Probability Question (1 Viewer)

frog0101 · Oct 23, 2018

Hi,
I have been unable to successfully solve this question:
A biased coin was flipped such that: P(T)/P(H)=q, where P(T) is the probability of flipping tails, P(H) is the probability of flipping heads, and q is a positive integer. Find the probability of flippings n heads before m tails as m approaches infinity.

I attempted to find an expression in terms of q for P(T) and P(H), however to no avail. The answer given is 1.

InteGrand · Oct 23, 2018

frog0101 said:
Hi,
I have been unable to successfully solve this question:
A biased coin was flipped such that: P(T)/P(H)=q, where P(T) is the probability of flipping tails, P(H) is the probability of flipping heads, and q is a positive integer. Find the probability of flippings n heads before m tails as m approaches infinity.

I attempted to find an expression in terms of q for P(T) and P(H), however to no avail. The answer given is 1.

$\noindent To get an expression for $P(T)$ and $P(H)$ in terms of $q$, note that $P(T) = 1-P(H)$. Substitute this into $\frac{P(T)}{P(H)} = q$ and solve for $P(H)$. However, we don't actually need to know exactly what $P(H)$ is in terms of $q$ to do this question. It is enough to know that $P(H)$ is strictly between $0$ and $1$, which is guaranteed by the fact that $\frac{P(T)}{P(H)}$ is equal to a positive number.$

$\noindent Now, if we denote $p = P(H)\in (0,1)$, the probability of having \textbf{exactly} $n$ heads before $m$ tails is$

$P(\text{exactly }n\text{ heads before }m\text{ tails}) = \binom{n+m-1}{n}p^{n}(1-p)^{m},$

$\noindent which you can show combinatorially as an exercise (and would need to prove in the HSC if using it).$

$\noindeint For any fixed $n$, since $p\in (0,1)$, it turns out that \textbf{\underline{this expression tends to $0$ as $m\to \infty$}}. If the problem meant find the probability of obtaining \emph{at least} $n$ heads before $m$ tails, the expression for this is$

$\begin{align*} P(\text{at least }n\text{ heads before }m\text{ tails}) &= 1-P(\text{less than }n\text{ heads before }m\text{ tails})\\ &= 1 - \sum_{k=0}^{n-1} P(\text{exactly }k\text{ heads before }m\text{ tails}).\end{align*}$

$\noindent Assuming you've managed to prove the statement in bold and underlined above (which you should do as an exercise!), we know that as $m\to \infty$, we have$

$P(\text{exactly }k\text{ heads before }m\text{ tails})\to 0$

$\noindent for any $k$, whence (using the fact that the limit of a finite sum is the sum of the limits)$

$\lim\limits_{m\to\infty}P(\text{at least }n\text{ heads before }m\text{ tails}) =1 - \sum_{k=0}^{n-1}0 = 1,$

$\noindent so the answer is 1.$

Probability Question (1 Viewer)

frog0101

Member

InteGrand

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)