An experiment consists of tossing three biased coins simultaneously and writing down the outcomes observed. The probabilities of the two outcomes are as follows:
$$
\begin{eqnarray}
P(\text{heads}) &=& \frac{2}{3} \\
P(\text{tails}) &=& \frac{1}{3}
\end{eqnarray}
$$
With these probabilities, what is:
a) the sample space?
b) the probability of an even number of heads?
c) the probability that at least one coin shows heads?
I know that answer $a$ is the sample space $s$ for the eight total outcomes, but I'm really stuck on $b$ and $c$.
Best Answer
For $b,$(if I understood correctly):
You mean what is the possibility of having $2$ heads:
If you need to get exactly $2$ of them, then you need $2$ heads and $1$, the chances of that are $(\frac23)^2 * \frac13 = \frac{4}{27}$. You need to multiply this with $3$ since cases can change.
For having $0$: You need both of them to be tail: $(\frac13)^3 = \frac{1}{27}$
For $c,$:
If you want at least one head, then you can write: What is the possibility that it will never show one head? Then tail has to happen every time, and tail has $\frac13$ chance of happening, and you have get it $3$ times after an other, the probability of that is ($\frac13) ^ 3 = \frac{1}{27}$(you can multiply them, since coin flips are independent events). So the chances of you having at least one head is 1-$\frac{1}{27} = \frac{26}{27}$.