Solution : Probability of getting score 9 in a single throw = \(4\over 36\) = \(1\over 9\) Probability of getting score 9 exactly in double throw = \(^3C_2\) \(\times\) \(({1\over 9})^2\) \(\times\) \(8\over 9\) = \(8\over 243\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence …
Solution : According to given condition, 6.80 = \((6-a)^2 + (6-b)^2 + (6-8)^2 + (6-5)^2 + (6-10)^2\over 5\) \(\implies\) 34 = \((6-a)^2 + (6-b)^2\) + 4 + 1 + 16 \(\implies\) \((6-a)^2 + (6-b)^2\) = 13 \(\implies\) \((6-a)^2 + (6-b)^2\) = 13 = \(3^2\) + \(2^2\) \(\implies\) a = 3 and b = 4 Similar …
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 …
