2 votes 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 asked Nov 15, 2018 Ayush Upadhyaya 1.4k views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply akash.dinkar12 commented Nov 15, 2018 reply Follow Share I m getting 6... 0 votes 0 votes Ayush Upadhyaya commented Nov 15, 2018 reply Follow Share Yeah 6 is given in the key. 0 votes 0 votes utk0203 commented Nov 15, 2018 reply Follow Share Integrate from 1 to 2 which evaluates to 1 ,Therefore lambda=6 0 votes 0 votes Deepanshu commented Nov 15, 2018 reply Follow Share akash.dinkar12 plzz add answer .. i am also getting diff. 0 votes 0 votes akash.dinkar12 commented Nov 15, 2018 reply Follow Share see this.... 2 votes 2 votes rekhameena commented Nov 15, 2018 reply Follow Share I am getting 6...i think you have done some calculation mistake. 1 votes 1 votes srestha commented Nov 15, 2018 reply Follow Share yes.. 0 votes 0 votes Ayush Upadhyaya commented Nov 16, 2018 reply Follow Share Yeah I was doing calculation mistake. Answer is 6. 0 votes 0 votes Please log in or register to add a comment.