[FONT="]can someone help me with simple algebra concerning this probability problem ?
I don't understand how we got P(DlN)=x , P(DlL)=2x , P(DlH)=4x.[/FONT] [FONT="] [/FONT] [FONT="]like.. if P(DlN) = x , shouldn't P(DlL)=1/2x and P(DlH)=1/4 ???[/FONT]
You are given $P(D|L) = 2P(D|N) = \frac 1 2 P(D|H)$. So if $P(D|N) =x$, put that in the middle so you have $P(D|L) = 2x = \frac 1 2 P(D|H)$. That tells you $P(D|L) = 2x$ on the left and $2x = \frac 1 2 P(D|H)$ on the right. Multiply both sides of that by $2$ to get $4x = P(D|H)$.