Question : If C and D are two events such that C \(\subset\) D and P(D) \(\ne\) 0, then the correct statement among the following is (a) P(C/D) \(\ge\) P(C) (b) P(C/D) < P(C) (c) P(C/D) = \(P(D)\over P(C)\) (d) P(C/D) = P(C) Solution : As P(C/D) = \(P(C \cap D)\over P(D)\) = \(P(C)\over P(D)\) …
Solution : Here, n = 5 and r \(\ge\) 1 \(\therefore\) p(X = r) = \(^nC_r\) \(p^{n-r}\) \(q^r\) P(X \(\ge\) 1) = 1 – P(X = 0) = 1 – \(^5C_0 . p^5 . q^0\) \(\ge\) \(31\over 32\) [Given] \(\implies\) \(p^5\) \(\le\) 1 – \(31\over 32\) = \(1\over 32\) \(\therefore\) p \(\le\) \(1\over 2\) and …
Solution : Let us assume that the coordinates of the center of the circle are C(h,k) and its radius is r. Now, since the circle touches X-axis at (1,0), hence its radius should be equal to ordinate of center. \(\implies\) r = k Hence, the equation of circle is \((x – h)^2 + (y – …
Solution : \({\sigma^2}\) = \(\sum(x_i – \bar{x})^2\over n\) \(\bar{X}\) = \(\sum x_i\over n\) = \(2 + 4 + 6 + 8 + ….. + 100\over 50\) = 51 \({\sigma^2}\) = \(2^2 + 4^2 + …. + 100^2\over 50\) – \(51^2\) = 833 Similar Questions The mean and variance of a random variable X having a …
The variance of first 50 even natural numbers is Read More »
Solution : Given P(A \(\cup\) B)’ = 1/6, P(A \(\cap\) B) = 1/4 and P(A)’ = 1/4 \(\therefore\) P(A \(\cup\) B) = 1 – P(A \(\cup\) B)’ = 1 – \(1\over 6\) = \(5\over 6\) and P(A) = 1 – P(A)’ = 1 – \(1\over 4\) = \(3\over 4\) P(A \(\cup\) B) = P(A) + …
Solution : The number of ways to n people on circular table is (n-1)! So, first we fix position of men, the number of ways to sit men = 5! Now, women can sit in the gaps between men, there are 6 gaps between 5 mens, So, women can sit in \(^6P_5\) ways Hence, total …
Solution : The number of choices available to him = \(^5C_4\) x \(^8C_6\) + \(^5C_5\) x\(^8C_5\) = 5.4.7 + 8.7 = 140 + 56 = 196 Similar Questions The number of ways in which 6 men and 5 women can dine at a round table, if no two women are to sit together, is given …
Solution : The number of ways of distributing n identical things in m different boxes is \(^{n-1}C_{m-1}\) So, Required number of ways = \(^{8-1}C_{3-1}\) = \(^7C_2\) = 21 Similar Questions A student is to answer 10 out of 13 questions in an examination such that he must choose atleast 4 from the first five questions. …
Solution : Total number of ways in which all the letters can be arranged in alphabetical order = 6! There are two vowels (A, E) in the word ‘GARDEN’. Total number of ways in which these two vowels can be arranged = 2! \(\therefore\) Total number of required ways = \(6!\over 2!\) = 360 Similar …
Solution : In the word SACHIN’ order of alphabets is A, C, H, I, N and S. Number of words start with A = 5!, so with C, H, I, N. Now, words start with S and after that ACHIN are in ascending order of position, so 5.5! = 600 words are in dictionary before, …
