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EAPCET Mathematics · Intermediate second year · Paper IIA

Probability

Probability covers classical and axiomatic probability, the addition and multiplication theorems, conditional probability, independent events and Bayes’ theorem.

Subject
Mathematics
Taught in
Intermediate second year · Paper IIA
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)
  • P(A′)=1−P(A)P(A') = 1 - P(A)
  • P(A∣B)=P(A∩B)P(B)P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}
  • AA and BB are independent when P(A∩B)=P(A) P(B)P(A \cap B) = P(A)\,P(B)
  • Bayes: P(Ei∣A)=P(Ei) P(A∣Ei)∑jP(Ej) P(A∣Ej)P(E_{i} \mid A) = \dfrac{P(E_{i})\,P(A \mid E_{i})}{\sum_{j} P(E_{j})\,P(A \mid E_{j})}
  • P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)+P(A∩B∩C)P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)
  • P(exactly one of A,B)=P(A)+P(B)−2P(A∩B)P(\text{exactly one of } A, B) = P(A) + P(B) - 2P(A \cap B)
  • P(A∩B′)=P(A)−P(A∩B)P(A \cap B') = P(A) - P(A \cap B) and P(A′∩B′)=1−P(A∪B)P(A' \cap B') = 1 - P(A \cup B)
  • P(at least two of A,B,C)=P(A∩B)+P(B∩C)+P(C∩A)−2P(A∩B∩C)P(\text{at least two of } A, B, C) = P(A \cap B) + P(B \cap C) + P(C \cap A) - 2P(A \cap B \cap C)
  • P(exactly two of A,B,C)=P(A∩B)+P(B∩C)+P(C∩A)−3P(A∩B∩C)P(\text{exactly two of } A, B, C) = P(A \cap B) + P(B \cap C) + P(C \cap A) - 3P(A \cap B \cap C)
  • Total probability: P(A)=∑P(Bi) P(A∣Bi)P(A) = \sum P(B_{i})\,P(A \mid B_{i}) for mutually exclusive and exhaustive B1,…,BnB_{1}, \dots, B_{n}
  • Multiplication theorem: P(A∩B)=P(A) P(B∣A)=P(B) P(A∣B)P(A \cap B) = P(A)\,P(B \mid A) = P(B)\,P(A \mid B)
  • Odds in favour of E=P(E):P(E′)E = P(E) : P(E'); odds against E=P(E′):P(E)E = P(E') : P(E)

What this chapter covers

  • Classical and axiomatic definitions
  • Addition theorem
  • Conditional probability
  • Multiplication theorem and independent events
  • Total probability and Bayes’ theorem

How it is usually asked

  • Problems on dice, coins and cards
  • Conditional probability
  • Bayes’ theorem with urns or machines
  • Independent events

Understanding Probability

In the classical definition, when all outcomes of an experiment are equally likely, the probability of an event is the number of favourable outcomes divided by the total number of outcomes. The axiomatic view generalises this: probabilities are non-negative, the whole sample space has probability 1, and probabilities of mutually exclusive events add. From these come P(A′) = 1 − P(A) and the addition theorem P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Conditional probability P(A | B) = P(A ∩ B)/P(B) is the probability of A once we know B has happened; it shrinks the sample space to B. Rearranged, it gives the multiplication theorem P(A ∩ B) = P(B) P(A | B). Events are independent when knowing one does not change the other, which is exactly P(A ∩ B) = P(A) P(B).

When an outcome can arise through several mutually exclusive causes, the total probability theorem adds the contributions of each cause, and Bayes’ theorem reverses the question to find the probability of a cause given the outcome. EAPCET questions use dice, coins, cards, urns and machines, and most errors come from a wrongly counted sample space.

Worked example

EAPCET-style question

Two fair dice are thrown. Given that at least one die shows 5, the probability that the sum of the numbers is 8 is

  1. (1)211\dfrac{2}{11}
  2. (2)536\dfrac{5}{36}
  3. (3)118\dfrac{1}{18}
  4. (4)29\dfrac{2}{9}

Solution

  1. BB = at least one 5. The outcomes with no 5 number 5×5=255 \times 5 = 25, so BB has 36−25=1136 - 25 = 11 outcomes.
  2. A∩BA \cap B = sum 8 with a 5: (3,5)(3, 5) and (5,3)(5, 3), which is 2 outcomes.
  3. P(A∣B)=n(A∩B)n(B)=211P(A \mid B) = \dfrac{n(A \cap B)}{n(B)} = \dfrac{2}{11}.
  4. 536\tfrac{5}{36} is the unconditional probability of a sum of 8, which ignores the given condition.

Answer: 211\dfrac{2}{11}

Shortcuts and results to remember

  • P(A∩B)≤P(A)≤P(A∪B)≤P(A)+P(B)P(A \cap B) \le P(A) \le P(A \cup B) \le P(A) + P(B).
  • Least possible P(A∩B)=P(A)+P(B)−1P(A \cap B) = P(A) + P(B) - 1 when that is positive; for example, if 63% like one thing and 76% like another, at least 39% like both.
  • Mutually exclusive means P(A∩B)=0P(A \cap B) = 0; independent means P(A∩B)=P(A) P(B)P(A \cap B) = P(A)\,P(B). Do not mix them up.

More practice questions

Practice question 1

Bag 1 has 3 red and 2 black balls, bag 2 has 2 red and 3 black, bag 3 has 2 red and 2 black. A die is rolled: on a prime number bag 1 is used, on 6 bag 2, otherwise bag 3. A ball is drawn. The probability that it is red is

  1. (1)815\dfrac{8}{15}
  2. (2)12\dfrac{1}{2}
  3. (3)715\dfrac{7}{15}
  4. (4)35\dfrac{3}{5}

Solution

  1. P(bag 1)=36=12P(\text{bag 1}) = \tfrac{3}{6} = \tfrac{1}{2} (2, 3, 5), P(bag 2)=16P(\text{bag 2}) = \tfrac{1}{6}, P(bag 3)=26=13P(\text{bag 3}) = \tfrac{2}{6} = \tfrac{1}{3} (1, 4)
  2. P(red)=(12)(35)+(16)(25)+(13)(24)P(\text{red}) = \left(\tfrac{1}{2}\right)\left(\tfrac{3}{5}\right) + \left(\tfrac{1}{6}\right)\left(\tfrac{2}{5}\right) + \left(\tfrac{1}{3}\right)\left(\tfrac{2}{4}\right)
  3. =310+115+16=930+230+530= \tfrac{3}{10} + \tfrac{1}{15} + \tfrac{1}{6} = \tfrac{9}{30} + \tfrac{2}{30} + \tfrac{5}{30}
  4. =1630=815= \tfrac{16}{30} = \tfrac{8}{15}

Answer: 815\dfrac{8}{15}

Practice question 2

Two numbers are chosen at random from the first 20 natural numbers. The probability that their sum is even is

  1. (1)919\dfrac{9}{19}
  2. (2)1019\dfrac{10}{19}
  3. (3)12\dfrac{1}{2}
  4. (4)920\dfrac{9}{20}

Solution

  1. There are 10 odd and 10 even numbers; total selections 20C2=190{}^{20}C_{2} = 190
  2. Sum is even when both are odd or both are even: 10C2+10C2=45+45=90{}^{10}C_{2} + {}^{10}C_{2} = 45 + 45 = 90
  3. Probability =90190=919= \dfrac{90}{190} = \dfrac{9}{19}

Answer: 919\dfrac{9}{19}

Common mistakes to avoid

  • Treating mutually exclusive events as independent; if P(A) and P(B) are positive, mutually exclusive events cannot be independent.
  • Dividing by P(A) instead of P(B) for P(A | B); you divide by the probability of the event that is given.
  • Counting outcomes of two dice as unordered pairs; there are 36 equally likely ordered outcomes, not 21.
  • Computing “at least one” by adding cases; it is usually faster and safer as 1 − P(none).

How to prepare it

Write the size of the sample space before anything else — it prevents most counting errors.

Once the formulas feel familiar, test them the way the exam does: practise Probability questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Probability chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Probability part of the EAPCET Mathematics syllabus?
Yes. Probability is taught in Intermediate second year · Paper IIA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Probability for EAPCET?
Write the size of the sample space before anything else — it prevents most counting errors.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Probability?
Treating mutually exclusive events as independent; if P(A) and P(B) are positive, mutually exclusive events cannot be independent. Dividing by P(A) instead of P(B) for P(A | B); you divide by the probability of the event that is given. Counting outcomes of two dice as unordered pairs; there are 36 equally likely ordered outcomes, not 21. Computing “at least one” by adding cases; it is usually faster and safer as 1 − P(none).