2
$\begingroup$

I'm having some trouble with this problem

Suppose you flip a biased coin until a head appears. The coin has a $75%$ chance of coming up tails. Let $n$ be the number of flips that you need to do. What is the probability of the following events:

a) $n$ is at most $3$?

b) $n$ is even (note that a geometric series of the form $a + ar + ar^2 + ar^3 + ...$ is equal to $\frac{a}{1-r}$

We really haven't talked much about biased coins, or even how to use variables like $n$ in answers.

$\endgroup$
  • $\begingroup$ Have you done any work on Geometric Distributions? $\endgroup$ – Graham Kemp Apr 2 '15 at 4:40
  • $\begingroup$ For the first, we could have H, or TH, or TTH. Probability of $H$ is $1/4$. Probability of TH is $(3/4)(1/4)$. Probability of TTH is $(3/4)(3/4)(1/4)$. Add up. A fancier way: We have $n\gt 3$ if the first three tosses are tails, probability $(3/4)^3$. Subtract from $1$. (The first way should give you an idea of how to answer the second question.) $\endgroup$ – André Nicolas Apr 2 '15 at 4:41
1
$\begingroup$

The biased coin in your case only means that $P(T)=\frac{3}{4}$.
First note that to stop at a single Head in $n$ tosses, you need to get Tail in the first $n-1$ tosses and a Head in the last (that is $n^{th}$ toss).

$1.$ When $n\leq3$. You can get a head in these ways - $H,TH,TTH$.
(Here $TH$ implies first toss gives Tail and second toss gives Head)

So the probability is $$P(H) + P(TH) + P(THH)=\frac{1}{4}+\frac{3}{4}.\frac{1}{4}+\frac{3}{4}.\frac{3}{4}.\frac{1}{4}$$

$2.$ When $n$ is even.
Here you can see that your probablity will be given by $$P(TH)+P(TTTH)+P(TTTTTH)+\dots$$ Write the probabilities in a similar way to the first case and you will obtain an infinite G.P, the formula for sum to which is given in the question.

$\endgroup$
0
$\begingroup$

Suppose you flip a biased coin until a head appears. The coin has a 75% chance of coming up tails. Let n be the number of flips that you need to do. What is the probability of :

$n$ has a Geometric Distribution.

Let $q=0.75$ be the probability of getting a head. (We can call this a failure.)

Let $p=0.25$ be the probability of getting a tail. (We can call this a success.)

The probability of requiring $k$ trials until a success is: the probability of $k-1$ failures then one success:$$P(n=k) = q^{k-1}p$$

So the probability of requiring at most three trials is: $$\begin{align} \mathsf P(n\leq 3) & = \sum_{k=1}^3 \mathsf P(n=k) \\ & = \Big(1+q+q^2\Big)p \end{align}$$

And the probability of requiring any even number of trials, is the sum of all probabilities for even numbers.

$$\begin{align} \mathsf P(\operatorname{Even}(n)) & = \sum_{k=1}^\infty \mathsf P(n=2k) \\ & = p \sum_{k=1}^\infty q^{2k-1} \end{align}$$

$\endgroup$
0
$\begingroup$

For part (2), here is perhaps a simpler way. For every sequence $TT...TH$ that leads to an even $n$, there is a shorter sequence (less by a tail $T$) that leads to an odd $n$. Thus if $p_E$ is the probability of $n$ being even, then $p_E = (1-p_E)\frac34 \implies p_E = \frac37$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.