# Coin toss probability problem

1. Dec 16, 2009

### naspek

Hi all =)

Question...
A fair coin is tossed thrice. Supposed we denote a "head" turning up as 1 and
"tail" as 0. Given that the total on all three tosses is an odd number,
what is the probability that at first toss, we get a "head"?

i dont have any idea to start answering this question..

2. Dec 16, 2009

### rasmhop

Re: Probability

Ok our sample space is
$$\Omega = \{0,1\}^3 = \{(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)\}$$
We assume a fair coin so every outcome is equally likely which gives us the probability function:
$$p(x) = \frac{1}{8} \qquad \textrm{for all }x \in \Omega$$
Now we have two events $E_1,E_2$. Let $E_2$ denote the event that the total on all three tosses is an odd number. Let $E_1$ denote the event that on the first toss we get a head. What you want is the conditional probability $P(E_1|E_2)$. List all elements in $E_1$ and $E_2$ and you should be able to compute it using the formula:
$$P(E_1|E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$$