Let the probability of getting head for a biased coin be [(1)/(4)].

Question

Let the probability of getting head for a biased coin be $$\frac{1}{4}$$. It is tossed repeatedly until a head appears. Let $$\mathrm{N}$$

Accepted Answer

We have the quadratic equation $64{x}^{2}+5Nx+1=0$. 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 :

$(5N{)}^{2}-4	imes 64	imes 1<0$

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

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

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

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

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

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

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

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

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 $27$.

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