0 votes 0 votes Y is distributed uniformly in [1,6] 3x^2+6xY+3Y+6=0 P(Y) for 2 real roots ? jatin khachane 1 asked Feb 4, 2019 jatin khachane 1 974 views answer comment Share Follow See all 15 Comments See all 15 15 Comments reply Aman Juyal commented Feb 4, 2019 reply Follow Share I think here you have to find b^2-4ac>=0 for range of real root and then after getting range of y apply uniform distrubution ie(X2-X1)/6-1 0 votes 0 votes jatin khachane 1 commented Feb 4, 2019 reply Follow Share what is final answer 0 votes 0 votes Aman Juyal commented Feb 4, 2019 reply Follow Share I think i got 6-2/6-1=4/5=0.8 0 votes 0 votes xariniov9 commented Feb 4, 2019 reply Follow Share The interval was open (1, 6) and not [1, 6] 0 votes 0 votes jatin khachane 1 commented Feb 4, 2019 reply Follow Share what will be answer ?? @Shivam Kasat 0 votes 0 votes Aman Juyal commented Feb 4, 2019 reply Follow Share here random variable is continuous .it doesn't matter is it closed or not in continuous distribution . and in range there would be infinite point so if not 2 so 2.0000000004 would be included in answer 1 votes 1 votes jatin khachane 1 commented Feb 4, 2019 reply Follow Share this ques was for how many marks 0 votes 0 votes Aman Juyal commented Feb 4, 2019 reply Follow Share 3(3Y+6 )=9(Y+2)not Y+6 0 votes 0 votes Duffer commented Feb 4, 2019 reply Follow Share You did a mistake , it's y^2 - y -2 , so y>=2 is sufficient condition 0 votes 0 votes Balaji Jegan commented Feb 4, 2019 reply Follow Share Upto 1 digit or 2 digits, if it is upto 1 digit, then 4/5 is equal to 5/6 0 votes 0 votes jatin khachane 1 commented Feb 4, 2019 reply Follow Share upto 1 digit i think it was mentioned so ans is 0.8 ?? 0 votes 0 votes Balaji Jegan commented Feb 4, 2019 reply Follow Share 0.8 is ryt. 0 votes 0 votes VikramRB commented Feb 5, 2019 reply Follow Share For real root condition, I got 1 value of y between 1 and 6 as 2. Now since it is continuous distribution, probability would (area of between 2-6)/(total area1-6) 0 votes 0 votes sai charan chakrala commented Feb 5, 2019 reply Follow Share This was asked for how many marks? 0 votes 0 votes Meet2698 commented Feb 11, 2019 reply Follow Share f(y)= { 1/5 1<=y<=6 ; 0 otherwise;} 3x^2 +6xY + (3Y+6) =0 For real roots, b^2 - 4ac =0; therefore, Y^2 - Y - 2 >=0 i.e Y>=2; P(Y) = $\int_{2}^{6}$f(Y) = (1/5)*(6-2) =4/5 =0.8 Is the above way the right way to solve ? 0 votes 0 votes Please log in or register to add a comment.