Question

I loss three coins together. The possible outcomes are no heads, $1$ head, $2$ heads and $3$ heads. So, I say that, probability of no heads is $\frac{1}{4}$. What is wrong with this conclusion?

Answer 1

Toss three coins getting heads in one $=\frac{1}{2}$ so in all three $=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$

So outcomes of $1$ head $\frac{3}{8}$ and $2$ head$=\frac{3}{8}$

and remaining $3$ heads $=1-\frac{1}{8}-\frac{3}{8}-\frac{3}{8}=\frac{1}{8}$

So, probability of getting no heads$\frac{1}{4}$ is wrong it would be $\frac{1}{8}$