Solution : The total number of ways in which numbers can be choosed = 25*25 = 625 The number of ways in which either players can choose same numbers = 25 \(\therefore\) Probability that they win a prize = \(25\over 625\) = \(1\over 25\) Thus, the probability that they will not win a prize in …
Solution : Since, total number of students = 100 and number of boys = 70 \(\therefore\) number of girls = (100 – 70) = 30 Now, the total marks of 100 students = 100*72 = 7200 And total marks of 70 boys = 70*75 = 5250 Total marks of 30 girls = 7250 – 5250 …
Solution : Since the probability of solving the problem by A, B and C is \(1\over 2\), \(1\over 3\) and \(1\over 4\) respectively. \(\therefore\) Probability that problem is not solved is = P(A’)P(B’)P(C’) = (\(1 – {1\over 2}\))(\(1 – {1\over 3}\))(\(1 – {1\over 4}\)) = \(1\over 2\)\(2\over 3\)\(3\over 4\) = \(1\over 4\) = Hence, the …
Solution : Given that, for binomial distribution mean, np = 4 and variance, npq = 2. \(\therefore\) q = 1/2, but p + q = 1 \(\implies\) p = 1/2 and n \(\times\) \(1\over 2\) = 4 \(\implies\) n = 8 We know that, P(X = r) = \(^nC_r p^r q^{n-r}\) \(\therefore\) P(X = 1) …
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 …
