Chapter 13 Probability
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Chapter 13 Probability

Probability Questions

NCERT Solutions for Class 12 Maths – Chapter 13 – PROBABILITY EXERCISE 13.1 Questions 1 to 15

Probability 15 Practice Questions

NCERT Solutions for Class 12 Maths – Chapter 13 – PROBABILITY EXERCISE 13.2 Questions 1 to 18

NCERT Solutions for Class 12 Maths – Chapter 13 – PROBABILITY EXERCISE 13.3 Questions 1 to 14

NCERT Solutions for Class 12 Maths – Chapter 13 – PROBABILITY Miscellaneous Exercise Questions 1 to 13

  1. Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E|F) and P(F|E)
  2. Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32
  3. If P(A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find (i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B)
  4. Evaluate P(A ∪ B), if 2P(A) = P(B) = 5/13 and P(A|B) = 2/5
  5. If P(A) = 6/11, P(B) = 5/11 and P(A ∪ B) = 7/11, find (i) P(A∩B) (ii) P(A|B) (iii) P(B|A)

    6. Determine P(E|F) in Exercises 6 to 9.

  6. A coin is tossed three times, where (i) E : head on third toss , F : heads on first two tosses (ii) E : at least two heads , F : at most two heads (iii) E : at most two tails , F : at least one tail
  7. Two coins are tossed once, where (i) E : tail appears on one coin, F : one coin shows head (ii) E : no tail appears, F : no head appears
  8. A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses
  9. Mother, father and son line up at random for a family picture E : son on one end, F : father in middle
  10. A black and a red dice are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
  11. A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5} Find (i) P(E|F) and P(F|E) (ii) P(E|G) and P(G|E) (iii) P((E ∪ F)|G) and P((E ∩ F)|G)
  12. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?
  13. An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
  14. Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
  15. Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

    16. In each of the Exercises 16 and 17 choose the correct answer:

  16. If P(A) = 1/2, P(B) = 0, then P(A|B) is
    (A) 0 (B) 1/2 (C) not defined (D) 1
  17. If A and B are events such that P(A|B) = P(B|A), then
    (A) A ⊂ B but A ≠ B (B) A = B (C) A ∩ B = φ (D) P(A) = P(B)

    18. Chapter 13 – PROBABILITY EXERCISE 13.1 Questions 1 to 17 Solutions

The salient features of the chapter are:

  • The conditional probability of an event E, given the occurrence of the event F, is given by P(EF) = P(E ∩ F)/P(F), where P(F) ≠ 0.
  • 0 ≤ P(EF) ≤ 1, P(E'F) = 1 - P(EF).
  • P((E ∪ F)G) = P(EG) + P(FG) - P((E ∩ F)G).
  • P(E ∩ F) = P(E) P(FE), where P(E) ≠ 0.
  • P(E ∩ F) = P(F) P(EF), where P(F) ≠ 0.
  • If E and F are independent, then P(E ∩ F) = P(E) P(F), P(EF) = P(E), P(F) ≠ 0, and P(FE) = P(F), P(E) ≠ 0.
  • Theorem of total probability: Let {E1, E2, ..., En} be a partition of a sample space and suppose that each of E1, E2, ..., En has nonzero probability. Let A be any event associated with S, then P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + ... + P(En) P(A|En).
  • Bayes' theorem: If E1, E2, ..., En are events which constitute a partition of sample space S, i.e. E1, E2, ..., En are pairwise disjoint and E1 ∪ E2 ∪ ... ∪ En = S and A be any event with nonzero probability, then...
                          P(E i )P(A|E i )
P(E i|A) = -----------------------------  
                       n 
                       ∑      P(E j )P(A|E j )
                       j=1