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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
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Required probability = 2 / 6 = 1/3 Hence, option B is correct.
A single letter is selected at random from the word 'PROBABILITY'. The probability that it is a vowel is –
PROBABILITY vowel = O, A, I, I Total case = 11C1 = 11 Favourable case = 4C1 = 4 Required probability = 4/11
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,
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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