Key Concepts and Formulas

To solve this problem, we will utilize two fundamental concepts from JEE Mathematics:

• Nature of Roots of a Quadratic Equation: For a quadratic equation of the form $ax^2 + bx + c = 0$ (where $a \neq 0$), the nature of its roots is determined by the discriminant, $D = b^2 - 4ac$.

  • If $D > 0$, there are two distinct real roots.
  • If $D = 0$, there is one real root (a repeated root).
  • If $D < 0$, there are no real roots (the roots are complex conjugates). In this problem, we are interested in the condition $D < 0$.

• Geometric Probability Distribution: This distribution models the number of independent Bernoulli trials required to achieve the first success. If $p$ is the probability of success in a single trial and $q = 1-p$ is the probability of failure, then the probability that the first success occurs on the $n$-th trial is given by: $P(N=n) = q^{n-1}p = (1-p)^{n-1}p, \quad \text{for } n = 1, 2, 3, \ldots$ This means there are $n-1$ failures followed by one success.

Step-by-Step Solution

Let's systematically break down the problem and solve each part.

Step 1: Determine the condition on $N$ for the quadratic equation to have no real roots.

The given quadratic equation is $64x^2 + 5Nx + 1 = 0$. We identify the coefficients by comparing it with the standard form $ax^2 + bx + c = 0$: $a = 64$ $b = 5N$ $c = 1$

For the equation to have no real roots, its discriminant $D$ must be strictly less than zero ($D < 0$). Using the discriminant formula $D = b^2 - 4ac$: $(5N)^2 - 4(64)(1) < 0$ $25N^2 - 256 < 0$

Now, we solve this inequality for $N$: $25N^2 < 256$ $N^2 < \frac{256}{25}$ Taking the square root of both sides. Since $N$ represents the number of tosses, it must be a positive integer ($N \ge 1$). Therefore, we consider the positive square root: $N < \sqrt{\frac{256}{25}}$ $N < \frac{16}{5}$ $N < 3.2$

Since $N$ must be a positive integer, the possible values for $N$ that satisfy this condition are: $N \in \{1, 2, 3\}$ These are the only integer values for $N$ for which the quadratic equation will have no real roots.

Step 2: Calculate the probabilities for each possible value of $N$.

The problem states that a biased coin is tossed repeatedly until a head appears, and $N$ is the number of tosses required. This is a classic geometric distribution scenario. The probability of getting a head (success) is given as $p = \frac{1}{4}$. The probability of getting a tail (failure) is $q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$.

We use the geometric distribution formula $P(N=n) = q^{n-1}p$ for the values of $N$ identified in Step 1:

• For $N=1$: The first toss is a Head. $P(N=1) = \left(\frac{3}{4}\right)^{1-1} \times \frac{1}{4} = \left(\frac{3}{4}\right)^0 \times \frac{1}{4} = 1 \times \frac{1}{4} = \frac{1}{4}$

• For $N=2$: The first toss is a Tail, and the second toss is a Head (TH). $P(N=2) = \left(\frac{3}{4}\right)^{2-1} \times \frac{1}{4} = \frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$

• For $N=3$: The first two tosses are Tails, and the third toss is a Head (TTH). $P(N=3) = \left(\frac{3}{4}\right)^{3-1} \times \frac{1}{4} = \left(\frac{3}{4}\right)^2 \times \frac{1}{4} = \frac{9}{16} \times \frac{1}{4} = \frac{9}{64}$

Step 3: Calculate the total probability that the equation has no real roots.

The event "the equation has no real root" occurs if $N=1$ OR $N=2$ OR $N=3$. Since these are mutually exclusive events (only one value of $N$ can occur), the total probability is the sum of their individual probabilities: $P(\text{no real root}) = P(N=1) + P(N=2) + P(N=3)$ Substituting the probabilities calculated in Step 2: $P(\text{no real root}) = \frac{1}{4} + \frac{3}{16} + \frac{9}{64}$ To sum these fractions, we find a common denominator, which is 64: $P(\text{no real root}) = \frac{1 \times 16}{4 \times 16} + \frac{3 \times 4}{16 \times 4} + \frac{9}{64}$ $P(\text{no real root}) = \frac{16}{64} + \frac{12}{64} + \frac{9}{64}$ $P(\text{no real root}) = \frac{16 + 12 + 9}{64}$ $P(\text{no real root}) = \frac{37}{64}$

Step 4: Determine $p$ and $q$ and calculate $q-p$.

The problem states that the probability is $\frac{p}{q}$, where $p$ and $q$ are coprime. From our calculation, we have: $\frac{p}{q} = \frac{37}{64}$ By comparing the numerator and denominator, we get: $p = 37$ $q = 64$

We must verify that $p$ and $q$ are coprime. 37 is a prime number. The prime factorization of 64 is $2^6$. Since 37 is not 2, 37 and 64 share no common prime factors, meaning they are coprime.

Finally, we calculate the value of $q-p$: $q-p = 64 - 37$ $q-p = 27$

Common Mistakes & Tips

• Inequality for Discriminant: Be meticulous with the inequality sign when applying the discriminant condition. "No real roots" explicitly means $D < 0$, not $D \le 0$.
• Nature of Random Variable N: Remember that $N$ represents a count (number of tosses), so it must be a positive integer ($1, 2, 3, \ldots$). When solving inequalities for $N$, only select valid integer values.
• Geometric Distribution Formula: Ensure the exponent for the probability of failure ($q$) is correctly $n-1$, corresponding to $n-1$ failures before the first success on the $n$-th trial.
• Coprime Check: Always confirm that the numerator and denominator of your final probability fraction are coprime. If they are not, simplify the fraction before identifying $p$ and $q$.

Summary

This problem elegantly combines concepts from quadratic equations and probability. We first used the discriminant condition ($D < 0$) to find the possible integer values of $N$ for which the quadratic equation has no real roots. Then, recognizing $N$ as a geometrically distributed random variable, we calculated the probability for each of these $N$ values. Summing these probabilities gave us the total probability $\frac{37}{64}$. Finally, by setting $p=37$ and $q=64$ and verifying their coprimality, we calculated $q-p$.

The final answer is $\boxed{27}$.