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

We have the quadratic equation $64x^2 + 5Nx + 1 = 0$. For it to have no real roots, the discriminant ($b^2 - 4ac$) should be less than 0. Here, $a = 64$, $b = 5N$, and $c = 1$. <br/><br/>This gives us : <br/><br/>$(5N)^2 - 4\times64\times1 < 0$ <br/><br/>$\Rightarrow 25N^2 < 256$ <br/><br/>$\Rightarrow N^2 < \frac{256}{25}$ <br/><br/>$\Rightarrow N < \sqrt{\frac{256}{25}} = \frac{16}{5}$ <br/><br/>Since $N$ must be an integer (as it represents the number of tosses), the possible values of $N$ are 1, 2, or 3. <br/><br/>The probability of getting the first head on the $n$-th toss (given the probability of getting a head is $1/4$) is given by the geometric distribution formula, $(1 - p)^{n-1}\times p$. <br/><br/>So, the probability for our specific values of $N$ is: <br/><br/>$P(N=1) = (1 - 1/4)^{1-1}\times(1/4) = 1/4$ <br/><br/>$P(N=2) = (1 - 1/4)^{2-1}\times(1/4) = 3/4 \times 1/4 = 3/16$ <br/><br/>$P(N=3) = (1 - 1/4)^{3-1}\times(1/4) = (3/4)^2 \times 1/4 = 9/64$ <br/><br/>Therefore, the total probability (p/q) is : <br/><br/>$p/q = P(N=1) + P(N=2) + P(N=3)$ <br/><br/>$= 1/4 + 3/16 + 9/64$ <br/><br/>$= 16/64 + 12/64 + 9/64$ <br/><br/>$= 37/64$ <br/><br/>So, $p = 37$, $q = 64$ and $q-p = 64 - 37 = 27$. <br/><br/>Therefore, $q-p$ is equal to $27$.