Some probabilities are defined on infinite sample spaces

Question

Some probabilities are defined on infinite sample spaces. For example, we could flip a coin until tails occurs. From everyday experience we might expect this to take a small number of flips, but one could be extremely unlucky. Now consider how long we might have to flip a coin until a string of 100 consecutive tails occurs. Again, from everyday experience we would expect this to take longer; just by the definition, it will take at least 100 flips of the coin, and probably much much more.

In the following, the basic unit of the experiment is the trial. Trials are identical and independent, much like flipping a coin. Also, we will consider only two possible outcomes from our trial: success and failure. Let the probability of success (written s for brevity) be p and let the probability of failure (f ) be q . Determine an algebraic identity between p and q .

Find the probabilities of the following strings of trials.

Trials

• f
• f
• f
• s
• f

gives a probability of?

while

Trial

• s
• s
• f
• s
• s
• s
• s
• s
• s
• f
• s
• s

gives a probability of ?

Now consider a string of 10 trials, and we seek the first substring of three consecutive successes. Investigate why the status of the first substring of three trials with regard to passing/failing this consecutive successes test is informative of the likelihood that the substring of trials 2 through 4 passes this consecutive successes test by filling in the following probabilities.

first 3 of 10 consecutive trials

• f
• f
• f

gives probability that trials 2 through 4 pass this test?

first 3 of 10 consecutive trials

• s
• s
• f

gives probability that trials 2 through 4 pass this test?

Answer 1

The first question is asking for an equation using p and q. We can consider p and q as a probability distribution since there are only two elements in the sample space {s,f}. 

 

1. That being said what must be true of any probability distribution?

 

2. Next you are asked the probability of fffsf. Since the trials are independent the probabilities here can be multiplied giving q x q x q x p x q = q⁴p¹

 

3. For the next set try this one on your own using (2) above as a guide.

 

4. I believe the next part is asking if we get fff for the first 3 trials what is the probability trials 2-4 will be sss. So if I am reading the question correctly the probability should be 0 since trials 2 and 3 are already f and trials 2-4 cannot be sss.

 

5. Use (4) as a guide to answer this.

 

Any questions please add a comment.