SOLUTION: Matt flipped two fairs coins together. He continued doing so until at least one of the coins turns up heads. What is the probability that both are heads on the last flip? please ex

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Question 583695: Matt flipped two fairs coins together. He continued doing so until at least one of the coins turns up heads. What is the probability that both are heads on the last flip? please explain
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Matt flipped two fairs coins together. He continued doing so until at least one of the coins turns up heads. What is the probability that both are heads on the last flip? please explain
This could theoretically go on forever and Matt never get anything but tails.
So it is a infinite geometric series of probabilities.

He will always either get HH, HT, TH, or TT, each with probability of 1%2F4.

To succeed, Matt must get HH on the first throw or get only TT successively
a bunch of times and then HH on the final throw. 

So we want the probability of:

1. HH 1st time with probability 1%2F4
OR
2. TT 1st time and HH 2nd time with probability  1%2F4·1%2F4 = %281%2F4%29%5E2
OR
3. TT 1st 2 times and HH 3rd time with probability %281%2F4%29%5E2·1%2F4 = %281%2F4%29%5E3
OR
4. TT 1st 3 times and HH 4th time with probability %281%2F4%29%5E3·1%2F4 = %281%2F4%29%5E4
OR
...

OR means "add", so the desired probability is the sum of this infinite
geometric series:

%281%2F4%29 + %281%2F4%29%5E2 + %281%2F4%29%5E3 + %281%2F4%29%5E4 + ...

the formula for which is

S%5Binfinity%5D = a%5B1%5D%2F%281-r%29 where a%5B1%5D = r = 1%2F4

S%5Binfinity%5D = %281%2F4%29%2F%281-1%2F4%29 = %281%2F4%29%2F%283%2F4%29 = expr%281%2F4%29expr%284%2F3%29 = expr%281%2Fcross%284%29%29expr%28cross%284%29%2F3%29 = 1%2F3

Edwin