Solution : Total number of cases = \(^9C_3\) = 84 Number of favourable cases = \(^3C_1\).\(^4C_1\).\(^2C_1\) = 24 \(\therefore\)  P = \(24\over 84\) = \(2\over 7\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that in a match series […]

Solution : \(P({A’ \cap B’\over C})\) = \(P(A’ \cap B’ \cap C)\over P(C)\) = \(P(C) – P(A \cap C) – P(B \cap C) + P(A \cap B\cap C)\over P(C)\)   ……..(i) Given, \(P(A \cap B\cap C)\)  = 0 and A, B and C are pairwise independent. \(\therefore\)  \(P(A \cap C)\) = P(A).P(C) and \(P(B \cap C)\)

Solution : Probability of guessing a correct answer, p = \(1\over 3\) and probability of guessing a wrong answer, q  = \(2\over 3\) So, the probability of guessing 4 or more correct answers is = \(^5C_4\) \(({1\over 3})^4\). \(2\over 3\) + \(^5C_5\) \(({1\over 3})^5\) = \(5.2\over {3^5}\) + \(1\over {3^5}\) = \(11\over {3^5}\) Similar Questions

Question : If C and D are two events such that C \(\subset\) D and P(D) \(\ne\) 0, then the correct statement among the following is (a) P(C/D) \(\ge\) P(C) (b) P(C/D) < P(C) (c) P(C/D) = \(P(D)\over P(C)\) (d) P(C/D) = P(C) Solution : As P(C/D) = \(P(C \cap D)\over P(D)\) = \(P(C)\over P(D)\) 

Solution : Here, n = 5 and r \(\ge\) 1 \(\therefore\)   p(X = r) = \(^nC_r\) \(p^{n-r}\) \(q^r\) P(X \(\ge\) 1) = 1 – P(X = 0) = 1 – \(^5C_0 . p^5 . q^0\) \(\ge\) \(31\over 32\)   [Given] \(\implies\)   \(p^5\) \(\le\) 1 – \(31\over 32\) = \(1\over 32\) \(\therefore\)  p \(\le\) \(1\over 2\) and