Solution : The probability that Mr A selected the lossing horse = \(4\over 5\) \(\times\) \(3\over 4\) = \(3\over 5\) The probability that Mr A selected the winning horse = 1 – \(3\over 5\) = \(2\over 5\) Similar Questions Consider rolling two dice once. Calculate the probability of rolling a sum of 5 or an […]
Solution : Sample space of rolling two dice = 36 Now, Event of sum of 5 = { (1,4) (2,3) (3,2) (4,1) } = 4 \(\implies\) probability of getting sum of 5 is 4/36 = 1/9 Now, Event of even sum = { (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6)
Solution : Median of new set remains the same as of the original set. 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 Let \(x_1\), \(x_2\), ….. , \(x_n\), be n observations such that \(\sum{x_i}^2\) = 400 and \(\sum{x_i}\)
Solution : Given that, mean = 4 \(\implies\) np = 4 And Variance = 2 \(\implies\) npq = 2 \(\implies\) 4q = 2 \(\implies\) q = \(1\over 2\) \(\therefore\) p = 1 – q = 1 – \(1\over 2\) = \(1\over 2\) Also, n = 8 Probability of 2 successes = P(X = 2) =
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 \(\therefore\) \(\sum{x_i}^2\over n\) –
