Solution : A = {4. 5. 6} and B = {1, 2, 3, 4} \(A \cap B\) = 4 \(\therefore\) \(P(A \cup B)\) = P(A) + P(B) – \(A \cap B\) \(\implies\) \(P(A \cup B)\) = 3/6 + 4/6 – 1/6 = 1 Similar Questions The probability of India winning a test match against the …
Solution : We know that, P(A/B) = \(P(A \cap B)\over P(B)\) …….(i) and P(B/A) = \(P(B \cap A)\over P(A)\) ……….(ii) \(\therefore\) P(B) = \(P(B/A).P(A)\over P(A/B)\) = \(1\over 3\) 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 …
Solution : \(\therefore\) Variance = [\({1\over n}\sum{(x_i)^2}\)] – \((\bar{x})^2\) = \(1\over n\)[\(2^2 + 4^2 + ….. + (2n)^2\)] – \((n+1)^2\) = \(1\over n\)\(2^2 [ 1^2 + 2^2 + ….. + n^2]\) – \((n+1)^2\) = \(4\over n\) \(n(n + 1)(2n + 1)\over 6\) – \((n+1)^2\) = \((n + 1)[(2n + 1) – 3(n + 1)\over 3\) …
Find the variance of first n even natural numbers ? Read More »
Solution : S = { 00, 01, 02, ……, 49 } Let A be the event that sum of the digits on the selected ticket is 8, then A = { 08, 17, 26, 35, 44 } Let B be the event that the product of the digits is zero. B = { 00, 01, …
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 : Correct mean = old mean + 2 = 30 + 2 = 32 As standard deviation is independent of change of origin. Hence, it remains same. Similar Questions The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is The median …
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 : Median of a, 2a, 3a, 4a, ….. . 50a is \(25a + 26a\over 2\) = 25.5a Mean deviation = \(\sum{|x_i – Median|}\over N\) \(\implies\) 50 = \(1\over 50\) {2|a|.(0.5 + 1.5 + …… + 24.5)] \(\implies\) 2500 = 2|a|. \(25\over 2\) (25) \(\implies\) |a| = 4 Similar Questions The mean and variance of …
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 : All the students of a class performed poorly in mathematics. The teacher decided to give grace marks of 10 to each of the students. Which statistical measure will not change even after the grace marks were given ? (a) Mean (b) Median (c) Mode (d) Variance Solution : If initially all marks were …
