Mathematics Class 11
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🎓 Class 11 Mathematics CBSE Theory Ch 14 — Probability ⏱ ~15 min
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Targeting Class 11 level in Probability, with Advanced difficulty.

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End-of-Chapter Exercises

Exercise 14.1 — Sample spaces and events
1. A coin is tossed three times. Describe the sample space.
\(S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}\), 8 outcomes.
2. A die is thrown. Describe the sample space, write the event "a prime number" and "a number greater than 4".
\(S=\{1,2,3,4,5,6\}\). Prime: \(\{2,3,5\}\). Greater than 4: \(\{5,6\}\).
3. A coin is tossed and a die is thrown. Describe S.
\(S=\{H1,H2,\ldots,H6,T1,T2,\ldots,T6\}\), 12 outcomes.
4. A coin is tossed; if it shows a head we throw a die, otherwise toss the coin again. Describe S.
\(S=\{H1,H2,H3,H4,H5,H6,TH,TT\}\), 8 outcomes.
5. From a class of 30, one student is selected at random. List the events: "selected student is a girl", "selected student is a boy", "selected student is left-handed".
Sample space \(S=\) {30 students}. Each event is the appropriate subset; mutually exclusive: girl/boy. Left-handedness can overlap with girl or boy.
7. One die is thrown twice. Sample space?
\(S=\{(i,j):1\le i,j\le 6\}\), \(|S|=36\).
Exercise 14.2 — Probability calculations
1. Which of the following can be a probability function for the sample space \(\{\omega_1,\ldots,\omega_6\}\)? (Various tables of values given.)
A valid probability function requires (i) all values \(\ge 0\), (ii) sum equals 1. Check each row of the table given in the textbook against these two rules; reject any row with negative entries or with a sum \(\ne 1\).
2. A coin is tossed twice; what is the probability of "at least one tail"?
\(S=\{HH,HT,TH,TT\}\). Event = \(\{HT,TH,TT\}\), size 3. \(P=3/4\). (Or 1 − P(HH) = 1 − 1/4 = 3/4.)
3. A die is thrown. Find P(prime number), P(number > 4), P(non-zero number).
P(prime) = 3/6 = 1/2. P(>4) = 2/6 = 1/3. P(non-zero) = 6/6 = 1 (every face is non-zero).
4. A card is drawn from a deck. Find P(card is an ace) and P(card is black).
P(ace) = 4/52 = 1/13. P(black) = 26/52 = 1/2.
5. A fair coin and a fair die are tossed. Find probabilities of (i) head and a 3, (ii) head and an even number, (iii) tail or a 5.
|S|=12. (i) {H3} = 1/12. (ii) {H2,H4,H6} = 3/12 = 1/4. (iii) Tail (6 outcomes) ∪ "5" on die (H5,T5 — 2 outcomes), but T5 was already counted; |T| ∪ |"5"| = 6 + 1 = 7. So 7/12.
6. There are 4 men and 6 women on an organising committee. One member is chosen at random. P(man chosen)?
P(man) = 4/10 = 2/5.
7. A fair die is rolled. \(E\) = "score divisible by 3". Compute \(P(E)\) and \(P(E')\).
E = {3, 6}, P(E) = 2/6 = 1/3. P(E') = 1 − 1/3 = 2/3.
8. Two dice rolled. Find P(sum even), P(sum is multiple of 3), P(sum is at most 5).
|S|=36. Sum even: 18 outcomes (sums 2,4,6,8,10,12 with counts 1,3,5,5,3,1) → 18/36 = 1/2. Sum multiple of 3: sums 3,6,9,12 with counts 2,5,4,1 → 12/36 = 1/3. Sum ≤ 5: counts 1,2,3,4 → 10/36 = 5/18.
11. P(A) = 0.42, P(B) = 0.48, P(A ∩ B) = 0.16. Find P(not A), P(not B), P(A ∪ B).
P(A') = 0.58. P(B') = 0.52. P(A∪B) = 0.42 + 0.48 − 0.16 = 0.74.
14. P(A) = 5/13, P(B) = 7/13, P(A ∩ B) = 4/13. Find P(A ∪ B), P(A' ∩ B').
P(A∪B) = 5/13 + 7/13 − 4/13 = 8/13. P(A'∩B') = P((A∪B)') = 1 − 8/13 = 5/13.
Miscellaneous Exercise on Chapter 14
1. A box contains 10 red, 20 blue, 30 green marbles. 5 marbles are drawn. P(all blue)?
|S| = \(\binom{60}{5}=5\,461\,512\). Favourable = \(\binom{20}{5}=15\,504\). P = 15504/5461512 = 34/11977 ≈ 0.0028.
2. 4 cards drawn from a 52-deck. P(at least one is a king)?
P(no king) = \(\binom{48}{4}/\binom{52}{4}=194580/270725\). P(at least one king) = 1 − 194580/270725 = 76145/270725 ≈ 0.2813.
3. A die is loaded so that even faces are twice as likely as odd faces. P(getting an even number greater than 2)?
Let odd-face probability = p. Even-face = 2p. Sum: 3p + 6p = 9p = 1, so p = 1/9. Even > 2 = {4, 6}, P = 2(2p) = 4p = 4/9.
5. A and B are two events. P(A) = 1/4, P(B) = 1/2, P(A ∩ B) = 1/8. Find P(not A and not B).
P(A∪B) = 1/4 + 1/2 − 1/8 = 5/8. P(A'∩B') = 1 − P(A∪B) = 3/8.
7. In an entrance test, P(failing in math) = 0.30, P(failing in English) = 0.20, P(failing in both) = 0.10. P(failing in either)?
P(M ∪ E) = 0.30 + 0.20 − 0.10 = 0.40.
Activity: Build the Birthday-Paradox Probability
L4 Analyse
Materials: Calculator.
Predict: In a group of 23 people, what is the probability that at least 2 share a birthday? Most people guess very low — say 5%. The truth surprises everyone.
  1. Use the complement: P(all 23 birthdays distinct).
  2. P(distinct) = (365/365)·(364/365)·(363/365)·…·(343/365), 23 factors.
  3. Compute step by step: this product ≈ 0.4927.
  4. So P(at least one match) ≈ 1 − 0.4927 = 0.5073, just over 50%!
  5. For 50 people, P(match) ≈ 0.97. The famous "birthday paradox".
The intuition fails because we're not asking "does someone share MY birthday?" (which would be ≈ 23/365 ≈ 6%) but "does ANY pair share?" — and there are \(\binom{23}{2}=253\) pairs. Each pair has probability 1/365 of matching, so the expected number of matches ≈ 253/365 ≈ 0.7 — high enough that real matches are common.

Consolidation Competency-Based Questions

Scenario: A weather app's forecast for tomorrow says: P(rain) = 0.4, P(thunder) = 0.2, P(rain AND thunder) = 0.15.
Q1. P(rain or thunder)
L3 Apply
Answer: 0.4 + 0.2 − 0.15 = 0.45.
Q2. P(neither rain nor thunder)
L3 Apply
Answer: 1 − 0.45 = 0.55.
Q3. (T/F) "If P(A) = 0.7 and P(B) = 0.5, then A and B cannot be mutually exclusive." Justify.
L5 Evaluate
True. If mutually exclusive, P(A ∪ B) = 1.2 — impossible since probabilities are ≤ 1. So they MUST overlap.
Q4. Apply: in a group of 5 students, what is the probability that at least 2 share the same birth month? Use the complement method.
L4 Analyse
Solution: P(all distinct months) = (12·11·10·9·8)/12⁵ = 95040/248832 ≈ 0.382. P(match) ≈ 0.618 — about 62%.
Q5. Design: a quality-control engineer flags batches if the probability of "≥ 1 defective in 50 items" exceeds 0.5. The defect rate of an individual item is p. What is the largest p the manufacturer can have while staying under the threshold?
L6 Create
Solution: P(no defective) = (1−p)^{50}; we want P(≥ 1 defective) = 1 − (1−p)^{50} ≤ 0.5, i.e. (1−p)^{50} ≥ 0.5. Take 50th root: 1−p ≥ 0.5^{1/50} ≈ 0.9862, so p ≤ 0.0138 (≈ 1.4%). Manufacturer must keep defect rate under ~1.4%.

Consolidation Assertion–Reason

Assertion (A): If P(A) = 0.6 and P(B) = 0.5, then P(A ∩ B) ≥ 0.1.
Reason (R): P(A ∪ B) ≤ 1, and P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). 1 ≥ 0.6 + 0.5 − P(A∩B) ⇒ P(A∩B) ≥ 0.1.
Assertion (A): The probability of an event always lies between 0 and 1.
Reason (R): Probability counts a fraction (favourable / total), and a fraction of a non-negative count by a positive total is between 0 and 1.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). R captures the classical case; the axiomatic statement is the same conclusion derived from axioms 1, 2, 3.

Chapter Summary

Key formulas
  • Sample space \(S\): set of all outcomes of a random experiment.
  • Event: any subset of \(S\). Types: impossible (∅), sure (S), simple (singleton), compound, complementary (\(A'\)).
  • Mutually exclusive: \(A \cap B = \varnothing\). Exhaustive: \(A_1\cup\cdots\cup A_n=S\).
  • Probability axioms (Kolmogorov): \(P(A)\ge 0\), \(P(S)=1\), \(P(\bigcup_i A_i)=\sum_i P(A_i)\) for pairwise disjoint events.
  • Equally likely: \(P(A)=|A|/|S|\).
  • Complement: \(P(A')=1-P(A)\).
  • Addition theorem: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\); for mutually exclusive events the last term is 0.
  • De Morgan / "neither": \(P(A'\cap B')=1-P(A\cup B)\).

Historical Note

Probability theory likely many other branches of mathematics, evolved out of practical consideration. It had its origin in the 16th century when an Italian physician and mathematician Jerome Cardan (1501–1576) wrote the first book on the subject "Book on Games of Chance" (Liber de Ludo Aleae). It was published in 1663 after his death.

In 1654, a gambler Chevalier de Mere approached the well-known French Philosopher and Mathematician Blaise Pascal (1623–1662) for certain dice problem. Pascal became interested in these problems and discussed with famous French Mathematician Pierre de Fermat (1601–1665). Both Pascal and Fermat solved the problem independently. Besides Pascal and Fermat, outstanding contributions to probability theory were also made by Christian Huygens (1629–1665), a Dutchman, J. Bernoulli (1654–1705), De Moivre (1667–1754), a Frenchman Pierre Laplace (1749–1827), the Russian P.L. Chebyshev (1821–1897), A.A. Markov (1856–1922) and A.N. Kolmogorov (1903–1987). Kolmogorov is credited with the axiomatic theory of probability. His book Foundations of Probability published in 1933, introduced probability as a set function and is considered a classic.

Frequently Asked Questions

What is the probability of getting a head when a fair coin is tossed?
P(H) = 1/2, since the sample space {H, T} has 2 equally likely outcomes.
What is the probability of drawing an ace from a 52-card deck?
There are 4 aces among 52 cards, so P(ace) = 4/52 = 1/13.
What is the chapter summary of Class 11 Maths Chapter 14?
A random experiment has a sample space S of all possible outcomes. An event is a subset of S. Probability is a function P satisfying P(A) ≥ 0, P(S) = 1, and P(A ∪ B) = P(A) + P(B) for mutually exclusive events.
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