Maths Questions

If \({\sum}_{r=1}^{n‎} T_r\) = \(n\over 8\) (n + 1)(n + 2)(n + 3), then find \({\sum}_{r=1}^{n‎} \)\(1\over T_r\)

Solution : \(\because\) \(T_n\) = \(S_n – S_{n-1}\) = \({\sum}_{r=1}^{n‎} T_r\) – \({\sum}_{r=1}^{n‎ – 1} T_r\) = \(n(n+1)(n+2)(n+3)\over 8\) – \((n-1)(n)(n+1)(n+2)\over 8\) = \(n(n+1)(n+2)\over 8\)[(n+3) – (n-1)] = \(n(n+1)(n+2)\over 8\)(4) \(T_n\) = \(n(n+1)(n+2)\over 2\) \(\implies\) \(1\over T_n\) = \(2\over n(n+1)(n+2)\) = \((n+2)-n\over n(n+1)(n+2)\) = \(1\over n(n+1)\) – \(1\over (n+1)(n+2)\) Let \(V_n\) = \(1\over n(n+1)\) \(\therefore\) […]

If \({\sum}_{r=1}^{n‎} T_r\) = \(n\over 8\) (n + 1)(n + 2)(n + 3), then find \({\sum}_{r=1}^{n‎} \)\(1\over T_r\) Read More »

Let A = [x: x \(\in\) R, |x| < 1]; B = [x : x \(\in\) R, |x - 1| \(\ge\) 1] and A \(\cup\) B = R - D, then the set D is

Solution : A = [x: x \(\in\) R,-1 < x < 1] B = [x : x \(\in\) R, x – 1 \(\le\) -1 or x – 1 \(\ge\) 1] [x: x \(\in\) R, x \(\le\) 0 or x \(\ge\) 2] \(\therefore\) A \(\cup\) B = R – D where D = [x : x

Let A = [x: x \(\in\) R, |x| < 1]; B = [x : x \(\in\) R, |x - 1| \(\ge\) 1] and A \(\cup\) B = R - D, then the set D is Read More »

If N is a set of first 10 natural numbers and a relation R is defined as a + 2b = 10 where a, b \(\in\) N. find inverse of R.

Solution : R = {(2, 4), (4, 3), (6, 2), (8, 1)} \(R^{-1}\) = {(4, 2), (3, 4), (2, 6), (1, 8)} Similar Questions If A = {x,y}, then the power set of A is If aN = {ax : x \(\in\) N}, then the set 6N \(\cap\) 8N is equal to Let A =

If N is a set of first 10 natural numbers and a relation R is defined as a + 2b = 10 where a, b \(\in\) N. find inverse of R. Read More »

Three groups A, B, C are contesting for positions on the board of directors of a company. The probabilities of their winning are 0.5, 0.3, 0.2 respectively. If the group A wins, the probability of introducing a new product is 0.7 and the corresponding probabilities for group B and C are 0.6 and 0.5 respectively. Find the probability that the new product will be introduced.

Solution : Given P(A) = 0.5, P(B) = 0.3 and P(C) = 0.2 \(\therefore\) P(A) + P(B) + P(C) = 1 then events A, B, C are exhaustive. If P(E) = Probability of introducing a new product, then as given P(E|A) = 0.7, P(E|B) = 0.6 and P(E|C) = 0.5 = 0.5 \(\times\) 0.7 +

Three groups A, B, C are contesting for positions on the board of directors of a company. The probabilities of their winning are 0.5, 0.3, 0.2 respectively. If the group A wins, the probability of introducing a new product is 0.7 and the corresponding probabilities for group B and C are 0.6 and 0.5 respectively. Find the probability that the new product will be introduced. Read More »

A bag contains 4 red and 4 blue balls. Four balls are drawn one by one from the bag, then find the probability that the drawn balls are in alternate color.

Solution : \(E_1\) : Event that first drawn ball is red, second is blue and so on. \(E_2\) : Event that first drawn ball is blue, second is red and so on. \(\therefore\)  P(\(E_1\)) = \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\) and \(\therefore\)  P(\(E_2\)) = \(4\over 8\) \(\times\) \(4\over 7\)

A bag contains 4 red and 4 blue balls. Four balls are drawn one by one from the bag, then find the probability that the drawn balls are in alternate color. Read More »

From a group of 10 persons consisting of 5 lawyers, 3 doctors and 2 engineers, four persons are selected at random. The probability that selection contains one of each category is

Solution : n(S) = \(^{10}C_4\) = 210 n(E)= \(^5C_2 \times ^3C_1 \times ^2C_1\) + \(^5C_1 \times ^3C_2 \times ^2C_1\) + \(^5C_1 \times ^3C_1 \times ^2C_2\) = 105 \(\therefore\) P(E) = \(105\over 210\) = \(1\over 2\) Similar Questions How many different words can be formed by jumbling the letters in the word ‘MISSISSIPPI’ in which no

From a group of 10 persons consisting of 5 lawyers, 3 doctors and 2 engineers, four persons are selected at random. The probability that selection contains one of each category is Read More »