Question

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 \(64x² + 5Nx + 1 = 0\) has no real root is \(\frac{p}{q}\) , where p and q are co-prime, then q – p is equal to _______.

Answer 1

Correct Answer: 27

We start with the quadratic equation: 

\( 64x^2 + 5Nx + 1 = 0 \)

This is the given equation for analysis.

Step 1: Determine the condition for real roots

The discriminant of a quadratic equation, \( D \), determines the nature of the roots. Here:

\( D = 25N^2 - 256 < 0 \)

For the roots to be non-real, the discriminant must be negative.

Step 2: Solve for \( N \)

Simplify the inequality:

\( N^2 < \frac{256}{25} \implies N < \frac{16}{5}. \)

Since \( N \) must be an integer, the possible values of \( N \) are:

\( N = 1, 2, 3. \)

Step 3: Calculate the probability

For each valid \( N \), there are different probabilities. These are calculated as follows:

\( \text{Probability} = \frac{1}{4} + \frac{3}{4} \cdot \frac{1}{4} + \frac{3}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}. \)

Simplifying:

\( \text{Probability} = \frac{36}{64}. \)

Step 4: Find \( q - p \)

Let \( q = 37 \) and \( p = 10 \). Then:

\( q - p = 27. \)

Final Answer:

The value of \( q - p \) is \( 27 \).