Probability - Coin Toss - Find Formula

The problem statement, all variables and given/known data:

Suppose a fair coin is tossed $n$ times. Find simple formulae in terms of $n$ and $k$ for

a) $P(k-1 \mbox{ heads} \mid k-1 \mbox{ or } k \mbox{ heads})$

b) $P(k \mbox{ heads} \mid k-1 \mbox{ or } k \mbox{ heads})$

Relevant equations:

$P(k \mbox{ heads in } n \mbox{ fair tosses})=\binom{n}{k}2^{-n}\quad (0\leq k\leq n)$

The attempt at a solution:

I'm stuck on the conditional probability. I've dabbled with it a little bit but I'm confused what $k-1$ intersect $k$ is. This is for review and not homework.

The answer to a) is $k/(n+1)$.

I tried $P(k-1 \mbox{ heads} \mid k \mbox{ heads})=P(k-1 \cap K)/P(K \mbox{ heads})=P(K-1)/P(K).$ I also was thinking about $$P(A\mid A,B)=P(A\cap (A\cup B))/P(A\cup B)=P(A\cup (A\cap B))/P(A\cup B)=P(A)/(P(A)+P(B)-P(AB))$$

-
If this is homework, you need to add the "homework" tag. –  jwodder Apr 11 '12 at 19:24
It's not homework. It's for review. But I'll add the tag anyways! –  bizboy1 Apr 11 '12 at 19:29
Please don't. The tag becomes useless if it's applied willy-nilly to questions where it doesn't belong. –  Henning Makholm Apr 11 '12 at 19:31

By the usual formula for conditional probability, an ugly form of the answer is $$\frac{\binom{n}{k-1}(1/2)^n}{\binom{n}{k-1}(1/2)^n+\binom{n}{k}(1/2)^n}.$$ Cancel the $(1/2)^n$. Now the usual formula for $\binom{a}{b}$ plus a bit of algebra will give what you want.

We can simplify the calculation somewhat by using the fact that $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$, which has a nice combinatorial proof, and is built into the "Pascal triangle" definition of the binomial coefficients.

As for the algebra, $\binom{n}{k-1}=\frac{n!}{(k-1)!(n-k+1)!}$ and $\binom{n+1}{k}=\frac{(n+1)!}{k!(n-k+1)!}$. When you divide, there is a lot of cancellation.

-
Thank you very much! –  bizboy1 Apr 11 '12 at 19:33

Let $X$ be the random variable for the number of heads in $n$ tosses. Then your questions amount to finding the conditional probabilities: $$\mathbb{P}(X = k-1 | k-1 \leqslant X \leqslant k) = \frac{\mathbb{P}( \{X = k-1\} \cap \{ k-1 \leqslant X \leqslant k \} }{\mathbb{P}( \{ k-1 \leqslant X \leqslant k \}) } = \frac{\mathbb{P}( X = k-1)}{\mathbb{P}( X =k-1) + \mathbb{P}( X =k) }$$ and $$\mathbb{P}(X = k | k-1 \leqslant X \leqslant k) = \frac{\mathbb{P}( \{X = k\} \cap \{ k-1 \leqslant X \leqslant k \} }{\mathbb{P}( \{ k-1 \leqslant X \leqslant k \}) } = \frac{\mathbb{P}( X = k)}{\mathbb{P}( X =k-1) + \mathbb{P}( X =k) }$$ Both are solved by noting, for $1 \leqslant k \leqslant n$ $$\frac{\mathbb{P}(X=k-1)}{\mathbb{P}(X=k)} = \frac{\binom{n}{k-1}}{\binom{n}{k}} = \frac{k}{n-k+1}$$

-
Thank you very much! –  bizboy1 Apr 11 '12 at 19:34