Class 11 : Maths (In English) – Lesson 14. Probability
EXPLANATION & SUMMARY
✨ Detailed Explanation
🔹 Introduction
🔵 Probability is the branch of mathematics that measures the likelihood or chance of an event occurring.
🟢 It assigns a value between 0 and 1:
0 → Impossible event
1 → Certain (sure) event
💡 Example: When tossing a fair coin,
➡️ P(Head) = 1/2, P(Tail) = 1/2
🔹 Random Experiment
🧠 An experiment that when repeated under identical conditions can give different outcomes is called a random experiment.
💡 Examples:
✔️ Tossing a coin
✔️ Rolling a die
✔️ Drawing a card from a deck
🔹 Sample Space (S)
🟢 The set of all possible outcomes of a random experiment.
💡 Examples:
Toss 1 coin → S = {H, T}
Roll 1 die → S = {1, 2, 3, 4, 5, 6}
📌 n(S) = number of outcomes
🔹 Event (E)
🔴 Any subset of the sample space is an event.
💡 Example: Rolling an even number → E = {2, 4, 6}
🔹 Types of Events
🟢 Simple Event: Contains one outcome (e.g., {3})
🟡 Compound Event: Contains more than one outcome (e.g., {2, 4, 6})
🔴 Impossible Event: No outcome satisfies condition → P = 0
🔵 Certain Event: All outcomes satisfy → P = 1
🔹 Equally Likely Outcomes
🧠 When all outcomes have same chance of occurring.
💡 Example: In rolling a fair die, 1 to 6 are equally likely.
🔹 Mutually Exclusive Events
🟠 Two events are mutually exclusive if they cannot occur together.
💡 Example: “Getting 2” and “Getting 5” in one die roll.
🔹 Exhaustive Events
🔵 Events that together cover entire sample space.
💡 Example: {1,2,3,4,5,6} in a die is exhaustive.
🔹 Favourable Outcomes
🟢 Outcomes which satisfy the condition of an event.
💡 Example: Even numbers → {2,4,6}, m = 3
🔹 Classical Definition of Probability
If:
total outcomes = n
favourable outcomes = m
then
➡️ P(E) = m / n
(valid when outcomes are equally likely)
💡 Example: Rolling a die, E = “even number”
→ m = 3, n = 6 → P(E) = 3/6 = 1/2
🔹 Properties of Probability
✔️ (i) 0 ≤ P(E) ≤ 1
✔️ (ii) P(S) = 1
✔️ (iii) P(ϕ) = 0
✔️ (iv) P(A′) = 1 − P(A)
🔹 Complementary Events
🟢 If A is an event, then A′ means “A does not occur.”
💡 Example: Toss coin → A = {Head}, A′ = {Tail}
➡️ P(A) + P(A′) = 1
🔹 Union and Intersection of Events
🧠
Union (A ∪ B): A or B or both occur
Intersection (A ∩ B): Both A and B occur
💡 Addition Theorem:
➡️ P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
If A and B are mutually exclusive → P(A ∩ B) = 0
➡️ P(A ∪ B) = P(A) + P(B)
🔹 Complement Rule
✔️ P(A′) = 1 − P(A)
🔹 Independent Events (Preview)
Events A and B are independent if occurrence of one does not affect the other.
➡️ P(A ∩ B) = P(A) × P(B)
🔹 Empirical / Experimental Probability
If an event occurs f times out of n trials:
➡️ P(E) = f / n
💡 Example: 100 tosses, 55 heads → P(Head) ≈ 0.55
🔹 Geometrical Probability
When outcomes are points in line, area, or volume:
➡️ P(E) = (Favourable measure) / (Total measure)
💡 Example: Random point on 10 cm line in 4 cm segment → P = 4/10 = 0.4
🔹 Examples
1️⃣ Coin Toss:
P(H) = 1/2, P(T) = 1/2
2️⃣ Die Roll:
P(Even) = 3/6 = 1/2
P(Prime) = 3/6 = 1/2
3️⃣ Card Deck:
P(Heart) = 13/52 = 1/4
P(Face card) = 12/52 = 3/13
4️⃣ Bag of Balls:
If 3 red, 5 black → P(Red) = 3/8, P(Black) = 5/8
🔹 Key Formulas
✔️ P(E) = m / n
✔️ P(S) = 1
✔️ P(ϕ) = 0
✔️ P(A′) = 1 − P(A)
✔️ P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
✔️ P(A ∩ B) = P(A) × P(B) (if independent)
🔹 Important Notes
✏️ Note 1: Probability never negative.
✏️ Note 2: Total probability of all exhaustive and mutually exclusive events = 1.
✏️ Note 3: If P(A) > P(B), A more likely.
🔹 Common Mistakes
⚠️ Treating unequal outcomes as equally likely
⚠️ Ignoring complement
⚠️ Adding probabilities without mutual exclusivity check
💡 Concept Tip:
If outcomes are equally likely, use classical definition; otherwise use experimental.
🌟 Applications
✔️ Weather forecasting
✔️ Gambling, cards, coins
✔️ Risk analysis
✔️ Quality control
✔️ Genetics & data analysis
🧾 Summary (~300 words)
🔹 Definition:
Probability = Likelihood of event, value between 0 and 1
🔹 Experiment: Process with uncertain outcomes
🔹 Random Experiment: Exact outcome unpredictable
🔹 Sample Space (S): All outcomes
🔹 Event (E): Subset of S
🔹 Formula:
➡️ P(E) = m / n (m = favourable outcomes, n = total outcomes)
🔹 Properties:
✔️ 0 ≤ P(E) ≤ 1
✔️ P(S) = 1
✔️ P(ϕ) = 0
✔️ P(A′) = 1 − P(A)
🔹 Addition Rule:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
🔹 Mutually Exclusive:
P(A ∪ B) = P(A) + P(B)
🔹 Independent Events:
P(A ∩ B) = P(A) × P(B)
🔹 Examples:
Coin toss, die roll, deck of cards
🔹 Applications:
Games, weather, risk, decision-making
📝 Quick Recap
🔹 Probability value: 0 ≤ P(E) ≤ 1
🔹 P(E) = m/n
🔹 P(S) = 1, P(ϕ) = 0
🔹 P(A′) = 1 − P(A)
🔹 P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
🔹 Mutually exclusive ⇒ P(A ∩ B) = 0
🔹 Independent ⇒ P(A ∩ B) = P(A)×P(B)
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QUESTIONS FROM TEXTBOOK
📄 Exercise 14.1
🔵 Question 1:
A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”.
Are E and F mutually exclusive?
🟢 Answer:
✳️ Sample space S = {1, 2, 3, 4, 5, 6}
✳️ E = {4}
✳️ F = {2, 4, 6}
✳️ Intersection E ∩ F = {4} ≠ φ
➡️ Since E ∩ F ≠ φ, events E and F are not mutually exclusive.
✔️ Final Answer: ❌ Not mutually exclusive.
🔵 Question 2:
A die is thrown. Describe the following events:
✳️ S = {1, 2, 3, 4, 5, 6}
(i) A: number less than 7
➡️ A = {1, 2, 3, 4, 5, 6}
(ii) B: number greater than 7
➡️ B = φ
(iii) C: multiple of 3
➡️ C = {3, 6}
(iv) D: number less than 4
➡️ D = {1, 2, 3}
(v) E: even number greater than 4
➡️ E = {6}
(vi) F: number not less than 3
➡️ F = {3, 4, 5, 6}
🧠 Now Find:
1️⃣ A ∪ B = {1, 2, 3, 4, 5, 6} ∪ φ = {1, 2, 3, 4, 5, 6}
2️⃣ A ∩ B = {1, 2, 3, 4, 5, 6} ∩ φ = φ
3️⃣ B ∪ C = φ ∪ {3, 6} = {3, 6}
4️⃣ E ∩ F = {6} ∩ {3, 4, 5, 6} = {6}
5️⃣ D ∩ E = {1, 2, 3} ∩ {6} = φ
6️⃣ A − C = {1, 2, 3, 4, 5, 6} − {3, 6} = {1, 2, 4, 5}
7️⃣ D − E = {1, 2, 3} − {6} = {1, 2, 3}
8️⃣ E ∩ F′ = {6} ∩ (S − F) = {6} ∩ {1, 2} = φ
9️⃣ F′ = S − F = {1, 2}
🔵 Question 3
An experiment involves rolling a pair of dice and recording the numbers that come up.
Describe the following events:
A: the sum is greater than 8
B: 2 occurs on either die
C: the sum is at least 7 and a multiple of 3
Also, find which pairs of these events are mutually exclusive.
🟢 Answer:
✳️ Sample space size = 36 (ordered pairs (i, j), where i, j ∈ {1, 2, 3, 4, 5, 6})
🔹 Event A: sum > 8
➡️ Possible sums: 9, 10, 11, 12
➡️ A = { (3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6) }
🔹 Event B: 2 occurs on either die
➡️ B = { (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (1,2), (3,2), (4,2), (5,2), (6,2) }
🔹 Event C: sum ≥ 7 and multiple of 3
➡️ Multiples of 3 ≥ 7 are 9 and 12
➡️ C = { (3,6), (4,5), (5,4), (6,3), (6,6) }
🧠 Check mutual exclusivity:
1️⃣ A ∩ B: any outcome in both?
A contains sums > 8, B contains outcomes with 2 on any die.
No sum > 8 can have a 2 (max 2+6=8).
➡️ A ∩ B = φ ✅ A and B are mutually exclusive.
2️⃣ A ∩ C:
C outcomes (sum 9,12) ⊂ A outcomes (sum >8).
➡️ A ∩ C ≠ φ ❌ Not mutually exclusive.
3️⃣ B ∩ C:
B has outcomes with 2; C has sums 9 or 12.
No combination with 2 gives 9 or 12.
➡️ B ∩ C = φ ✅ B and C are mutually exclusive.
✔️ Final:
Mutually exclusive pairs:
➡️ (A, B) and (B, C)
🔵 Question 4
Three coins are tossed once.
Let
A: “three heads show”
B: “two heads and one tail show”
C: “three tails show”
D: “a head shows on the first coin”
(i) Which events are mutually exclusive?
(ii) Which events are simple?
(iii) Which events are compound?
🟢 Answer:
✳️ Sample space (S):
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
🔹 A = {HHH}
🔹 B = {HHT, HTH, THH}
🔹 C = {TTT}
🔹 D = {HHH, HHT, HTH, HTT} (first coin = H)
(i) Mutually exclusive:
Events with no common outcomes:
➡️ A & B: φ ✅
➡️ A & C: φ ✅
➡️ B & C: φ ✅
➡️ D overlaps with A, B → ❌
✔️ Mutually exclusive: (A, B), (A, C), (B, C)
(ii) Simple events: contain 1 outcome
➡️ A = {HHH}, C = {TTT} ✔️
(iii) Compound events: more than one outcome
➡️ B = {HHT, HTH, THH}, D = {HHH, HHT, HTH, HTT} ✔️
🔵 Question 5
Three coins are tossed. Describe:
(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events which are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive.
(v) Three events which are mutually exclusive but not exhaustive.
🟢 Answer:
✳️ Sample space (S):
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
(i) 🔹 E₁: “exactly two heads” = {HHT, HTH, THH}
🔹 E₂: “exactly one head” = {HTT, THT, TTH}
➡️ No overlap → Mutually exclusive
(ii) 🔹 E₁: “no head” = {TTT}
🔹 E₂: “exactly one head” = {HTT, THT, TTH}
🔹 E₃: “two or more heads” = {HHT, HTH, THH, HHH}
➡️ Disjoint and cover all outcomes → Mutually exclusive & exhaustive
(iii) 🔹 E₁: “at least one head” = {HHH, HHT, HTH, THH, HTT, THT, TTH}
🔹 E₂: “at least one tail” = {HHT, HTH, THH, HTT, THT, TTH, TTT}
➡️ Overlap exists → Not mutually exclusive
(iv) 🔹 E₁: “exactly two heads” = {HHT, HTH, THH}
🔹 E₂: “exactly one head” = {HTT, THT, TTH}
➡️ Disjoint, but do not cover {HHH, TTT} → Mutually exclusive, not exhaustive
(v) 🔹 E₁: “no head” = {TTT}
🔹 E₂: “exactly one head” = {HTT, THT, TTH}
🔹 E₃: “exactly two heads” = {HHT, HTH, THH}
➡️ Disjoint, missing {HHH} → Mutually exclusive but not exhaustive
🔵 Question 6
Two dice are thrown.
A: getting an even number on the first die
B: getting an odd number on the first die
C: getting the sum ≤ 5
Let S = {(i, j) : i, j ∈ {1,2,3,4,5,6}} (36 outcomes).
🟢 Describe the events
✳️ (i) A′
➡️ Complement of A = “first die not even” = “first die odd”
➡️ A′ = B
✳️ (ii) not B
➡️ B′ = “first die not odd” = “first die even”
➡️ not B = A
✳️ (iii) A or B
➡️ “first die even or odd” = whole space
✔️ A ∪ B = S
✳️ (iv) A and B
➡️ “first die even and odd” simultaneously
✔️ A ∩ B = φ
✳️ (v) A but not C
➡️ A ∩ C′ = “first die even” and “sum ≥ 6”
➡️ Outcomes (list):
(2,4), (2,5), (2,6);
(4,2), (4,3), (4,4), (4,5), (4,6);
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
✳️ (vi) B or C
➡️ B ∪ C = “first die odd” or “sum ≤ 5”
➡️ Equivalently: all 36 outcomes except those in part (v) (i.e., S − (A ∩ C′)).
➡️ Count = 36 − 14 = 22 outcomes.
✳️ (vii) B and C
➡️ B ∩ C = “first die odd” and “sum ≤ 5”
➡️ From C = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}
➡️ Keep first-odd: {(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)}
✳️ (viii) A ∩ B′ ∩ C′
➡️ B′ = A ⇒ A ∩ B′ = A
➡️ So A ∩ B′ ∩ C′ = A ∩ C′, same set as in part (v).
🔵 Question 7
(Refer to Q6.) State True/False with reason.
🟢 Answers with reasons
✳️ (i) A and B are mutually exclusive.
➡️ A ∩ B = φ (first die can’t be even and odd).
✔️ True
✳️ (ii) A and B are mutually exclusive and exhaustive.
➡️ Disjoint and A ∪ B = S.
✔️ True
✳️ (iii) A = B′.
➡️ Complements on parity of first die.
✔️ True
✳️ (iv) A and C are mutually exclusive.
➡️ Counterexamples: (2,1), (2,2), (2,3), (4,1) ∈ A ∩ C.
❌ False
✳️ (v) A and B′ are mutually exclusive.
➡️ B′ = A ⇒ A ∩ B′ = A ≠ φ.
❌ False
✳️ (vi) A′, B′, C are mutually exclusive and exhaustive.
➡️ A′ and B′ already cover S and overlap with C (e.g., (1,1) ∈ A′ ∩ C).
❌ False (not mutually exclusive; exhaustiveness already achieved by A′ and B′).
📄 Exercise 14.2
🔵 Question 1:
Which of the following cannot be a valid assignment of probabilities for outcomes of sample space
S = {ω₁, ω₂, ω₃, ω₄, ω₅, ω₆, ω₇}?
🟢 Answer:
✳️ Condition for valid probability assignment:
1️⃣ Each probability P(ωᵢ) must satisfy 0 ≤ P(ωᵢ) ≤ 1
2️⃣ Total sum = Σ P(ωᵢ) = 1
🔹 Option (a)
P(ω₁)=0.1, P(ω₂)=0.01, P(ω₃)=0.05, P(ω₄)=0.03, P(ω₅)=0.01, P(ω₆)=0.2, P(ω₇)=0.6
➡️ Sum = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1.00 ✔️
➡️ All ≥ 0 and ≤ 1 ✔️
✅ Valid
🔹 Option (b)
Each = 1/7
➡️ Sum = 7 × (1/7) = 1 ✔️
➡️ All ≥ 0 and ≤ 1 ✔️
✅ Valid
🔹 Option (c)
P = 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7
➡️ Sum = 2.8 ❌
➡️ Must equal 1 → ❌ Invalid
➡️ Also, values >1 not allowed ❌
🚫 Invalid
🔹 Option (d)
P = −0.1, 0.2, 0.3, 0.4, −0.2, 0.1, 0.3
➡️ Negative probabilities ❌
🚫 Invalid
🔹 Option (e)
P = 1/14, 2/14, 3/14, 4/14, 5/14, 6/14, 15/14
➡️ Sum = (1+2+3+4+5+6+15)/14 = 36/14 = 2.57 ❌
➡️ Also 15/14 > 1 ❌
🚫 Invalid
✔️ Final Answer: Options (c), (d), and (e) ❌ cannot be valid assignments.
🔵 Question 2:
A coin is tossed twice. What is the probability that at least one tail occurs?
🟢 Answer:
✳️ Sample space (S): {HH, HT, TH, TT}
✳️ Total outcomes = 4
✳️ Event (E): “At least one tail” = {HT, TH, TT}
✔️ n(E) = 3
➡️ P(E) = n(E)/n(S) = 3/4
✔️ Final: P = 3/4
🔵 Question 3:
A die is thrown. Find the probability of:
S = {1, 2, 3, 4, 5, 6}, n(S) = 6
(i) Prime number appears
➡️ E = {2, 3, 5}, n(E) = 3
✔️ P = 3/6 = 1/2
(ii) Number ≥ 3
➡️ E = {3, 4, 5, 6}, n(E) = 4
✔️ P = 4/6 = 2/3
(iii) Number ≤ 1
➡️ E = {1}, n(E) = 1
✔️ P = 1/6
(iv) Number > 6
➡️ E = φ, n(E) = 0
✔️ P = 0
(v) Number < 6
➡️ E = {1, 2, 3, 4, 5}, n(E) = 5
✔️ P = 5/6
🔵 Question 4:
A card is selected from a pack of 52 cards.
(a) How many points in sample space?
✔️ n(S) = 52
(b) Probability card is an Ace of Spades
➡️ Only 1 card = Ace of Spades
✔️ P = 1/52
(c) Probability card is:
(i) An Ace → 4 Aces total
✔️ P = 4/52 = 1/13
(ii) A black card → 26 black cards (spades + clubs)
✔️ P = 26/52 = 1/2
🔵 Question 5:
A fair coin with faces marked 1 and 6, and a fair die are tossed.
Find probability that the sum =
✳️ Sample space: coin → {1, 6}; die → {1, 2, 3, 4, 5, 6}
→ total outcomes = 12
(i) Sum = 3
Pairs: (1,2)
✔️ P = 1/12
(ii) Sum = 12
Pairs: (6,6)
✔️ P = 1/12
🔵 Question 6:
There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?
🟢 Answer:
✳️ Step 1 ➤ Total members = 4 (men) + 6 (women) = 10
✳️ Step 2 ➤ Favourable outcomes (women) = 6
➡️ P(woman) = 6/10 = 3/5
✔️ Final Answer: 3/5
🔵 Question 7:
A fair coin is tossed four times, and a person wins ₹1 for each head and loses ₹1.50 for each tail that turns up.
From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.
🟢 Answer:
✳️ Step 1 ➤ Let the number of heads = H and number of tails = T
Since 4 tosses are made ➡️ H + T = 4
✳️ Step 2 ➤ Gain for each head = ₹1, loss for each tail = ₹1.50
So, total money after 4 tosses =
M = 1 × H − 1.5 × T
✳️ Step 3 ➤ Replace T = 4 − H
➡️ M = H − 1.5(4 − H)
➡️ M = H − 6 + 1.5H
➡️ M = 2.5H − 6
🧠 Step 4 ➤ Possible values of H = 0, 1, 2, 3, 4
🔹 If H = 0, M = 2.5(0) − 6 = −6
🔹 If H = 1, M = 2.5(1) − 6 = −3.5
🔹 If H = 2, M = 2.5(2) − 6 = −1.0
🔹 If H = 3, M = 2.5(3) − 6 = +1.5
🔹 If H = 4, M = 2.5(4) − 6 = +4.0
Hence, possible amounts of money are:
−6, −3.5, −1, +1.5, +4
✳️ Step 5 ➤ Each toss has 2 outcomes ⇒ Total outcomes = 2⁴ = 16
Probability of getting H heads = P(H) = C(4, H) / 16
🔵 When H = 0 → P = C(4,0)/16 = 1/16
🟢 When H = 1 → P = C(4,1)/16 = 4/16
🟡 When H = 2 → P = C(4,2)/16 = 6/16
🔴 When H = 3 → P = C(4,3)/16 = 4/16
🟠 When H = 4 → P = C(4,4)/16 = 1/16
✔️ Step 6 ➤ Final Results:
🔹 Amount = −6 → Probability = 1/16
🔹 Amount = −3.5 → Probability = 4/16
🔹 Amount = −1 → Probability = 6/16
🔹 Amount = +1.5 → Probability = 4/16
🔹 Amount = +4 → Probability = 1/16
🎯 Final Answer:
Different amounts = −6, −3.5, −1, +1.5, +4
Probabilities = 1/16, 4/16, 6/16, 4/16, 1/16
✔️ Sum of probabilities = 1 ✅ (Hence correct)
🔵 Question 8:
Three coins are tossed once. Find the probability of getting:
(i) 3 heads (ii) 2 heads (iii) at least 2 heads (iv) at most 2 heads
(v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) at most two tails
🟢 Answer:
✳️ Step 1 ➤ Total outcomes = 2³ = 8
✳️ Step 2 ➤ Probabilities:
(i) 3 heads → 1/8
(ii) 2 heads → 3/8
(iii) at least 2 heads → (2H + 3H) = 3/8 + 1/8 = 1/2
(iv) at most 2 heads → 1 − 1/8 = 7/8
(v) no head (all tails) → 1/8
(vi) 3 tails → 1/8
(vii) exactly two tails → 3/8
(viii) no tail (all heads) → 1/8
(ix) at most two tails → 7/8
✔️ Final Answer:
(i) 1/8 (ii) 3/8 (iii) 1/2 (iv) 7/8 (v) 1/8 (vi) 1/8 (vii) 3/8 (viii) 1/8 (ix) 7/8
🔵 Question 9:
If 2/11 is the probability of an event A, what is the probability of the event ‘not A’?
🟢 Answer:
✳️ Step 1 ➤ Formula: P(A′) = 1 − P(A)
✳️ Step 2 ➤ P(A′) = 1 − 2/11 = 9/11
✔️ Final Answer: 9/11
🔵 Question 10:
A letter is chosen at random from the word ASSASSINATION. Find the probability that the letter is
(i) a vowel (ii) a consonant
🟢 Answer:
✳️ Step 1 ➤ Total letters = 13
✳️ Step 2 ➤ Vowels = A(3) + I(2) + O(1) = 6
✳️ Step 3 ➤ Consonants = 13 − 6 = 7
✳️ Step 4 ➤
(i) P(vowel) = 6/13
(ii) P(consonant) = 7/13
✔️ Final Answer:
(i) 6/13 (ii) 7/13
🔵 Question 11:
In a lottery, a person chooses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize.
What is the probability of winning the prize in the game?
[Hint: Order of the numbers is not important]
🟢 Answer:
✳️ Step 1 ➤ Total natural numbers = 20
Each ticket has 6 different numbers (order not important)
Hence,
🔹 Total possible selections = n(S) = C(20, 6)
✳️ Step 2 ➤ To win, all 6 chosen numbers must match with fixed 6 numbers
So,
🔹 Favourable cases = 1 (only one set matches exactly)
✳️ Step 3 ➤ Therefore,
P(winning) = Favourable / Total
➡️ P = 1 / C(20, 6)
✳️ Step 4 ➤ Compute
C(20, 6) = 20! / (6! × 14!) = (20 × 19 × 18 × 17 × 16 × 15) / (6 × 5 × 4 × 3 × 2 × 1) = 38760
Hence,
🧮 P = 1 / 38760
🎯 Final Answer: Probability of winning = 1 / 38760
🔵 Question 12:
Check whether the following probabilities P(A) and P(B) are consistently defined:
(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8
🟢 Answer:
✳️ Step 1 ➤ We know
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Also,
➡️ 0 ≤ P(A ∩ B) ≤ min[P(A), P(B)]
🔹 (i)
P(A) = 0.5, P(B) = 0.7
Given P(A ∩ B) = 0.6
Check condition:
min[P(A), P(B)] = 0.5
But P(A ∩ B) = 0.6 > 0.5 ❌ Not possible
So, ❌ Not consistently defined
🔹 (ii)
P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8
Use formula:
P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
➡️ P(A ∩ B) = 0.5 + 0.4 − 0.8 = 0.1
Check:
0 ≤ 0.1 ≤ min(0.5, 0.4) ✅ True
So, ✅ Consistently defined
🔵 Question 13:
Fill in the blanks in the following table:
P(A) P(B) P(A ∩ B) P(A ∪ B)
( i ) 1/3 1/5 1/15 …
( ii ) 0.35 … 0.25 0.6
( iii ) 0.5 0.35 … 0.7
🟢 Answer:
✳️ Formula → P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
🔹 (i)
➡️ P(A ∪ B) = 1/3 + 1/5 − 1/15
➡️ = 5/15 + 3/15 − 1/15 = 7/15
✔️ P(A ∪ B) = 7/15
🔹 (ii)
➡️ P(B) = P(A ∪ B) − P(A) + P(A ∩ B)
➡️ = 0.6 − 0.35 + 0.25 = 0.50
✔️ P(B) = 0.50
🔹 (iii)
➡️ P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
➡️ = 0.5 + 0.35 − 0.7 = 0.15
✔️ P(A ∩ B) = 0.15
🔵 Question 14:
Given P(A) = 3/5 and P(B) = 1/5, find P(A or B) if A and B are mutually exclusive events.
🟢 Answer:
✳️ For mutually exclusive events → P(A ∩ B) = 0
➡️ P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
➡️ = 3/5 + 1/5 − 0 = 4/5
✔️ P(A ∪ B) = 4/5
🔵 Question 15:
If P(E) = 1/4, P(F) = 1/2, and P(E ∩ F) = 1/8, find:
(i) P(E or F)
(ii) P(not E and not F)
🟢 Answer:
(i) ✳️ P(E ∪ F) = P(E) + P(F) − P(E ∩ F)
➡️ = 1/4 + 1/2 − 1/8 = 2/8 + 4/8 − 1/8 = 5/8
✔️ P(E ∪ F) = 5/8
(ii) ✳️ P(E′ ∩ F′) = 1 − P(E ∪ F)
➡️ = 1 − 5/8 = 3/8
✔️ P(E′ ∩ F′) = 3/8
🔵 Question 16:
Events E and F are such that P(not E or not F) = 0.25. State whether E and F are mutually exclusive.
🟢 Answer:
✳️ P(not E or not F) = P(E′ ∪ F′)
✳️ By De Morgan’s law → P((E ∩ F)′) = 0.25
➡️ 1 − P(E ∩ F) = 0.25
➡️ P(E ∩ F) = 0.75
✳️ For mutually exclusive events, P(E ∩ F) = 0
✔️ Since 0.75 ≠ 0, E and F are not mutually exclusive
🔵 Question 17:
A and B are events such that P(A) = 0.42, P(B) = 0.48, and P(A ∩ B) = 0.16. Determine:
(i) P(not A)
(ii) P(not B)
(iii) P(A or B)
🟢 Answer:
(i) ✳️ P(A′) = 1 − P(A) = 1 − 0.42 = 0.58
✔️ P(A′) = 0.58
(ii) ✳️ P(B′) = 1 − P(B) = 1 − 0.48 = 0.52
✔️ P(B′) = 0.52
(iii) ✳️ P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
➡️ = 0.42 + 0.48 − 0.16 = 0.74
✔️ P(A ∪ B) = 0.74
🔵 Question 18:
In Class XI, 40% of the students study Mathematics, 30% study Biology, and 10% study both. If a student is selected at random, find the probability that the student studies Mathematics or Biology.
🟢 Answer:
✳️ Let M = Mathematics, B = Biology
✳️ P(M) = 0.40, P(B) = 0.30, P(M ∩ B) = 0.10
✳️ P(M ∪ B) = P(M) + P(B) − P(M ∩ B)
➡️ = 0.40 + 0.30 − 0.10 = 0.60
✔️ P(M ∪ B) = 0.60
🔵 Question 19:
In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95.
What is the probability of passing both?
🟢 Answer:
✳️ Step 1: Let
P(A) = Probability of passing first exam = 0.8
P(B) = Probability of passing second exam = 0.7
P(A ∪ B) = Probability of passing at least one = 0.95
✳️ Step 2: Formula → P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
✳️ Step 3: Substitute
0.95 = 0.8 + 0.7 − P(A ∩ B)
➡️ P(A ∩ B) = 1.5 − 0.95 = 0.55
✔️ P(A ∩ B) = 0.55
🔵 Question 20:
The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1.
If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?
🟢 Answer:
✳️ Step 1: Let
P(E) = 0.75, P(E ∩ H) = 0.5, P(neither) = 0.1
✳️ Step 2:
P(E ∪ H) = 1 − P(neither) = 1 − 0.1 = 0.9
✳️ Step 3: Formula → P(E ∪ H) = P(E) + P(H) − P(E ∩ H)
➡️ 0.9 = 0.75 + P(H) − 0.5
➡️ P(H) = 0.9 − 0.25 = 0.65
✔️ P(H) = 0.65
🔵 Question 21:
In a class of 60 students, 30 opted for NCC, 32 opted for NSS, and 24 opted for both NCC and NSS.
If one of these students is selected at random, find the probability that
(i) the student opted for NCC or NSS
(ii) the student has opted neither NCC nor NSS
(iii) the student has opted NSS but not NCC
🟢 Answer:
✳️ Step 1: Let
n = 60, n(A) = 30 (NCC), n(B) = 32 (NSS), n(A ∩ B) = 24
✳️ Step 2:
(i) P(A ∪ B) = [n(A) + n(B) − n(A ∩ B)] / n
➡️ = (30 + 32 − 24)/60 = 38/60 = 19/30
✔️ Probability = 19/30
(ii) P(neither) = 1 − P(A ∪ B)
➡️ = 1 − 19/30 = 11/30
✔️ Probability = 11/30
(iii) NSS but not NCC = B − A
➡️ n(B − A) = n(B) − n(A ∩ B) = 32 − 24 = 8
➡️ P(B − A) = 8/60 = 2/15
✔️ Probability = 2/15
————————————————————————————————————————————————————————————————————————————
OTHER IMPORTANT QUESTIONS FOR EXAMS
CBSE STYLE MODEL PAPER
ESPECIALLY FROM THIS CHAPTER ONLY
🧭 Section A – Very Short / Objective Type (1 Mark each)
🔵 Question 1:
What is the range of probability of any event?
🟢 Answer:
Range of probability = [0, 1]
✔️ Minimum 0 (impossible event), Maximum 1 (certain event)
🔵 Question 2:
If P(A) = 0.65, what is P(Ā)?
🟢 Answer:
P(Ā) = 1 − P(A) = 1 − 0.65 = 0.35
🔵 Question 3:
An event that is sure to occur is called _______.
🟢 Answer:
Certain event (Probability = 1)
🔵 Question 4:
Two events are said to be independent if
🟢 Answer:
P(A ∩ B) = P(A) × P(B)
🔵 Question 5:
If two events are mutually exclusive, what is P(A ∩ B)?
🟢 Answer:
P(A ∩ B) = 0
🔵 Question 6 (MCQ):
The probability of getting a head when a coin is tossed is
1️⃣ 0
2️⃣ 1/2
3️⃣ 1
4️⃣ 2
🟢 Answer: 2️⃣ 1/2
🔹 Section B – 2 Marks Each
🔵 Question 7:
A die is thrown once. Find the probability of getting
(a) a prime number (b) a number greater than 4
🟢 Answer:
Total outcomes = 6 → {1, 2, 3, 4, 5, 6}
(a) Prime = {2, 3, 5} → n(E) = 3
P = 3/6 = 1/2
(b) >4 = {5, 6} → n(E) = 2
P = 2/6 = 1/3
🔵 Question 8:
Two coins are tossed. Find probability of getting
(a) two heads (b) at least one head
🟢 Answer:
Sample S = {HH, HT, TH, TT}, n(S)=4
(a) E = {HH} → P = 1/4
(b) E = {HH, HT, TH} → P = 3/4
🔵 Question 9:
A bag has 5 red and 3 blue balls. One ball is drawn. Find the probability that it is
(a) red (b) blue
🟢 Answer:
Total = 8
(a) Red = 5 → P = 5/8
(b) Blue = 3 → P = 3/8
🔵 Question 10:
If P(A) = 0.4, P(B) = 0.5, P(A ∪ B) = 0.7, find P(A ∩ B).
🟢 Answer:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
0.7 = 0.4 + 0.5 − x ⇒ x = 0.2
✔️ P(A ∩ B) = 0.2
🔵 Question 11:
If A and B are independent with P(A)=0.3, P(B)=0.4, find
(a) P(A ∩ B) (b) P(A ∪ B)
🟢 Answer:
(a) P(A ∩ B) = 0.3 × 0.4 = 0.12
(b) P(A ∪ B) = 0.3 + 0.4 − 0.12 = 0.58
🔵 Question 12:
A card is drawn from a standard 52-card pack. Find probability of drawing
(a) a king (b) a red card
🟢 Answer:
(a) Kings = 4 → P = 4/52 = 1/13
(b) Red = 26 → P = 26/52 = 1/2
🔹 Question 13:
Two dice are thrown together. Find the probability of getting
(a) sum = 10 (b) sum > 10 (c) sum < 5
🟢 Answer:
Total outcomes = 36
(a) Sum = 10 → (4,6), (5,5), (6,4) → 3 outcomes → P = 3/36 = 1/12
(b) Sum > 10 → (5,6), (6,5), (6,6) → 3 outcomes → P = 3/36 = 1/12
(c) Sum < 5 → (1,1), (1,2), (2,1), (1,3), (3,1), (2,2) → 6 outcomes → P = 6/36 = 1/6
🔹 Question 14:
A card is drawn from a pack. Find the probability that it is
(a) red king (b) a face card (c) not an ace
🟢 Answer:
Total cards = 52
(a) Red kings = 2 → P = 2/52 = 1/26
(b) Face cards = 12 → P = 12/52 = 3/13
(c) Not an ace = 52 − 4 = 48 → P = 48/52 = 12/13
🔹 Question 15:
A bag has 4 white and 6 black balls. Two balls are drawn together. Find the probability that both are black.
🟢 Answer:
Total = 10 → combinations = 10C2 = 45
Black = 6 → ways = 6C2 = 15
P = 15/45 = 1/3
🔹 Question 16:
A box has 3 red and 2 blue balls. One ball is drawn at random, replaced, and drawn again. Find the probability of
(a) two red (b) one red one blue
🟢 Answer:
P(R) = 3/5, P(B) = 2/5
(a) Two red = P(R) × P(R) = (3/5) × (3/5) = 9/25
(b) One red one blue = 2 × P(R) × P(B) = 2 × (3/5) × (2/5) = 12/25
🔹 Question 17:
From 1 to 50, a number is selected. Find probability that
(a) it is a multiple of 5 (b) multiple of 7 (c) multiple of both 5 and 7
🟢 Answer:
Total = 50
(a) Multiples of 5 = 10 → P = 10/50 = 1/5
(b) Multiples of 7 = 7 → P = 7/50
(c) Multiples of 35 = 1 → P = 1/50
🔹 Question 18:
If P(A) = 0.5, P(B) = 0.4, P(A ∩ B) = 0.2, find
(a) P(A ∪ B) (b) P(Ā ∩ B) (c) P(A ∩ B̄)
🟢 Answer:
(a) P(A ∪ B) = 0.5 + 0.4 − 0.2 = 0.7
(b) P(Ā ∩ B) = P(B) − P(A ∩ B) = 0.4 − 0.2 = 0.2
(c) P(A ∩ B̄) = P(A) − P(A ∩ B) = 0.5 − 0.2 = 0.3
🔹 Question 19:
A and B are two events with P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3. Check if A and B are independent.
🟢 Answer:
Check: P(A) × P(B) = 0.6 × 0.5 = 0.3 = P(A ∩ B)
✔️ Hence, A and B are independent
🔹 Question 20:
Three coins are tossed. Find probability of
(a) exactly 2 heads (b) at least 2 heads (c) at most 2 heads
🟢 Answer:
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} → n(S)=8
(a) Exactly 2 heads = {HHT, HTH, THH} → 3/8
(b) At least 2 heads = {HHT, HTH, THH, HHH} → 4/8 = 1/2
(c) At most 2 heads = 8 − 1 = 7 → 7/8
🔹 Question 21:
Two dice are thrown. Find probability of getting a doublet (same number on both dice).
🟢 Answer:
Doublets = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} → 6 outcomes
P = 6/36 = 1/6
🔹 Question 22:
If E and F are events such that P(E) = 0.7, P(F) = 0.5, and P(E ∩ F) = 0.4, find
(a) P(E ∪ F) (b) P(Ē ∩ F̄) (c) P(Ē ∪ F̄)
🟢 Answer:
(a) P(E ∪ F) = 0.7 + 0.5 − 0.4 = 0.8
(b) P(Ē ∩ F̄) = 1 − P(E ∪ F) = 1 − 0.8 = 0.2
(c) P(Ē ∪ F̄) = 1 − P(E ∩ F) = 1 − 0.4 = 0.6
🔵 Question 23:
In a box there are 5 red, 4 blue and 3 green balls. Three balls are drawn one by one without replacement. Find the probability that all three are of different colours.
🟢 Answer:
🧩 Total balls = 5 + 4 + 3 = 12
➡️ Favourable patterns (distinct colours) = permutations of (R, B, G) in 3 draws = 3! = 6 sequences
➡️ Compute probability for one ordering, then multiply by 6
✨ For ordering R → B → G:
➤ P(R first) = 5/12
➤ P(B second | R first) = 4/11
➤ P(G third | R,B drawn) = 3/10
➤ P(RBG) = (5/12) × (4/11) × (3/10)
✨ Since 6 permutations are equally likely:
➤ Required P = 6 × [(5/12) × (4/11) × (3/10)]
➤ Simplify: 6 × (5×4×3)/(12×11×10)
➤ = (6×60)/(1320)
➤ = 360/1320
➡️ Final answer: 3/11
🔵 Question 24:
A die is thrown three times. Find the probability that exactly two different numbers appear.
🟢 Answer:
💡 “Exactly two different numbers” means outcomes like a,a,b in any order with a ≠ b.
➡️ Choose the pair of numbers: 6C2 = 15
➡️ Choose which number repeats thrice? Pick the repeated one among the two: 2 ways
➡️ Arrange positions: patterns of a,a,b → number of permutations = 3 (positions of single b)
➡️ Favourable outcomes = 15 × 2 × 3 = 90
➡️ Total outcomes = 6^3 = 216
➡️ Probability = 90/216 = 15/36
➡️ Final answer: 5/12
🟡 OR
🔵 Alternative Question 24:
Two dice are thrown 4 times. Find the probability that the sum 7 appears exactly once.
🟢 Answer (OR):
➤ P(sum 7 in one throw) = 6/36 = 1/6
➤ P(not 7) = 5/6
➤ Exactly once in 4 throws: 4C1 × (1/6)^1 × (5/6)^3
➤ = 4 × (1/6) × (125/216)
➤ = 500/1296
➡️ Final answer: 125/324
🔵 Question 25:
A box contains 7 bulbs of which 3 are defective. Two bulbs are drawn one by one without replacement. Find the probability that at least one bulb is defective.
🟢 Answer:
➡️ Use complement: P(at least one defective) = 1 − P(both good)
➤ Good bulbs = 4, total = 7
➤ P(both good) = (4/7) × (3/6) = 12/42 = 2/7
➡️ Required P = 1 − 2/7 = 5/7
➡️ Final answer: 5/7
🔵 Question 26:
Events A and B satisfy P(A) = 0.5, P(B) = 0.4 and P(A ∪ B) = 0.7. Find:
(i) P(A ∩ B), (ii) P(Ā ∩ B), (iii) P(A ∣ B).
🟢 Answer:
➤ P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
➤ = 0.5 + 0.4 − 0.7
➤ = 0.2
➤ P(Ā ∩ B) = P(B) − P(A ∩ B) = 0.4 − 0.2 = 0.2
➤ P(A ∣ B) = P(A ∩ B) / P(B) = 0.2 / 0.4 = 1/2
➡️ Final answer: (i) 0.2, (ii) 0.2, (iii) 1/2
🟡 OR
🔵 Alternative Question 26:
Given P(E) = 0.6, P(F) = 0.5 and P(E ∩ F) = 0.3. Are E and F independent? Also find P(Ē ∪ F̄).
🟢 Answer (OR):
➤ Test independence: P(E)P(F) = 0.6×0.5 = 0.3 = P(E ∩ F) ✔️ independent
➤ P(Ē ∪ F̄) = 1 − P(E ∩ F) = 1 − 0.3 = 0.7
➡️ Final answer: Independent; P(Ē ∪ F̄) = 0.7
🔵 Question 27:
Three machines A, B, C produce 25%, 35%, 40% of total items respectively. Their defect rates are 5%, 4%, 2% respectively. An item picked at random is found defective. Find the probability that it was produced by machine A.
🟢 Answer (Bayes’ theorem):
➤ P(A) = 0.25, P(B) = 0.35, P(C) = 0.40
➤ P(D ∣ A) = 0.05, P(D ∣ B) = 0.04, P(D ∣ C) = 0.02
➤ P(D) = Σ P(machine)×P(D ∣ machine)
➤ P(D) = 0.25×0.05 + 0.35×0.04 + 0.40×0.02
➤ = 0.0125 + 0.014 + 0.008
➤ = 0.0345
➤ P(A ∣ D) = [P(A) P(D ∣ A)] / P(D)
➤ = (0.25×0.05) / 0.0345
➤ = 0.0125 / 0.0345
➡️ Final answer: 25/69 ≈ 0.3623
🔵 Question 28:
From a standard deck, two cards are drawn without replacement. Find the probability that both cards are of the same suit.
🟢 Answer:
➡️ Method 1 (sequential):
➤ P(first any) = 1
➤ P(second same suit ∣ first) = 12/51
➤ Required P = 12/51 = 4/17
➡️ Method 2 (counting check):
➤ Favourable = 4 × C(13,2)
➤ Total = C(52,2)
➤ Probability = [4×78]/1326 = 312/1326 = 4/17
➡️ Final answer: 4/17
🔵 Question 29:
A box contains 8 white and 4 black balls. Two balls are drawn without replacement.
(i) If the first is white, find the probability that the second is black.
(ii) Also, find the unconditional probability that the second is black.
(iii) OR: If two cards are drawn from a deck without replacement, find the probability that at least one is an ace.
🟢 Answer:
(i) Conditional probability
➡️ Total balls = 12
➡️ After first white: remaining = 11 balls (7 white, 4 black)
➡️ P(second black ∣ first white) = 4/11
✔️ Answer: 4/11
(ii) Unconditional probability
Use total probability:
P(second black) =
= P(W first) × P(B second ∣ W first) + P(B first) × P(B second ∣ B first)
= (8/12) × (4/11) + (4/12) × (3/11)
= (32 + 12)/132
= 44/132
= 1/3
✔️ Answer: 1/3
(iii) OR: Two cards drawn from a 52-card deck
Find P(at least one ace)
➡️ Use complement: P(at least one ace) = 1 − P(no ace in both)
➡️ Non-aces = 48
P(no ace) = C(48, 2) / C(52, 2)
= (48×47) / (52×51)
= 564 / 663
So,
P(at least one ace) = 1 − 564/663 = 99/663 = 33/221 ≈ 0.1493
✔️ Answer: 33/221
🔵 Question 30:
The probability that a student passes Mathematics is 0.7 and that she passes Physics is 0.6. The probability that she passes at least one of the subjects is 0.9. Find the probability that she passes:
(i) both subjects, (ii) exactly one subject, (iii) Mathematics but not Physics.
🟢 Answer:
➤ Let M = pass Maths, P = pass Physics
➤ P(M) = 0.7, P(P) = 0.6, P(M ∪ P) = 0.9
➤ P(M ∩ P) = P(M) + P(P) − P(M ∪ P) = 0.7 + 0.6 − 0.9 = 0.4
➤ P(exactly one) = P(M ∪ P) − P(M ∩ P) = 0.9 − 0.4 = 0.5
➤ P(M ∩ P̄) = P(M) − P(M ∩ P) = 0.7 − 0.4 = 0.3
➡️ Final answer: (i) 0.4, (ii) 0.5, (iii) 0.3
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