19yes, fantastic solution nishant da !
If p,q,r,s are probablities of raining at 4 diff places at some fixed moment.then find max and min value of 3(p^2+q^2+r^2+s^2)-2(p+q+r+s)+4
-
UP 0 DOWN 0 1 7
7 Answers
62is this not the same as maximizing and minimzing
4x(3p2-2p+1)??
where p lies between 0 and 1
so the minima will be at 6p-2, so p=1/3
Maxima is at p=1, the value is 8
Minima is 8/3
I dont know if i have misunderstood the question but this seems to be correct to me :)
1Top notch solution, so i'm guessing p=q=r=s is not a necessary condition for it to be solved this way. Ok, then why not? Will it only be a special case?
62no it is not a necessary condition....
but it will so happen that for a maxima to exist, they will be all equal to 1 and for minima they will be all equal to 1/3