Probability is the mathematical measure of how likely an event is to occur. The scale runs from 0 (impossible) to 1 (certain). Everything in between is a fraction.
Here is the core formula — and it never changes:
Think of it like this: you are standing at a chowk (intersection) with 10 roads. Three of those roads lead to Delhi. If you pick a road blindfolded, the probability of reaching Delhi is 3/10. The "favourable" roads are 3; the total roads are 10.
Key vocabulary you must lock in:
Boundaries of probability:
A probability answer greater than 1 or negative is always wrong — eliminate such options immediately in the exam hall.
The complement rule deserves its own line because UP Police PYQs use it repeatedly:
This flips a hard counting problem into a simple subtraction. You will see this pattern in coins, cards, and committee questions. The moment you read "at least one," reach for the complement.
Standard sample spaces you should know cold:
| Experiment | Total Outcomes | |---|---| | 1 coin toss | 2 | | 2 coin tosses | 4 | | 3 coin tosses | 8 | | 1 die roll | 6 | | 2 dice rolls | 36 | | 1 card from 52-card deck | 52 |
The entire difficulty of probability questions lies in correctly counting favourable outcomes. The formula is simple; the counting is where mistakes happen. Work through every question in two steps: (1) find total outcomes, (2) find favourable outcomes. Never mix the two steps.
These are the most common type in UP Police Constable. A bag has objects of different types — count everything first.
Total = sum of all object counts. Favourable = count of objects matching the condition.
Example structure: "Basket has 6 apples, 3 mangoes, 5 oranges. Find P(orange)."
Total = 14. Favourable (orange) = 5. P = 5/14.
For "neither A nor B" conditions: favourable = everything except A and B = total − count(A) − count(B).
For coin tosses, total outcomes = .
For "at least one head in tosses":
For 2 tosses: . For 3 tosses: .
A standard deck has 52 cards. Know this structure:
P(Ace) = 4/52 = 1/13. Memorise this directly — it appears in PYQs.
P(Face card) = 12/52 = 3/13.
P(Red card) = 26/52 = 1/2.
These are pure counting questions dressed as probability.
Days of the week: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday = 7 days.
Days starting with 'S': Saturday, Sunday = 2. Days starting with 'W': Wednesday = 1. Total starting with 'S' or 'W' = 3. P = 3/7.
Months of the year = 12. Months starting with 'J': January, June, July = 3. P = 3/12 = 1/4.
The formula solves two types of questions directly:
Direct: "P(event) = 0.55. Find P(event does not happen)." → . Done in 3 seconds.
Indirect (at least one): "P(at least one woman in committee)" → Calculate P(no women), subtract from 1.
These use .
Structure:
Example: 5 men + 4 women, committee of 3.
| Expression | Value | |---|---| | | 84 | | | 10 | | | 6 | | | 15 | | | 52 |
When two events A and B can both happen:
In counting form: favourable(A or B) = favourable(A) + favourable(B) − favourable(both A and B).
Use this for "perfect square OR multiple of 6" type questions. Never add without subtracting the overlap.
Any time a question asks "at least one [X]", flip it: calculate P(zero X) and subtract from 1.
Example — 2 coins, at least one head:
For 3 coins, direct listing takes 60s+ (8 outcomes to scan). Complement takes 10s: P(no heads) = 1/8, answer = 7/8. Standard method: 60s, shortcut: 10s.
Memorise these five card fractions — they cover 90% of card PYQs:
When you see a card question, match it to one of these before you start calculating. Standard calculation: 20s. Memory recall: 3s.
For "neither A nor B" questions, do not add favourable counts — subtract the excluded ones from the total.
Basket: 6 apples, 3 mangoes, 5 oranges. P(neither apple nor mango)?
Wrong approach (common mistake): add apple + mango = 9 and put it as numerator. Correct approach: Total = 14. Excluded = 6 + 3 = 9. Favourable = 14 − 9 = 5. P = 5/14.
This saves you from inverting the fraction by accident. Step count: subtract once vs. risk of selecting wrong numerator. Saves 1 error-correction cycle ≈ 30s.
For "day starting with letter X" questions, run through the week in this fixed order: Mon, Tue, Wed, Thu, Fri, Sat, Sun. Mark matches, count them.
Days starting with 'S': Sat, Sun → 2. Days starting with 'W': Wed → 1. Combined = 3, P = 3/7.
Never rely on memory alone — write the 7 days and tick the matching ones. This takes 12s and eliminates miscounts. Standard guesswork approach with errors: 45s + correction.
Before writing your final answer, check: is the denominator equal to the total sample space you stated? If not, you have made an arithmetic error somewhere.
For basket (total 14): denominator must be 14. For 2-coin toss (total 4): denominator must be 4. For 52-card deck: denominator must be 52 (or a simplified factor of 52).
This single check catches fraction inversion errors and wrong-total errors. It adds 5s and saves you from picking the wrong option. When options include both 5/14 and 9/14, this check tells you immediately which denominator is valid.
Read the question. Identify the experiment type in 5 seconds.
Is it a basket/bag question? → Total = sum of all items. Favourable = items matching condition (or total minus excluded items for "neither" questions). Write fraction, simplify.
Is it a coin/die question? → Use for coins, for dice. If "at least one X", use complement: .
Is it a card question? → Match to one of five memorised fractions. If not a direct match, total = 52, count favourables.
Is it a days/months/letters question? → Write the full list (7 days or 12 months), tick the qualifying ones, count.
Is it a committee/selection question? → Use . If "at least one woman", compute .
Is it a direct complement question? → . One subtraction.
Sanity check every answer: Is ? Is the denominator correct? Eliminate any option greater than 1 — it cannot be a probability.
Why this question: The simplest basket-type. Sets up the "favourable from total" framework cleanly.
Solving path: Total fruits = 6 + 3 + 5 = 14. "Neither apple nor mango" = only oranges qualify. Oranges = 5. P = 5/14. Scan options — 5/14 is option A. Done in 15 seconds. The trap here is option B (6/14) which is P(apple), not P(orange). Do not pick the first even-sounding fraction.
Why this question: Introduces the complement method for "at least one" in a combinatorics setting. High-value question type — harder than it looks at first glance.
Solving path: Total ways to choose 3 from 9 people = . Ways with zero women (all 3 from 5 men) = . P(no women) = 10/84. P(at least one woman) = . Option B. Notice option A (84/37) is greater than 1 — eliminate immediately.
Why this question: The most classic probability question. Every aspirant should solve this in under 10 seconds using the complement.
Solving path: 2 coin tosses → total outcomes = 4. Complement: P(no heads) = P(TT) = 1/4. P(at least one head) = 1 − 1/4 = 3/4. Option B. Option D (3/2) is greater than 1 — eliminate before reading other options.
Why this question: Tests your knowledge of the 7 days and letter-matching. Simple but students miscounting 'S' days drop marks here.
Solving path: Write the 7 days: Mon, Tue, Wed, Thu, Fri, Sat, Sun. Starting with 'W': Wednesday (1). Starting with 'S': Saturday, Sunday (2). Total qualifying = 3. P = 3/7. Option A. Common mistake: forgetting Sunday starts with 'S' and writing 2/7 (option D).
Why this question: The purest complement question. Tests whether you know directly.
Solving path: P(event does not happen) = 1 − 0.55 = 0.45. Option D. No setup needed. If you spent more than 8 seconds on this, you are over-thinking it. The trap is option A (0.30), which has no mathematical basis — it is there to catch guessers.
Why this question: A foundational card question. Tests whether you know the deck structure.
Solving path: 4 aces in 52 cards. P(Ace) = 4/52 = 1/13. Option B. This should take 5 seconds if you have memorised card fractions. Option D (1/52) is the probability of drawing one specific ace — a common confusion. Option A (3/13) is P(face card) misremembered.
Writing favourable count as denominator instead of total. In the basket question, writing P = 5/9 (oranges over non-orange items) instead of 5/14 (oranges over all fruits). Always use the complete sample space as the denominator.
Forgetting Sunday when counting 'S' days. Students list Saturday and stop. Sunday also starts with 'S'. Write out all 7 days explicitly — do not rely on recall alone.
Not applying complement for "at least one." Directly trying to count "at least one woman in committee" requires adding P(exactly 1 woman) + P(exactly 2 women) + P(exactly 3 women) — three separate calculations. The complement (1 − P(all men)) requires one calculation. Not using the complement costs 90+ seconds on this question type.
Accepting options greater than 1 as answers. In the committee question, option A is 84/37 which is greater than 1. A probability can never exceed 1. Train yourself to eliminate such options in under 2 seconds without even reading the rest.
Confusing values. and are the two values that appear in the committee PYQ. If you compute correctly but as 15 (which is ), your final answer is wrong. Practice these values until they are immediate.
Using P(A and B) instead of P(A or B) for inclusion-exclusion questions. When a question says "perfect square or multiple of 6," you need to add both counts and subtract the overlap. Writing just P(perfect square) + P(multiple of 6) without subtracting the shared elements (multiples of 36) inflates your answer. The word "or" always triggers inclusion-exclusion.