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Probability Test - 4
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Probability Test - 4
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  • Question 1/3
    1 / -0

    A number is chosen randomly from the set {1, 2, 3, …15}. If chosen number is a two digit number, then the probability that the number is a prime numbers, is

    Solutions

    Required probability = 2 / 6 = 1/3 
    Hence, option B is correct.

  • Question 2/3
    1 / -0

    A single letter is selected at random from the word 'PROBABILITY'. The probability that it is a vowel is –

    Solutions

    PROBABILITY 
    vowel = O, A, I, I 
    Total case = 11C1 = 11 
    Favourable case = 4C1 = 4 
    Required probability = 4/11

  • Question 3/3
    1 / -0

    Let A, B, C be three mutually independent events. Consider the two statements S1 and S2 
    S1 : A and B ∪ C are independent 
    S2 : A and B ∩ C are independent 
    Then,

    Solutions

    We are given that

    P(A ∩ B) = P(A) P(B)
    P(B ∩ C) = P(B) P(C), P(C ∩ A) = P(C) P(A)
    And P( A ∩ B ∩ C)=P(A) P(B) P(C)
    =P(A) P(B) P(C) = P(A) P(B ∩ C).
    ⟹ A and B ∩ C are independent. Therefore, S2 is true.
    Also P[(A ∩ (B ∪ C)]= P[(A ∩ B) ∪ (A ∩ C)]
    =P(A ∩B) + P(A ∩ C) –P[A ∩ B) ∩ (A ∩ C)]
    =P(A ∩ B) + P(A ∩ C) – P(A ∩ B ∩ C)

    =P(A) P(B) +P(A) P(C) – P(A) P(B) P(C)
    =P(A) [P(B) +P(C)-P(B) P(C)]
    =P(A) [P(B) + P(C) –P(B ∩ C)] = P(A) P(B ∪ C)
    ∴ A and B ∪ C are independent.

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