Solution : Let a, ar, \(ar^2\) are in GP (r > 1) According to the question, a, 2ar, \(ar^2\) in AP. \(\implies\) 4ar = a + \(ar^2\) \(\implies\) \(r^2\) – 4r + 1 = 0 \(\implies\) r = \(2 \pm \sqrt{3}\) Hence, r = \(2 + \sqrt{3}\) [ \(\because\) AP is increasing] Similar Questions […]
Solution : \(K(10)^9\) = \((10)^9\) + \(2(11)^1(10)^8\) + \(3(11)^2(10)^7\) + …… + \(10(11)^9\) K = 1 + 2\(({11\over 10})\) + 3\(({11\over 10})^2\) + ….. + 10\(({11\over 10})^9\) ……(i) \(({11\over 10})\)K = 1\(({11\over 10})\) + 2\(({11\over 10})^2\) + 3\(({11\over 10})^3\) + ….. + 10\(({11\over 10})^{10}\) …..(ii) On subtracting equation (ii) from (i),
Solution : Let the probability of getting a head be p and not getting a head be q. Since, head appears first time in an even throw 2 or 4 or 6. \(\therefore\) \(2\over 5\) = qp + \(q^3\)p + \(q^5\)p + …… \(\implies\) \(2\over 5\) = \(qp\over {1 – q^2}\) Since q = 1-
Solution : Let \(A_1\), \(A_2\), \(A_3\) be the events of match winning in first, second and third matches respectively and whose probabilities are P(\(A_1\)) = P(\(A_2\)) = P(\(A_2\)) = \(1\over 2\) \(\therefore\) Required Probability = P(\(A_1\))P(\(A_2’\))P(\(A_3\)) + P(\(A_1’\))P(\(A_2\))P(\(A_3\)) = \(({1\over 2})^3\) + \(({1\over 2})^3\) = \(1\over 8\) + \(1\over 8\) = \(1\over 4\) Similar Questions
Solution : Probability of getting success, p = \(1\over 6\) and probability of failure, q = \(5\over 6\) Now, we must get 2 sixes in seven throws, so probability is \(^7C_2\)\(({1\over 6})^2\)\(({5\over 6})^5\) and the probability that 8th throw is \(1\over 6\). \(\therefore\) Required Probability = \(^7C_2\)\(({1\over 6})^2\)\(({5\over 6})^5\)\(1\over 6\) = \(^7C_2\times 5^5\over {6^8}\) Similar
Question : If A and B are two mutually exclusive events, then (a) P(A) < P(B’) (b) P(A) > P(B’) (c) P(A) < P(B) (d) None of these Solution : Since, A and B are two mutually exclusive events. \(\therefore\) A \(\cap\) B = \(\phi\) \(\implies\) Either A \(\subseteq\) B’ or B \(\subseteq\) A’
If A and B are two mutually exclusive events, then Read More »
Solution : The total number of ways in which numbers can be choosed = 25*25 = 625 The number of ways in which either players can choose same numbers = 25 \(\therefore\) Probability that they win a prize = \(25\over 625\) = \(1\over 25\) Thus, the probability that they will not win a prize in
Solution : Since, total number of students = 100 and number of boys = 70 \(\therefore\) number of girls = (100 – 70) = 30 Now, the total marks of 100 students = 100*72 = 7200 And total marks of 70 boys = 70*75 = 5250 Total marks of 30 girls = 7250 – 5250
Solution : Since the probability of solving the problem by A, B and C is \(1\over 2\), \(1\over 3\) and \(1\over 4\) respectively. \(\therefore\) Probability that problem is not solved is = P(A’)P(B’)P(C’) = (\(1 – {1\over 2}\))(\(1 – {1\over 3}\))(\(1 – {1\over 4}\)) = \(1\over 2\)\(2\over 3\)\(3\over 4\) = \(1\over 4\) = Hence, the
Solution : Given that, for binomial distribution mean, np = 4 and variance, npq = 2. \(\therefore\) q = 1/2, but p + q = 1 \(\implies\) p = 1/2 and n \(\times\) \(1\over 2\) = 4 \(\implies\) n = 8 We know that, P(X = r) = \(^nC_r p^r q^{n-r}\) \(\therefore\) P(X = 1)
