Coin toss probability

A

alaynad4

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Sep 26, 2014
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I am having trouble with the problem. The first part I understand (I believe). The second part is a trick question I feel like. See for yourself: You are going to toss a fair coin repeatedly in successive trials until you observe a head for the first time, and then you are going to stop. Compute the probabilities P("at least 3 tosses are required before the experiment terminates") and P("the process never stops"), respectively. HELP!!
 
alaynad4 said:
I am having trouble with the problem. The first part I understand (I believe). The second part is a trick question I feel like. See for yourself: You are going to toss a fair coin repeatedly in successive trials until you observe a head for the first time, and then you are going to stop. Compute the probabilities P("at least 3 tosses are required before the experiment terminates") and P("the process never stops"), respectively. HELP!!
Click to expand...

It is not clear to me as to exactly do understand.
If \(\displaystyle \mathcal{P}(T=k)\) is the probability that the trials terminate upon the kth trial( first head).

Is it true that \(\displaystyle \mathcal{P}(T=k)=2^k~?\)

The probability that it never terminates is \(\displaystyle 1-\sum\limits_{k = 1}^\infty {\mathcal{P}(T=k)}~?\)
 
pka said:
It is not clear to me as to exactly do understand.
If \(\displaystyle \mathcal{P}(T=k)\) is the probability that the trials terminate upon the kth trial( first head).

Is it true that \(\displaystyle \mathcal{P}(T=k)=2^k~?\)

The probability that it never terminates is \(\displaystyle 1-\sum\limits_{k = 1}^\infty {\mathcal{P}(T=k)}~?\)
Click to expand...

For the first part I had (1/2)*(1/2)*(1/2) and obviously got that the probability was 1/8. The last part I was stumped on. What you put makes sense to me though...
 
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