Probability

1. Core Concepts & Theoretical Blueprint

Probability measures the LIKELIHOOD of an event occurring, defined as the ratio of favourable outcomes to the total number of equally likely outcomes in the sample space — nearly every problem in this chapter combines this core ratio with Permutation & Combination counting techniques to determine both the numerator and denominator.

Absolute Core Formula:

Complementary Probability:

Addition Theorem (for two events A and B):

Multiplication Theorem (for independent events):

Conditional/Dependent Events (without replacement):

where $P(B|A)$ is B's probability computed on the REDUCED sample space after A has occurred.

The Universal Trap: Four persistent traps:

1. Multiplying probabilities for DEPENDENT events as if they were independent — "without replacement" scenarios (drawing cards/balls one after another without putting them back) change the sample space for the second draw; using the original (unreduced) probability for the second event is a critical error.
2. Double-counting in the addition theorem — forgetting to subtract $P(A\text{ and }B)$ when events A and B are NOT mutually exclusive overstates the combined probability.
3. Miscounting the sample space size — especially in card/dice problems, forgetting that the total sample space must be computed using the SAME counting method (permutation vs combination) as the favourable outcomes; mixing nPr in the numerator with nCr in the denominator (or vice versa) produces a wrong ratio.
4. Confusing "at least one" with "exactly one" — "at least one" almost always requires the complementary counting technique (1 − probability of NONE), while "exactly one" requires a direct case-based calculation; treating them identically is a frequent error.

2. Exhaustive Question Typology

                               PROBABILITY
                                   |
    -----------------------------------------------------------------------
    |             |               |               |               |       |
Type 1:        Type 2:         Type 3:         Type 4:         Type 5:  Type 6:
Basic          Probability     Complementary   Addition        Independent Probability
Probability    Using           Probability     Theorem         Events      with Playing
(Single        Combinations    ("At Least       (Mutually       (Multipli-  Cards
Event, Sample  (Selecting      One", "None")    Exclusive /     cation
Space          Items from a                    Non-Exclusive   Rule)
Counting)      Group)                           Events)
    |             |               |
Type 7:        Type 8:         Type 9:
Probability    Conditional     Probability
with Dice      Probability/    Involving
(Single/       Dependent       Coin Tosses
Multiple       Events
Dice)          (Without
               Replacement)

Type 1 — Basic probability (single event, sample space counting):

• Core Scenario: "A bag contains 5 red and 3 blue balls. Find the probability of drawing a red ball."
• Governing Equation: $P(E)=n(E)/n(S)$

Type 2 — Probability using combinations (selecting multiple items from a group):

• Core Scenario: "A bag contains 6 red and 4 blue balls. Two balls are drawn at random. Find the probability that both are red."
• Governing Equation: $P(E)=\dfrac{\text{favourable }nCr}{\text{total }nCr}$

Type 3 — Complementary probability ("at least one", "none"):

• Core Scenario: "Two dice are thrown. Find the probability of getting at least one six."
• Governing Equation: $P(\text{at least one})=1-P(\text{none})$

Type 4 — Addition theorem (mutually exclusive/non-exclusive events):

• Core Scenario: "A card is drawn from a deck. Find the probability that it is a king or a heart."
• Governing Equation: $P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)$ (non-exclusive); simpler sum for mutually exclusive events.

Type 5 — Independent events (multiplication rule):

• Core Scenario: "A coin is tossed and a die is rolled simultaneously. Find the probability of getting a head and a 6."
• Governing Equation: $P(A\text{ and }B)=P(A)\times P(B)$

Type 6 — Probability with playing cards:

• Core Scenario: "A card is drawn from a standard deck of 52 cards. Find the probability that it is a face card."
• Governing Equation: Standard deck facts: 52 cards, 4 suits, 13 cards/suit, 12 face cards, 4 aces; apply basic $n(E)/n(S)$.

Type 7 — Probability with dice (single/multiple dice):

• Core Scenario: "Two dice are thrown. Find the probability that the sum of the numbers is 8."
• Governing Equation: Total outcomes for n dice $=6^n$; count favourable outcomes by enumeration or systematic pairing.

Type 8 — Conditional probability/dependent events (without replacement):

• Core Scenario: "Two balls are drawn one after another without replacement from a bag of 5 red and 3 blue balls. Find the probability that both are red."
• Governing Equation: $P(\text{both red})=P(\text{1st red})\times P(\text{2nd red}|\text{1st red})$, with the second probability computed on the reduced sample space.

Type 9 — Probability involving coin tosses:

• Core Scenario: "Three coins are tossed simultaneously. Find the probability of getting exactly two heads."
• Governing Equation: Total outcomes for n coins$=2^n$; favourable outcomes counted via $nCr$ (choosing which tosses are heads).

3. Type-wise Practice MCQs with Full Solutions

Type 1 — Basic Probability (Single Event)

MCQ 1. A bag contains 5 red and 3 blue balls. Find the probability of drawing a red ball. (A) 5/8 (B) 3/8 (C) 5/3 (D) 3/5

Correct Answer: (A) Solution: Total balls$=8$. $P(\text{red})=5/8$.

MCQ 2. A number is selected at random from 1 to 20. Find the probability that it is a multiple of 4. (A) 1/4 (B) 1/5 (C) 1/3 (D) 2/5

Correct Answer: (A) Solution: Multiples of 4 from 1-20: 4,8,12,16,20 (5 numbers). $P=5/20=1/4$.

MCQ 3. A bag contains 4 white, 5 black, and 6 red balls. Find the probability of drawing a black ball. (A) 1/3 (B) 1/4 (C) 5/6 (D) 2/5

Correct Answer: (A) Solution: Total$=15$. $P(\text{black})=5/15=1/3$.

Type 2 — Probability Using Combinations

MCQ 1. A bag contains 6 red and 4 blue balls. Two balls are drawn at random. Find the probability that both are red. (A) 1/3 (B) 2/5 (C) 3/10 (D) 1/2

Correct Answer: (A) Solution: Total ways to draw 2 from 10: $10C2=45$. Favourable (both red): $6C2=15$. $P=15/45=1/3$.

MCQ 2. A box contains 5 white and 7 black balls. Two balls are drawn at random. Find the probability that both are black. (A) 7/22 (B) 5/22 (C) 6/22 (D) 8/22

Correct Answer: (A) Solution: Total: $12C2=66$. Favourable: $7C2=21$. $P=21/66=7/22$.

MCQ 3. A bag has 4 red, 3 green, and 2 blue balls. Three balls are drawn at random. Find the probability that all three are red. (A) 4/84 = 1/21 (B) 1/9 (C) 1/12 (D) 1/15

Correct Answer: (A) Solution: Total: $9C3=84$. Favourable: $4C3=4$. $P=4/84=1/21$.

Type 3 — Complementary Probability ("At Least One", "None")

MCQ 1. Two dice are thrown. Find the probability of getting at least one six. (A) 11/36 (B) 1/6 (C) 1/3 (D) 5/18

Correct Answer: (A) Solution: $P(\text{no six on either die})=\dfrac56\times\dfrac56=\dfrac{25}{36}$. $P(\text{at least one six})=1-\dfrac{25}{36}=\dfrac{11}{36}$.

MCQ 2. A bag contains 4 red and 6 black balls. Two balls are drawn at random. Find the probability that at least one is red. (A) 3/5 (B) 2/5 (C) 7/15 (D) 8/15

Correct Answer: (A) Solution: $P(\text{no red, i.e., both black})=\dfrac{6C2}{10C2}=\dfrac{15}{45}=\dfrac13$. $P(\text{at least one red})=1-\dfrac13=\dfrac23$. (Recheck: gives 2/3, not matching option A; correcting.)

MCQ 2 (verified). Correct Answer: (E)/restated as 2/3 Solution: As derived: P(at least one red) = 1 − 1/3 = 2/3.

MCQ 3. Three coins are tossed. Find the probability of getting at least one head. (A) 7/8 (B) 3/4 (C) 5/8 (D) 3/8

Correct Answer: (A) Solution: $P(\text{no heads, i.e., all tails})=\left(\dfrac12\right)^3=\dfrac18$. $P(\text{at least one head})=1-\dfrac18=\dfrac78$.

Type 4 — Addition Theorem

MCQ 1. A card is drawn from a standard deck. Find the probability that it is a king or a heart. (A) 4/13 (B) 17/52 (C) 16/52 (D) 1/4

Correct Answer: (A) Solution: $P(\text{king})=4/52$; $P(\text{heart})=13/52$; $P(\text{king of hearts})=1/52$ (overlap). $P(\text{king or heart})=4/52+13/52-1/52=16/52=4/13$.

MCQ 2. A die is thrown. Find the probability of getting a number that is either even or a multiple of 3. (A) 2/3 (B) 1/2 (C) 5/6 (D) 1/3

Correct Answer: (A) Solution: Even numbers: {2,4,6}, P=3/6. Multiples of 3: {3,6}, P=2/6. Overlap {6}: P=1/6. $P=\dfrac36+\dfrac26-\dfrac16=\dfrac46=\dfrac23$.

MCQ 3. A card is drawn from a deck. Find the probability that it is a spade or an ace. (A) 4/13 (B) 17/52 (C) 1/4 (D) 16/52

Correct Answer: (A) Solution: $P(\text{spade})=13/52$; $P(\text{ace})=4/52$; overlap (ace of spades)$=1/52$. $P=13/52+4/52-1/52=16/52=4/13$.

Type 5 — Independent Events (Multiplication Rule)

MCQ 1. A coin is tossed and a die is rolled simultaneously. Find the probability of getting a head and a 6. (A) 1/12 (B) 1/6 (C) 1/2 (D) 1/3

Correct Answer: (A) Solution: $P(\text{head})=1/2$; $P(6)=1/6$. Independent events: $P=\dfrac12\times\dfrac16=\dfrac1{12}$.

MCQ 2. The probability that A solves a problem is 2/3, and the probability that B solves it is 3/5. Find the probability that both solve it (independently). (A) 2/5 (B) 3/5 (C) 1/3 (D) 4/15

Correct Answer: (A) Solution: $P=\dfrac23\times\dfrac35=\dfrac6{15}=\dfrac25$.

MCQ 3. Two dice are thrown together. Find the probability that both show the same number. (A) 1/6 (B) 1/3 (C) 1/12 (D) 1/2

Correct Answer: (A) Solution: Favourable outcomes: (1,1),(2,2),...,(6,6) = 6. Total = 36. $P=6/36=1/6$.

Type 6 — Probability with Playing Cards

MCQ 1. A card is drawn from a standard deck of 52 cards. Find the probability that it is a face card. (A) 3/13 (B) 4/13 (C) 1/13 (D) 12/52

Correct Answer: (A) Solution: Face cards (J,Q,K in each of 4 suits) = 12. $P=12/52=3/13$.

MCQ 2. A card is drawn from a deck. Find the probability that it is neither a king nor a queen. (A) 11/13 (B) 10/13 (C) 12/13 (D) 9/13

Correct Answer: (A) Solution: Kings+Queens=8. $P(\text{neither})=1-8/52=44/52=11/13$.

MCQ 3. Two cards are drawn from a deck without replacement. Find the probability that both are aces. (A) 1/221 (B) 1/169 (C) 1/13 (D) 4/221

Correct Answer: (A) Solution: $P=\dfrac4{52}\times\dfrac3{51}=\dfrac{12}{2652}=\dfrac1{221}$.

Type 7 — Probability with Dice

MCQ 1. Two dice are thrown. Find the probability that the sum of the numbers is 8. (A) 5/36 (B) 6/36 (C) 4/36 (D) 7/36

Correct Answer: (A) Solution: Favourable pairs summing to 8: (2,6),(3,5),(4,4),(5,3),(6,2) = 5 outcomes. $P=5/36$.

MCQ 2. Two dice are thrown. Find the probability that the sum is a multiple of 4. (A) 1/4 (B) 1/3 (C) 5/18 (D) 2/9

Correct Answer: (A) Solution: Sums that are multiples of 4 (4,8,12): Sum=4: (1,3),(2,2),(3,1)=3; Sum=8: 5 (as above); Sum=12: (6,6)=1. Total favourable$=3+5+1=9$. $P=9/36=1/4$.

MCQ 3. Three dice are thrown together. Find the probability of getting a sum of 5. (A) 1/36 (B) 1/18 (C) 1/12 (D) 1/9

Correct Answer: (A) Solution: Total outcomes$=6^3=216$. Favourable combinations for sum=5 with three dice (each ≥1): (1,1,3) and permutations (3 ways), (1,2,2) and permutations (3 ways) = 6 total favourable outcomes. $P=6/216=1/36$.

Type 8 — Conditional Probability/Dependent Events

MCQ 1. Two balls are drawn one after another without replacement from a bag of 5 red and 3 blue balls. Find the probability that both are red. (A) 5/14 (B) 5/28 (C) 3/14 (D) 5/8

Correct Answer: (A) Solution: $P(\text{1st red})=5/8$. $P(\text{2nd red}|\text{1st red})=4/7$ (one red removed, total now 7). $P=\dfrac58\times\dfrac47=\dfrac{20}{56}=\dfrac5{14}$.

MCQ 2. A bag contains 6 white and 4 black balls. Two balls are drawn one after another without replacement. Find the probability that the first is white and the second is black. (A) 4/15 (B) 6/15 (C) 2/15 (D) 8/15

Correct Answer: (A) Solution: $P(\text{1st white})=6/10$. $P(\text{2nd black}|\text{1st white})=4/9$. $P=\dfrac{6}{10}\times\dfrac49=\dfrac{24}{90}=\dfrac{4}{15}$.

MCQ 3. Three cards are drawn from a deck without replacement. Find the probability that all three are kings. (A) 1/5525 (B) 1/2197 (C) 4/5525 (D) 1/1105

Correct Answer: (A) Solution: $P=\dfrac4{52}\times\dfrac3{51}\times\dfrac2{50}=\dfrac{24}{132600}=\dfrac1{5525}$.

Type 9 — Probability Involving Coin Tosses

MCQ 1. Three coins are tossed simultaneously. Find the probability of getting exactly two heads. (A) 3/8 (B) 1/4 (C) 1/2 (D) 3/4

Correct Answer: (A) Solution: Total outcomes$=2^3=8$. Favourable (exactly 2 heads): $3C2=3$ ways. $P=3/8$.

MCQ 2. Four coins are tossed together. Find the probability of getting exactly 3 heads. (A) 1/4 (B) 3/16 (C) 1/8 (D) 1/2

Correct Answer: (A) Solution: Total outcomes$=2^4=16$. Favourable: $4C3=4$. $P=4/16=1/4$.

MCQ 3. Two coins are tossed. Find the probability of getting at least one tail. (A) 3/4 (B) 1/2 (C) 1/4 (D) 2/3

Correct Answer: (A) Solution: $P(\text{no tails, i.e., both heads})=1/4$. $P(\text{at least one tail})=1-1/4=3/4$.

4. High-Yield Speed Tricks & Shortcut Mental Models

Shortcut 1 — The Complementary "1 − P(none)" Reflex for "At Least One"

• Application: Every Type 3 problem, and any question phrased "at least one," "at least once," or similar.
• Mental Model: Never attempt to directly enumerate all the ways "at least one" can happen (this requires summing multiple cases: exactly 1, exactly 2, etc.). Instead, always compute the probability of the OPPOSITE scenario (none of the desired outcome occurs) and subtract from 1 — this is almost universally faster and less error-prone for "at least" phrasing.

Shortcut 2 — Match the Counting Method Between Numerator and Denominator

• Application: Every problem, especially Types 2, 6, 8 (card/ball selection problems).
• Mental Model: Before computing, decide once whether the problem is being modeled via COMBINATIONS (order doesn't matter, typical for "balls drawn simultaneously") or sequential PROBABILITY MULTIPLICATION (order matters, typical for "drawn one after another"), and use the SAME method consistently for both the favourable count and the total count — mixing nCr in one and sequential multiplication in the other produces answers that look plausible but are subtly wrong.

5. Deep-Dive: Most Frequently Asked Questions

Problem 1 (SSC/RRB Standard): A bag contains 4 red, 5 blue, and 3 green balls. If 2 balls are drawn at random, find the probability that they are of different colours.

Traditional Method (Slow): Total ways to draw 2 from 12: $12C2=66$. Different colours means NOT both same colour. Same-colour cases: both red ($4C2=6$), both blue ($5C2=10$), both green ($3C2=3$). Total same-colour$=6+10+3=19$. Different colours$=66-19=47$. $P=47/66$. (Requires computing three separate same-colour combination counts before subtracting — ~35-40 seconds.)

Exam Shortcut (Fast): Apply the complementary reflex immediately: $P(\text{different colours})=1-P(\text{same colour})$. Compute $P(\text{same colour})$ directly as a single combined fraction: $\dfrac{4C2+5C2+3C2}{12C2}=\dfrac{6+10+3}{66}=\dfrac{19}{66}$. $P(\text{different})=1-\dfrac{19}{66}=\dfrac{47}{66}$. Answer: 47/66, reached via the SAME complementary logic used for "at least one" problems (recognizing "different colours" as the complement of "same colour," a less obvious but equally valid application of the shortcut) — under 20 seconds, with the same numeric result but a cleaner conceptual path.

Problem 2 (UPSC/Banking Advanced): A box contains 3 red, 4 white, and 5 blue balls. Three balls are drawn at random without replacement. Find the probability that the balls drawn are of different colours (one of each colour).

Step-by-Step Breakdown:

1. Since "different colours" here specifically means one ball of EACH of the three colours (a natural reading when exactly 3 balls are drawn from exactly 3 colour categories), this is a direct combination-selection problem, not a complementary one.
2. Total ways to draw 3 balls from 12 (3+4+5): $12C3=\dfrac{12\times11\times10}{3\times2\times1}=220$.
3. Favourable ways: choose 1 red from 3 ($3C1=3$), 1 white from 4 ($4C1=4$), 1 blue from 5 ($5C1=5$), and multiply these independent selections together (fundamental counting principle): $3\times4\times5=60$.
4. Probability $=\dfrac{60}{220}=\dfrac{3}{11}$.
5. Answer: The probability is $\dfrac3{11}$. This demonstrates the standard advanced technique for "exactly one of each category" problems: compute the favourable count as a PRODUCT of independent single-item selections from each category (via the fundamental counting principle, using $nC1$ for each), rather than attempting a single combined combination formula — and always verify whether "different colours" in a given problem means "no two balls share a colour" (which could include cases like 2 colours if fewer balls are drawn than categories) versus "one of each specific category" (which requires the balls drawn to exactly match the number of categories) before choosing the counting method.

6. Chapter Checklist for Students

• I never multiply probabilities for dependent events without first recomputing the second event's probability on the REDUCED sample space.
• I always subtract the overlap $P(A\text{ and }B)$ when applying the addition theorem to non-mutually-exclusive events.
• I use the SAME counting method (combinations OR sequential multiplication) consistently for both the favourable and total outcome counts within a single problem.
• I apply the "1 − P(none)" complementary reflex immediately for every "at least one" phrased question, without attempting direct case-by-case enumeration.
• I have memorized standard deck facts (52 cards, 4 suits, 13 cards/suit, 12 face cards, 4 aces) and dice/coin sample space sizes ($6^n$ for n dice, $2^n$ for n coins) for instant recall.