Let the probability of getting head for a biased coin be $$\frac{1}{4}$$. It is tossed repeatedly until a head appears. Let $$N$$ be the number of tosses required. If the probability that the equation $$64x^2 + 5Nx + 1 = 0$$ has no real root is $$\frac{p}{q}$$, where $$p$$ and $$q$$ are co-prime, then $$q - p$$ is equal to _______

Let $$N$$ denote the number of tosses required to obtain the first head.

Since the probability of obtaining a head is $$\frac{1}{4}$$ and the probability of obtaining a tail is $$\frac{3}{4}$$, $$N$$ follows a geometric distribution with

$$P(N=n)=\left(\frac{3}{4}\right)^{n-1}\left(\frac{1}{4}\right)$$

The quadratic equation is

$$64x^2+5Nx+1=0$$

For the equation to have no real roots, its discriminant must be negative.

$$\Delta=(5N)^2-4(64)(1)<0$$

$$25N^2-256<0$$

$$25N^2<256$$

$$N^2<\frac{256}{25}$$

$$N^2<10.24$$

Since $$N$$ is a positive integer,

$$N\in{1,2,3}$$

Therefore, the required probability is

$$P(N\leq 3)=P(1)+P(2)+P(3)$$

$$P(1)=\left(\frac{3}{4}\right)^0\left(\frac{1}{4}\right)=\frac{1}{4}$$

$$P(2)=\left(\frac{3}{4}\right)^1\left(\frac{1}{4}\right)=\frac{3}{16}$$

$$P(3)=\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)=\frac{9}{64}$$

Hence,

$$P(N\leq 3)=\frac{1}{4}+\frac{3}{16}+\frac{9}{64}$$

$$P(N\leq 3)=\frac{16}{64}+\frac{12}{64}+\frac{9}{64}$$

$$P(N\leq 3)=\frac{37}{64}$$

Thus,

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

giving

$$p=37,\quad q=64$$

Therefore,

$$q-p=64-37=27$$

Hence, the correct answer is

$$27$$