
First of all, what valid options for $N$? The smallest value that $N$ could be is
. Is there a largest possible value of $N$?
The probability that the first $13,462,305$ flips result in a head before you see your first tail is
. Therefore, $N$ could be any nonnegative integer. It would just be extremely unlikely to be a large number.

What is the probability that the first coin flip is a tail, $T$?
Therefore, the probability that $N$ is zero is $f_N(0) = P(N=0)=$
.

What is the probability that the first coin flip is a head, $H$?
In terms of the random variable $N$, this probability is
$=＿$.

So far, we've introduced two events, which we labeled in terms of the random variable $N$.
 $N=0$: the event that the first coin flip was $T$, so that we obtained zero heads.
 $N \ge 1$: the event that the first coin flip was $H$, so that we know we obtained at least one head.
Since we keep flipping coins only if we obtain an $H$, we continue with more events only conditioned on the event $N \ge 1$. In that case, we consider two more events.
(Online, enter $\ge$ as >=
or as the symbol ≥
.)
The probabilities of these two events, conditioned on the event $N \ge 1$, are simple since they involve just one more coin flip.
From the definition of conditional probability, recall that $P(A,B) = P(A\,\,B)P(B)$. Substituting the event $N \ge 1$ for $B$ and either event $N=1$ or $N \ge 2$ for $A$, we can calculate that
But now, the notation is a bit silly. If we've obtained exactly one head ($N=1$), then obviously we've obtained at least one head ($N \ge 1$). Similarly, if we've obtained at least two heads ($N \ge 2$), then obviously we've obtained at least one head ($N \ge 1$). The events $N=1$ and $N \ge 2$ cannot occur without the event $N \ge 1$. Hence, $P(N=1)=P(N=1, N \ge 1)$ and $P(N \ge 2) = P(N \ge 2, N \ge 1)$. We summarize by giving the probabilities of obtaining exactly one head and of obtaining at least two heads.

We can repeat this procedure to include one more coin flip and calculate the probability that $N=2$ (if third coin flip is $T$) and that $N \ge 3$ (if third coin flip is $H$).
Calculate the probabilities of these events conditioned on the fact that we already flipped heads twice in a row.
Multiply by $P(N \ge 2)$ to determine the probabilities of obtaining exactly two heads or obtaining at least three heads.

Notice to get from $P(N \ge 1)$ to $P(N \ge 2)$ we multiply by $\frac{1}{2}$, the probability of getting one more $H$. Similarly, we multiply by $\frac{1}{2}$ to get from $P(N \ge 2)$ to $P(N \ge 3)$. In general, the event $N \ge n$ corresponds to flipping $H$ $n$ times in a row, so the probability $P(N \ge n)$ is the product of $n$ $\frac{1}{2}$'s. $P(N \ge n)$ is the exponential function
$P(N \ge n) = $
.
(Online enter $a^b$ as a^b
.)
To know that we obtained exactly $n$ heads, we need to flip a coin one more time and this time obtain a $T$. We need to multiply by one more $\frac{1}{2}$ (the probability of $T$) to determine the probability distribution function $f_N(n)$.
$f_N(n) = P(N=n) =$
(Online, you'll need to use parentheses after the ^
to put a n+1
in the exponent.)

Use the below applet to sketch the probability distribution $f_N(n)$ of the number $N$ of consecutive heads. The distribution continues for all nonnegative integers $n$, but we just plot $0 \le n \le 5$.
Feedback from applet
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