Study MaterialsNCERT SolutionsNCERT Solutions For Class 12 MathematicsNCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.2

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.2

Exercise 13.2 of NCERT Solutions for Class 12 Maths Chapter 13 – Probability covers the following topics:

    Fill Out the Form for Expert Academic Guidance!



    +91

    Verify OTP Code (required)


    I agree to the terms and conditions and privacy policy.

    1. Multiplication Theorem on Probability
    2. Independent Events: These are events where the probability of one occurring is not influenced by the occurrence of the other.

    This exercise contains problems focused on these topics, helping students understand and apply probability concepts in real-life scenarios.

    NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.2 – Free PDF Download

    Do you need help with your Homework? Are you preparing for Exams? Study without Internet (Offline)
    ×

      Download PDF for Free.
      Study without Internet (Offline)



      +91



      Live ClassesBooksTest SeriesSelf Learning



      Verify OTP Code (required)

      I agree to the terms and conditions and privacy policy.

      Exercise 13.2

      Access NCERT Solutions for Class 12 Maths Chapter Probability 13.2

      1. If P (A) = 3/5 and P (B) = 1/5, find P (A ∩ B) if A and B are independent events.

      Solution:

      Given P (A) = 3/5 and P (B) = 1/5

      As A and B are independent events.

      ⇒ P (A ∩ B) = P (A).P (B)

      = 3/5 × 1/5 = 3/25

      2. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

      Solution:

      Given a box of oranges.

      Let A, B and C denotes the events respectively that the first, second and third drawn orange is good.

      Now, P (A) = P (good orange in first draw) = 12/15

      Because the second orange is drawn without replacement, now the total number of good oranges will be 11, and the total oranges will be 14; that is the conditional probability of B, given that A has already occurred.

      Now, P (B/A) = P (good orange in second draw) = 11/14

      Because the third orange is drawn without replacement, now the total number of good oranges will be 10, and the total oranges will be 13; that is the conditional probability of C, given that A and B have already occurred.

      Now, P (C/AB) = P (good orange in third draw) = 10/13

      Thus the probability that all the oranges are good.

      ⇒ P (A ∩ B ∩ C) = 12/15 × 11/14 × 10/13 = 44/91

      Hence, the probability that a box will be approved for sale = 44/91

      3. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

      Solution:

      Given a fair coin and an unbiased die are tossed.

      We know that the sample space S:

      S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}

      Let A be the event head that appears on the coin:

      ⇒ A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}

      ⇒ P (A) = 6/12 = ½

      Now, Let B be the event 3 on the die

      ⇒ B = {(H, 3), (T, 3)}

      ⇒ P (B) = 2/12 = 1/6

      As, A ∩ B = {(H, 3)}

      ⇒ P (A ∩ B) = 1/12 …… (1)

      And P (A). P (B) = ½ × 1/6 = 1/12 …… (2)

      From (1) and (2) P (A ∩ B) = P (A). P (B)

      Therefore, A and B are independent events.

      4. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?

      Solution:

      The sample space for the dice will be

      S = {1, 2, 3, 4, 5, 6}

      Let A be the event, the number is even:

      ⇒ A = {2, 4, 6}

      ⇒ P (A) = 3/6 = ½

      Now, Let B be the event the number is red:

      ⇒ B = {1, 2, 3}

      ⇒ P (B) = 3/6 = 1/2

      As, A ∩ B = {2}

      ⇒ P (A ∩ B) = 1/6 …….. (1)

      And P (A). P (B) = ½ × ½ = ¼ ….. (2)

      From (1) and (2) P (A ∩ B) ≠ P (A). P (B)

      Therefore, A and B are not independent events.

      5. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
      A. 0
      B. 1/3
      C. 1/12
      D. 1/36

      Solution:

      D. 1/36

      Explanation:

      Given A pair of dice is rolled.

      Hence the number of outcomes = 36

      Let P (E) be the probability of getting an even prime number on each die.

      As we know, the only even prime number is 2.

      So, E = {2, 2}

      ⇒ P € = 1/36

      NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2 Tips

      NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.2 is an excellent resource prepared by experts to help students understand independent events and the multiplication theorem. These reliable guides enable students to solve problems independently and are essential for scoring well in exams.

      NCERT solutions aid in learning the properties, terms, and definitions related to independent events and their applications. The exemplar problems in Chapter 13 require a good understanding of conditional probability, making it crucial to grasp the concepts from the previous exercise thoroughly. Download the Infinity Learn NCERT Solutions PDF for free and start learning today.

      Chat on WhatsApp Call Infinity Learn