What is the sample space if a coin is tossed until the first time it comes up heads?
To determine the sample space, we simply need to refer to the definition. According to the link below, a sample space is the set of all possible outcomes. Therefore, we have to examine what outcomes are possible for this experiment!
We know that we will toss the coin until it comes up heads. This tells us what our sample space may look like! Let's just try a few examples. Suppose we have heads come up on the first toss. In this case, we stop immediately because we have a heads! Now, suppose it came up tails. Then we flip the coin again and repeat the process.
So, we can extend this pattern to see a possible sample space:
H, TH, TTH, TTTH, ...
We see that there is a string of `n` tails, where `n >= 0` and `n in ZZ`which is then followed by a single head. Any other pattern would have a coin toss after a head is reached, which is against the constraints of the problem!
A way to confirm this is the sample space? Let's add the probabilities of each case and see if all of the probabilities sum to 1. The probability of heads is `1/2`. The probability of getting a tails, then a heads is `1/2*1/2`.
Our summation of each probability is going to be the following:
`1/2 + 1/2*1/2 + 1/2*1/2*1/2 + ... = 1/2 + (1/2)^2 + (1/2)^3 + ...`` `
So, in summatino notation:
If you look at this, this is actually an infinite geometric sum! The way to determine the full sum is a well known formula:
`sum_(n=0)^oo r^n = 1/(1-r)`
Well, we simply need to get rid of the first term because our first term is n=1, so our sum looks like:
`(sum_(n=0)^oo (1/2)^n) - 1 = 1/(1-1/2) - 1 = 2-1 = 1`
There we go! The sum of the probabilities of our sample space will be 1, indicating we have the full sample space! Therefore, our answer is likely correct.
I hope that helps!