Coin is tossed $100$ times. For $1\leq i\leq 94$, let $A_i$ be the event that the $i^{th}$ toss is a tail and the following $6$ tosses are heads. Calculate the number of sample points in the event $A_i$ and hence determine its probability.
So the way I see it is $7$ coins have to be fixed for event $A_i$ to occur i.e. we get THHHHHH somewhere in the row of $100$, the T being one of the coins between and including coin $1$ and coin $94$. That means the other $93$ coins are unfixed for each $i$, so $93!$ possibilities where $A_i$ is satisfied. Overall we have $100!$ possible ways $100$ coin flips can turn out.
Probability = # (outcomes that satisfy Ai)/(Total outcomes) so is the answer $\frac{93!}{100!}$? The probability of six heads occuring in a row seems kinda small if this is the right answer…what am I doing wrong? I think I must be thinking about it the wrong way and would really appreciate it if someone could explain this to me. Thank you.
Best Answer
Your hypothesis is that in the last seven tosses you obtain a specific sequence, and you do not ask any other about the first 93 tosses.
You have to remember that for Bernoulli theorem the single extraction have not memory, so the probability to obtain THHHHHH in the last seven tosses is:
$$ P = \frac{1}{2^7} = \frac {1}{128} $$