Let the probability of getting head for a biased coin be 1 4 . It is tossed repeatedly until a head appears.…

Given a quadratic equation $64x^2+5nx+1=0$ has no real roots, so

$D<0$

$\Rightarrow 25N^2-4\times 64<0$

$\Rightarrow N<\frac{16}{5}\Rightarrow N=1,2,3$

And, For a biased coin, the probability of getting head is $P=\frac{1}{4},$ and probability of getting tail will be $Q=\frac{3}{4}$

So, required probability is

$\frac{1}{4}+\frac{3}{4}·\frac{1}{4}+{\left(\frac{{\displaystyle 3}}{{\displaystyle 4}}\right)}^2·\frac{{\displaystyle 1}}{{\displaystyle 4}}+{\left(\frac{{\displaystyle 3}}{{\displaystyle 4}}\right)}^3·\frac{{\displaystyle 1}}{{\displaystyle 4}}=\frac{p}{q}$

$\Rightarrow \frac{{\displaystyle \frac{1}{4}}\left(1-{\left({\displaystyle \frac{3}{4}}\right)}^3\right)}{1-\left(\frac{3}{4}\right)}=\frac{p}{q}$

$\Rightarrow 1-{\left(\frac{3}{4}\right)}^3=\frac{p}{q}$

$\Rightarrow 1-\frac{27}{64}=\frac{p}{q}$

$\Rightarrow \frac{p}{q}=\frac{37}{64}$

$q-p=27$