Let us assume E and E' be two events where
E = { (HHT) , (THT) }
and E' = { (HTT), (TTT), (HTH), (TTH), (HHH), (THH) }
Then P(E) = 2/8 = 1/4 and P(E') = 6/8 = 3/4
Now, according to the given condition in the problem at each chance we are tossing the coin 3 times , if the event E occurs then we will not get another chance to toss and if event E' occurs then we will get another chance to toss.
Now, the probability that I got only 1 chance to toss is 1/4 ( the event E occurred on the 1st chance)
The probability that I got 2 chances to toss is 3/4 * 1/4 ( the event E' occurred on the 1st chance and event E occurred on the 2nd chance)
So, if we go on like this we can find the probability that I got k chances to toss is (3/4)k * (1/4) (where the event E' occurred on first k-1 times after that event E occurred).
So, the probability that tossing would terminate within k chances is
1/4 + 1/4 * 3/4 + 1/4 * (3/4)2 + ..................... + 1/4 * (3/4)k
Now, in the problem it is not given that at most how many chances we can take. So, the number of chances taken by me can run up to infinity.
So, the probability that tossing would terminate is
1/4 + 1/4 * 3/4 + 1/4 * (3/4)2 + ...........................(Up to infinity)
Now, the above expression is the sum of infinite G.P series
We know that the sum of an infinite G.P. series is a/ (1-r) where a= the first term of infinite G.P. and r = common ratio
So, the sum of the above expression is (1/4) / (1- 3/4 ) = 1
So, the probability that the tossing would terminate is 1.
Hence our tossing will definitely terminate it cannot go forever.