# Finding probabilty of of getting heads of an unfair coin.

In three flips of an unfair coin the probability of getting 3 heads is the same as that of getting exactly 2 tails. What is the ratio of the probability of flipping a tail to the probability of flipping a head. Express your answer as a common fraction in simplest radical form.

Can someone explain what’s the general method for solving such unfair probability questions?

=================

=================

2

=================

Hint: Let pp denote the probability to flip a head, and qq denote the probability to flip a tail. Your first statement gives you an equation relating the two quantities (besides p+q=1p+q=1), which you can then simplify to give you the desired answer: take note that 1√3=√33.\frac{1}{\sqrt3}=\frac{\sqrt3}{3}.

Assume probability of tails is pp.

The probability of three heads is (1−p)3(1-p)^3. Probability of two tails is (32)p2(1−p)\binom{3}{2}p^2(1-p). Hence,
3p^2(1-p)=(1-p)^33p^2(1-p)=(1-p)^3

Therefore, \frac{p^2}{(1-p)^2}=\frac{1}{3}\Rightarrow \frac{p}{(1-p)}=\frac{\sqrt{3}}{3}\frac{p^2}{(1-p)^2}=\frac{1}{3}\Rightarrow \frac{p}{(1-p)}=\frac{\sqrt{3}}{3}

Then the probability of tails is p=\frac{1}{\sqrt{3}+1}=0.366p=\frac{1}{\sqrt{3}+1}=0.366 and the probability of heads = 1-p=0.6341-p=0.634
– msm
2 days ago