Solution : Given, probabilities of speaking truth are P(A) = \(4\over 5\) and P(B) = \(3\over 4\) And their corresponding probabilities of not speaking truth are P(A’) = \(1\over 5\) and P(B’) = \(1\over 4\) The probability that they contradict each other = P(A).P(B’) + P(B).P(A’) = \(4\over 5\) \(\times\) \(1\over 4\) + \(1\over 5\) …
Solution : In the 2n observations, half of them equals a and remaining half equal -a. Then, the mean of total 2n observations is equal to 0. \(\therefore\) SD = \(\sqrt{\sum(x – \bar{x})^2\over N}\) \(\implies\) 4 = \(\sum{x^2}\over 2n\) \(\implies\) 4 = \(2na^2\over 2n\) \(\implies\) \(a^2\) = 4 \(\therefore\) a = 2 Similar Questions The …
Solution : Now, P(X > 1.5) = P(2) + P(3) + …… \(\infty\) = 1 – [P(0) + P(1)] = 1 – \((e^{-2} + {e^{-2}(2)\over 1})\) = 1 – \(3\over e^2\) Similar Questions The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) …
A random variable X has poisson distribution with mean 2. Then, P(X > 1.5) equal to Read More »
Solution : \(\therefore\) Total number of cases = \(3^3\) Now, favourable cases = 3(as either all has applied for house 1 or 2 or 3) \(\therefore\) Required Probability = \(3\over 3^3\) = \(1\over 9\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence from match to …
Solution : Given that, mean = 21 and median = 22 Using the relation, Mode = 3 median – 2 mean \(\implies\) Mode = 3(22) – 2(21) = 66 – 42 = 24 Similar Questions The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X …
Question : Let \(x_1\), \(x_2\), ….. , \(x_n\), be n observations such that \(\sum{x_i}^2\) = 400 and \(\sum{x_i}\) = 80. Then, a possible value of among the following is (a) 12 (b) 9 (c) 18 (d) 15 Solution : Given \(\sum{x_i}^2\) = 400 and \(\sum{x_i}\) = 80 \(\because\) \(\sigma^2\) \(\ge\) 0 \(\therefore\) \(\sum{x_i}^2\over n\) – …
Solution : Required Probability = P(X = 0) + P(X = 1) = \(e^{-5}\over 0!\).\(5^0\) + \(e^{-5}\over 1!\).\(5^1\) = \(e^{-5}\) + 5\(e^{-5}\) = \(6\over {e^5}\) 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 India’s …
Solution : Since variance is independent of change of origin. Therefore, variance of observations 101, 102, …. , 200 is same as variance of 151, 152, ….. 250. \(\therefore\) \(V_A\) = \(V_B\) \(\implies\) \(V_A\over V_B\) = 1 Similar Questions The mean and variance of a random variable X having a binomial distribution are 4 and …
Solution : Let the number of boys and girls be x and y, respectively \(\therefore\) 52x + 42y = 50(x + y) \(\implies\) 52x + 42y = 50x + 50y \(\implies\) 2x = 8y \(\implies\) x = 4y \(\therefore\) Total number of students in the class = x + y = 4y + y = …
Solution : Let the events, A = Ist aeroplane hit the target B = 2nd aeroplane hit the target And their corresponding probabilities are P(A) = 0.3 and P(B) = 0.2 \(\implies\) P(A’) = 0.7 and P(B’) = 0.8 \(\therefore\) Required Probability = P(A’)P(B) + P(A’)P(B’)P(A’)P(B) + ……. = (0.7)(0.2) + (0.7)(0.8)(0.7)(0.2) + ……. = …
