QJEEM23Mac6b3: JEE Main 2023 Mathematics - Class 12 JEEM23Mac6b3 Solution: Let the probability of getting head for a biased coin be . It…

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 $c = 1$ .

This gives us :

(5N)^2 - 4\times64\times1 < 0

\Rightarrow 25N^2 < 256

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

\Rightarrow N < \sqrt{\frac{256}{25}} = \frac{16}{5}

Since $N$ must be an integer (as it represents the number of tosses), the possible values of $N$ are 1, 2, or 3.

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, $(1 - p)^{n-1}\times p$ .

So, the probability for our specific values of $N$ is:

$P(N=1) = (1 - 1/4)^{1-1}\times(1/4) = 1/4$

$P(N=2) = (1 - 1/4)^{2-1}\times(1/4) = 3/4 \times 1/4 = 3/16$

$P(N=3) = (1 - 1/4)^{3-1}\times(1/4) = (3/4)^2 \times 1/4 = 9/64$

Therefore, the total probability (p/q) is :

$p/q = P(N=1) + P(N=2) + P(N=3)$

$= 1/4 + 3/16 + 9/64$

$= 16/64 + 12/64 + 9/64$

$= 37/64$

So, $p = 37$ , $q = 64$ and $q-p = 64 - 37 = 27$ .

Therefore, $q-p$ is equal to $27$ .