2 2 votes Find the value of $\lambda$ such that function f(x) is valid probability density function $f(x)=\lambda (x-1)(2-x)$ for $1 \leq x \leq 2$ $=0$ otherwise My $\lambda$ is coming to be $- \frac{6}{5}$ Am I correct? Probability probability random-variable + – Ayush Upadhyaya 2.2k views answer comment Share Follow Print See all 8 Comments 8 8 Comments reply akash.dinkar12 commented Nov 15, 2018 reply Follow flag I m getting 6... 0 0 replyShare Ayush Upadhyaya commented Nov 15, 2018 reply Follow flag Yeah 6 is given in the key. 0 0 replyShare utk0203 commented Nov 15, 2018 reply Follow flag Integrate from 1 to 2 which evaluates to 1 ,Therefore lambda=6 0 0 replyShare Deepanshu commented Nov 15, 2018 reply Follow flag akash.dinkar12 plzz add answer .. i am also getting diff. 0 0 replyShare akash.dinkar12 commented Nov 15, 2018 reply Follow flag see this.... 2 2 replyShare rekhameena commented Nov 15, 2018 reply Follow flag I am getting 6...i think you have done some calculation mistake. 1 1 replyShare srestha commented Nov 15, 2018 reply Follow flag yes.. 0 0 replyShare Ayush Upadhyaya commented Nov 16, 2018 reply Follow flag Yeah I was doing calculation mistake. Answer is 6. 0 0 replyShare Please log in or register to add a comment.