAND & APPLY
Conditioning changes the denominator
P(A given B)=P(A and B)/P(B) when P(B)>0. In equally likely finite counting, retain only outcomes satisfying B, then count which of those also satisfy A. The conditioning event must have positive probability for this elementary formula to be defined.
The wording of the condition matters. At least one head and first toss head retain different outcome sets. Do not substitute a more specific observation for the one actually supplied.
The multiplication rule P(A and B)=P(A)×P(B given A) follows from conditional probability. It does not generally equal P(A)P(B) unless the events are independent.
ORIGINAL WORKED EXAMPLE
Two different coin conditions
Two independent fair coin tosses produce HH,HT,TH,TT with equal probability. Find the probability of two heads given at least one head, and given that the first toss is a head.
At least one head removes only TT, leaving HH,HT,TH. Exactly one of these three is HH, so probability 1/3.
First toss head leaves HH,HT, two equally likely outcomes. One is HH, so probability 1/2.
Both conditions have positive probability, but they are not the same information.
Answer The conditional probabilities are 1/3 and 1/2 respectively.
ORIGINAL PRACTICE · 1
Select uniformly from numbered cards 1–10. Given that the number is even, what is the probability it is greater than 6?
Reveal answer & reasoning
2/5.
The condition leaves 2,4,6,8,10. Favourable cards are 8 and 10. Use the conditional count 5, not the original 10.
06 / UNDERSTAND & APPLY
Independence is not mutual exclusion
Events are independent when learning one does not change the probability of the other, equivalently P(A and B)=P(A)P(B). Mutually exclusive events cannot occur together. If both have positive probability, exclusivity makes them dependent: observing one rules out the other.
Replacement and thorough remixing can restore a bag’s original composition between draws, supporting identical independent draw probabilities under a uniform-selection model. Without replacement, the composition changes and the second draw probability depends on the first result.
Independent repeated trials do not owe a balancing result. Three heads in a row do not make a fair independent coin more likely to show tails next; the next-head probability remains 1/2. Long-run tendencies do not impose short-run compensation.
ORIGINAL PRACTICE · 1
On one fair die roll, are the events even and odd independent?
Reveal answer & reasoning
No. They are mutually exclusive and each has positive probability.
Their intersection probability is 0, while the product of their probabilities is(1/2)(1/2)=1/4. Observing even makes odd impossible.
ORIGINAL PRACTICE · 2
A bag has three red and two blue balls. Draw uniformly, replace and remix, then draw again independently. What is the probability of two reds?
Reveal answer & reasoning
9/25.
Each draw has red probability 3/5 because replacement restores composition. Multiply(3/5)². Without replacement, the second red probability would instead be 2/4.
ORIGINAL PRACTICE · 3
An independent fair coin has just shown three heads. What is the probability the next toss is a head?
Reveal answer & reasoning
1/2.
The model states independence and fairness, so the earlier sequence does not change the next-toss probability. A compensating tail is not required.
07 / UNDERSTAND & APPLY
Multiply along paths, add disjoint paths
A tree makes sequential conditional probabilities explicit. Each node’s outgoing branches should sum to 1, and every complete path corresponds to a disjoint ordered outcome. Multiply the branch probabilities on a path to obtain its joint probability; add different paths when they satisfy the same event.
Without replacement, decrement both the total count and the count of the colour drawn. Label all branches, including outcomes you do not immediately need, so the model can be checked for completeness. A tree that keeps the original denominator on the second draw would model a different process.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Original uniform no-replacement model. Branch labels are conditional probabilities; leaf labels are joint probabilities.
ORIGINAL WORKED EXAMPLE
Two draws without replacement
A bag contains three red and two blue balls. Draw two balls uniformly in sequence without replacement. Find probabilities of RR,RB,BR,BB, exactly one red and at least one red.
First red probability 3/5. After red, two red and two blue remain: second red 2/4 and blue 2/4. Thus RR=3/10 and RB=3/10.
First blue probability 2/5. After blue, three red and one blue remain: second red 3/4 and blue 1/4. Thus BR=3/10 and BB=1/10.
Exactly one red consists of RB or BR, disjoint paths, so 3/10+3/10=3/5.
At least one red is the complement of BB:1−1/10=9/10. Check all four path probabilities sum 1.
Answer RR,RB,BR each have probability 3/10; BB has 1/10. Exactly one red is 3/5, and at least one red is 9/10.
08 / UNDERSTAND & APPLY
Audit the sample space
Before calculating, state what counts as one outcome, whether order matters, whether objects are distinguishable, how selection occurs and whether replacement happens. Then match the requested event exactly, including conditions such as at least or given.
After calculating, check the probability range, exhaustive totals and a simple alternative method where available. Symmetry can simplify a model only when the outcomes really are symmetric under its rules. A biased coin or nonuniform selection breaks the equal-count shortcut.
Keep a probability distinct from a guarantee. A 9/10 event can fail on the next trial, and an unlikely sequence is not impossible. These original finite models teach careful counting rather than forecasts of any real sequence.
KEEP THE BIG IDEAS
Quick revision
Order and distinguishability determine what you count.
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