Let $ABCD$ be a square shaped board. 4 equal rectangles are drawn into it. The length of the sides of the rectangles are $x$ and $y$, where $\frac{x}{y}$ = $3$. A dart is thrown towards the square shaped board $ABCD$. The probability of the dart hitting only the black block or small square $FEHG$ in the middle is $\frac{a}{b}$ where $a, b$ are coprimes. What is the value of $b-a$?
My Attempt:
As all rectangle are equal and congruent to one another, so $KE = x$ and $IH$ = $y$. Given that, $x$ = $3y$. So we get the lengthof $AB$ = $3y+y$ = $4y$ and the length of $FE$ = $4y-2y$ = $2y$.
I measured the probability by calculating with their area. So [$ABCD$] = ($4y$)$^\text{2}$ = $16y^\text{2}$ and [$FEHG$] = ($2y)^\text{2}$ = $4y^\text{2}$.
Hence the probability of hitting the dart towards the block area = $\frac{4y^\text{2}}{16y^\text{2}}$ = $\frac{1}{4}$
But I was wrong. Because $1$ and $4$ aren't prime numbers. So, I think my process is wrong. Then what is my mistake for figuring out the probability with measuring area? Moreover, how can I find out the probability except area?
SOURCE: BANGLADESH MATH OLYMPIAD
Any kind of mistake is highly excusable and thanks in advance.
You are correct in saying that $1$ and $4$ are not prime numbers - but they are coprime. That is to say $\gcd{(1,4)}=1$. That is what the question originally stated. The answer is then $4-1=3$.