Basic probability questions (game of heads or tails) I had my first probability course last week and I have a problem sheet to solve. Now probability is not very intuitive to me and so far I have not material to solve the following problems so my answers are essentially based on instinct:
We throw a biased coin: $P(heads) = p$. On the $n^{th}$ throw, what is the probability that:


*

*(1): heads appears for the first time

*(2): The number of heads is equal to the number of faces

*(3): heads has appeared exactly twice

*(4): heads has appeared at least twice
Here are my answers:
(1): probability = $(1-p)^{n-1}p$. Indeed, heads appeared once, hence $p^1$ and tails (which has a probability of $1-p$- appeared $n-1$ times.
(2): probability = $p^{n/2}(1-p)^{n/2}$ indeed, heads (which has a probability of p) appeared n/2 times and tails (which has a probability of $1-p$- appeared $n/2$ times.
(3): probability = $(1-p)^{n-2}p^2$ (same reasons as above)
(4): probability = $1-(1-p)^{n-1}$
 A: Don't forget - the order of heads/tails doesn't matter if all we're doing is counting how many times they show up.


*

*(1) Correct.

*(2) We should have:
$$
\begin{cases}
\binom{n}{n/2}p^{n/2}(1 - p)^{n/2} &\text{if $n$ is even} \\
0 &\text{if $n$ is odd}
\end{cases}
$$

*(3) We should have:
$$
\binom{n}{2}p^{2}(1 - p)^{n-2}
$$

*(4) We should have:
$$
1 - \binom{n}{0}p^{0}(1 - p)^{n-0} - \binom{n}{1}p^{1}(1 - p)^{n-1}
$$

A: 1: Looks correct. You're looking for a very specific sequence of events: n-1 times heads, and 1 times tails, so simply multiply the probabilities for those. (Assuming independence.)
2: Probably missing something. Order doesn't matter here, but your formula is for a specific order, so you are probably missing a $/2n!$
3: Similar to before, but again missing the divider to make it order-independent. You are probably missing a $/(2! * (n-2)!)$
4: Similar to 3, but add up with the results for heads exactly once or 0 times. You probably received cumulative probability tables for that.
