Important Notes: Probability
🔑 Fundamental Concepts
- Probability Range: For any event E,
0 ≤ P(E) ≤ 1. Probability can never be negative or greater than 1. - Complementary Events:
P(E) + P(not E) = 1. The probability of an event and its complement always sum to unity. - Impossible Event: An event that cannot happen has probability
0. - Certain Event: An event that is sure to happen has probability
1. - Sum of Elementary Events: The sum of probabilities of all elementary events in an experiment equals
1.
📐 Classical Probability Formula
The probability of an event E is defined as:
P(E) = Number of favourable outcomes / Total number of possible outcomes
🎲 Equally Likely Outcomes
Outcomes are equally likely when each has the same chance of occurring. Examples: fair coin toss (Head/Tail each 1/2), unbiased die (each face 1/6).
⚠️ Critical Insight: Elementary vs Compound Events
- When throwing two dice, there are 36 equally likely elementary outcomes (ordered pairs like (1,2), (2,1), etc.).
- The sums 2 through 12 are compound events with different probabilities. Do NOT assume each sum has probability 1/11.
- Always count elementary outcomes for correct probability calculation.
📊 Two Dice Probability Table
2
3
4
5
6
7
8
9
10
11
12
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
🧮 Probability Calculators
Basic Probability Calculator
Calculate P(E) = Favourable / Total outcomes.
Complementary Probability
Given P(E), find P(not E) = 1 − P(E).
Two Dice Sum Probability
Find the probability of getting a specific sum when two dice are thrown.
🎲 Interactive Visualizers
Coin Toss Simulator
H
T
Heads: 0 |
Tails: 0 |
Total: 0
Die Roll Simulator
1: 0
2: 0
3: 0
4: 0
5: 0
6: 0
Number Spinner (1-8)
?
Each number has probability 1/8
Questions 1 – 5
1
Complete the following statements:
(i) Probability of an event E + Probability of the event 'not E' = ______
✅ Correct Answer: 1
Complementary Events Rule: For any event E and its complement 'not E', the sum of their probabilities equals unity.
P(E) + P(not E) = 1
Example: If P(rain) = 0.3, then P(no rain) = 1 − 0.3 = 0.7. Together they cover all possibilities.
(ii) The probability of an event that cannot happen is ______. Such an event is called ______.
✅ Correct Answer: 0; Impossible event
An event that cannot happen under any circumstances has probability 0.
Such an event is called an impossible event. Example: Rolling a 7 on a standard die is impossible.
(iii) The probability of an event that is certain to happen is ______. Such an event is called ______.
✅ Correct Answer: 1; Sure event (or Certain event)
An event that is guaranteed to happen has probability 1.
Such an event is called a sure event or certain event. Example: Getting a number less than 7 when rolling a die is certain.
(iv) The sum of the probabilities of all the elementary events of an experiment is ______.
✅ Correct Answer: 1
Elementary events are the simplest outcomes of an experiment that cannot be broken down further.
The total probability space must cover all possible outcomes. Therefore, the sum equals 1.
Example: For a die, P(1)+P(2)+P(3)+P(4)+P(5)+P(6) = 1/6 × 6 = 1.
(v) The probability of an event is greater than or equal to ______ and less than or equal to ______.
✅ Correct Answer: 0 and 1
Probability is always bounded: 0 ≤ P(E) ≤ 1
0 represents impossibility, 1 represents certainty. No probability can be negative or exceed 1.
2
Which experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. (ii) A basketball shot. (iii) A true-false question. (iv) A baby is born (boy or girl).
✅ Correct Answer: Only (iii) and (iv)
(i) Car starting: NOT equally likely. Depends on battery, fuel, mechanical condition. P(start) ≠ P(not start).
(ii) Basketball shot: NOT equally likely. Depends on player skill, distance, defense. P(shoot) ≠ P(miss).
(iii) True-False question: Equally likely. Only two options with no bias → P(right) = P(wrong) = 1/2.
(iv) Baby gender: Equally likely. Biologically, P(boy) ≈ P(girl) ≈ 1/2.
3
Why is tossing a coin considered fair for deciding which team gets the ball?
✅ Correct Answer: Two equally likely outcomes, 50% each
A fair coin has exactly two possible outcomes: Heads and Tails.
Assuming the coin is unbiased: P(Heads) = P(Tails) = 1/2 = 0.5
Each team gets an equal and unbiased 50% probability of winning the toss. No team has an advantage.
4
Which of the following cannot be the probability of an event?
(A) 2/3 (B) −1.5 (C) 15% (D) 0.7
✅ Correct Answer: (B) −1.5
Probability must satisfy: 0 ≤ P(E) ≤ 1
(A) 2/3 ≈ 0.667 → Valid (between 0 and 1)
(B) −1.5 → INVALID — negative probability is impossible
(C) 15% = 0.15 → Valid
(D) 0.7 → Valid
5
If P(E) = 0.05, what is the probability of 'not E'?
✅ Correct Answer: 0.95
Using the complementary event formula: P(E) + P(not E) = 1
Given P(E) = 0.05:
P(not E) = 1 − 0.05 = 0.95
There is a 95% chance that event E will not occur.
Questions 6 – 10
6
A bag contains lemon flavoured candies only. Malini takes out one candy without looking. Find the probability of:
(i) an orange flavoured candy? (ii) a lemon flavoured candy?
✅ Correct Answer: (i) 0, (ii) 1
Given: The bag contains only lemon flavoured candies.
(i) Orange flavoured candy: P(orange) = 0/Total = 0 (Impossible event — no orange candies exist in the bag)
(ii) Lemon flavoured candy: P(lemon) = Total/Total = 1 (Certain event — every candy is lemon)
Since no other flavour exists, getting lemon is guaranteed.
7
In a group of 3 students, P(2 students not having same birthday) = 0.992. Find P(2 students have same birthday).
✅ Correct Answer: 0.008
Let E = event that 2 students have the same birthday
Let not E = event that 2 students do not have the same birthday
P(E) + P(not E) = 1
Given: P(not E) = 0.992
P(E) = 1 − 0.992 = 0.008
There is only a 0.8% chance that two students share the same birthday.
8
A bag contains 3 red balls and 5 black balls. A ball is drawn at random. Find P(red) and P(not red).
✅ Correct Answer: 3/8 and 5/8
Total balls = 3 (red) + 5 (black) = 8
(i) P(red): P(red) = 3/8 = 0.375
(ii) P(not red): P(not red) = 1 − 3/8 = 5/8 = 0.625
Alternatively: P(not red) = P(black) = 5/8
9
A box contains 5 red, 8 white and 4 green marbles. One marble is taken out at random. Find P(red), P(white), P(not green).
✅ Correct Answer: 5/17, 8/17, 13/17
Total marbles = 5 + 8 + 4 = 17
(i) P(red) = 5/17
(ii) P(white) = 8/17
(iii) P(not green) = P(red) + P(white) = (5+8)/17 = 13/17
Or: P(not green) = 1 − P(green) = 1 − 4/17 = 13/17
10
A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. Find P(50p coin) and P(not ₹5 coin).
✅ Correct Answer: 5/9 and 17/18
Total coins = 100 + 50 + 20 + 10 = 180
(i) P(50p coin): = 100/180 = 10/18 = 5/9
(ii) P(not ₹5 coin): = (100 + 50 + 20)/180 = 170/180 = 17/18
Or: 1 − P(₹5) = 1 − 10/180 = 170/180 = 17/18
Questions 11 – 15
11
A tank contains 5 male fish and 8 female fish. What is the probability that a randomly taken fish is male?
✅ Correct Answer: 5/13
Total fish = 5 (male) + 8 (female) = 13
Number of male fish = 5
P(male) = 5/13 ≈ 0.385
There is approximately a 38.5% chance of selecting a male fish.
12
A game spinner has numbers 1-8 (equally likely). Find P(8), P(odd), P(>2), P(<9).
✅ Correct Answer: 1/8, 1/2, 3/4, 1
Total outcomes = 8 (numbers 1 through 8)
(i) P(8) = 1/8 (only one favourable outcome)
(ii) P(odd) = {1,3,5,7} → 4/8 = 1/2
(iii) P(>2) = {3,4,5,6,7,8} → 6/8 = 3/4
(iv) P(<9) = {1,2,3,4,5,6,7,8} → 8/8 = 1 (Certain event)
13
A die is thrown once. Find P(prime), P(between 2 and 6), P(odd).
✅ Correct Answer: 1/2, 1/2, 1/2
Sample space = {1, 2, 3, 4, 5, 6} → Total = 6
(i) Prime numbers: {2, 3, 5} → 3 outcomes
P(prime) = 3/6 = 1/2
P(prime) = 3/6 = 1/2
(ii) Between 2 and 6: {3, 4, 5} → 3 outcomes
P(between 2 and 6) = 3/6 = 1/2
P(between 2 and 6) = 3/6 = 1/2
(iii) Odd numbers: {1, 3, 5} → 3 outcomes
P(odd) = 3/6 = 1/2
P(odd) = 3/6 = 1/2
14
One card is drawn from a well-shuffled deck of 52 cards. Which is correct for: (i) King of red, (ii) Face card, (iii) Red face card, (iv) Jack of hearts, (v) Spade, (vi) Queen of diamonds?
✅ Correct Answer: 1/26, 3/13, 3/26, 1/52, 1/4, 1/52
Total cards = 52
(i) King of red: King of Hearts + King of Diamonds = 2/52 = 1/26
(ii) Face card: 3 per suit × 4 suits = 12/52 = 3/13
(iii) Red face card: J,Q,K of Hearts + J,Q,K of Diamonds = 6/52 = 3/26
(iv) Jack of hearts: Only 1 card → 1/52
(v) Spade: 13 spades → 13/52 = 1/4
(vi) Queen of diamonds: Only 1 card → 1/52
15
Five cards (10, J, Q, K, A of diamonds) are shuffled. (i) P(Queen)? (ii) If queen is drawn and put aside, P(second is ace)? P(second is queen)?
✅ Correct Answer: 1/5, 1/4, 0
Total cards = 5 (10, J, Q, K, A of diamonds)
(i) P(Queen) = 1/5
(ii) After removing Queen: 4 cards remain (10, J, K, A)
P(second is ace) = 1/4
P(second is queen) = 0/4 = 0 (queen already removed)
This demonstrates conditional probability without replacement.
Questions 16 – 20
16
12 defective pens are mixed with 132 good ones. What is P(good pen)?
✅ Correct Answer: 11/12
Total pens = 12 (defective) + 132 (good) = 144
Good pens = 132
P(good) = 132/144 = 11/12 ≈ 0.917
There is a 91.7% chance of picking a good pen.
17
20 bulbs contain 4 defective. (i) P(defective)? (ii) If drawn bulb is not defective and not replaced, P(next is not defective)?
✅ Correct Answer: 1/5 and 15/19
Total bulbs = 20, Defective = 4, Good = 16
(i) P(defective) = 4/20 = 1/5
(ii) Without replacement: First drawn was NOT defective → 1 good bulb removed
Remaining: 15 good + 4 defective = 19 total
P(not defective) = 15/19 ≈ 0.789
18
A box has 90 discs numbered 1-90. Find P(two-digit), P(perfect square), P(divisible by 5).
✅ Correct Answer: 9/10, 1/10, 1/5
Total discs = 90
(i) Two-digit numbers: 10 to 90 inclusive = 81 numbers
P = 81/90 = 9/10
P = 81/90 = 9/10
(ii) Perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 = 9 numbers
P = 9/90 = 1/10
P = 9/90 = 1/10
(iii) Divisible by 5: 5, 10, 15, ..., 90 → 18 numbers
P = 18/90 = 1/5
P = 18/90 = 1/5
19
A die has faces: A, B, C, D, E, A. Find P(A) and P(D).
✅ Correct Answer: 1/3 and 1/6
Total faces = 6
Face frequency: A=2, B=1, C=1, D=1, E=1
(i) P(A) = 2/6 = 1/3
(ii) P(D) = 1/6
Even though there are 5 distinct letters, A appears twice, making it twice as likely as other letters.
20
A die is dropped randomly on a 3m × 2m rectangle. Find P(landing inside a circle with diameter 1m).
✅ Correct Answer: π/12
Area of rectangle = 3 × 2 = 6 m²
Area of circle: radius r = 0.5 m
A = πr² = π(0.5)² = π/4 m²
A = πr² = π(0.5)² = π/4 m²
P = (π/4) / 6 = π/12 ≈ 0.2618
This is a geometric probability problem where probability = Area(favourable) / Area(total).
Questions 21 – 25
21
144 ball pens: 20 defective, 124 good. Nuri buys if good. Find P(she buys) and P(she does not buy).
✅ Correct Answer: 31/36 and 5/36
Total pens = 144
Good pens = 124, Defective = 20
(i) P(she buys) = P(good) = 124/144 = 31/36
(ii) P(she does not buy) = P(defective) = 20/144 = 5/36
Check: 31/36 + 5/36 = 36/36 = 1 ✓
22
Complete the probability table for sum on 2 dice. Is the argument "11 outcomes, each with P=1/11" correct?
✅ Correct Answer: Table varies; Argument NOT correct
Complete Table:
2
1/36
3
2/36
4
3/36
5
4/36
6
5/36
7
6/36
8
5/36
9
4/36
10
3/36
11
2/36
12
1/36
Argument is NOT correct. There are 36 equally likely elementary outcomes (ordered pairs), not 11. The sums are compound events with different numbers of favourable pairs.
23
A coin is tossed 3 times. Hanif wins if all tosses are same (HHH or TTT). Calculate P(Hanif loses).
✅ Correct Answer: 3/4
Total possible outcomes when tossing a coin 3 times = 2³ = 8
Sample space: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Hanif wins: {HHH, TTT} → 2 outcomes
Hanif loses: All other outcomes → 6 outcomes
P(loses) = 6/8 = 3/4 = 0.75
Hanif has a 75% chance of losing the game.
24
A die is thrown twice. Find P(5 does not come up either time) and P(5 comes up at least once).
✅ Correct Answer: 25/36 and 11/36
Total outcomes = 6 × 6 = 36
(i) 5 does not come up either time:
Each die has 5 favourable outcomes (1,2,3,4,6)
P = (5/6) × (5/6) = 25/36
Each die has 5 favourable outcomes (1,2,3,4,6)
P = (5/6) × (5/6) = 25/36
(ii) 5 comes up at least once:
P = 1 − P(no 5) = 1 − 25/36 = 11/36
P = 1 − P(no 5) = 1 − 25/36 = 11/36
Or count directly: (5,1-6) + (1-6,5) − (5,5) = 6+6−1 = 11 outcomes
25
Which arguments are correct? (i) Two coins: 3 outcomes (HH, TT, one of each), each P=1/3. (ii) One die: 2 outcomes (odd/even), P(odd)=1/2.
✅ Correct Answer: (i) Not correct, (ii) Correct
(i) Two coins argument: NOT CORRECT
The actual equally likely outcomes are: HH, HT, TH, TT (4 outcomes).
P(HH) = 1/4, P(TT) = 1/4, P(one of each) = 2/4 = 1/2
The three listed outcomes are not equally likely.
The actual equally likely outcomes are: HH, HT, TH, TT (4 outcomes).
P(HH) = 1/4, P(TT) = 1/4, P(one of each) = 2/4 = 1/2
The three listed outcomes are not equally likely.
(ii) Die argument: CORRECT
Odd = {1,3,5}, Even = {2,4,6}. Each has exactly 3 outcomes.
P(odd) = 3/6 = 1/2 ✓
Odd = {1,3,5}, Even = {2,4,6}. Each has exactly 3 outcomes.
P(odd) = 3/6 = 1/2 ✓
NCERT Class 10 Mathematics — Chapter 14 Probability
Exercise 14.1 — Complete Interactive Solutions