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A biased coin is tossed repeatedly until the first "tail" occurs. The expected number of tosses required to produce the first tail is estimated as $T$. Assuming this is true, find the probability of at least two tails in $3T$ tosses.

This is a sample question I found for the GRE. I have no idea how to approach this problem. I know you can find the probability of a getting a tails by using the geometric distribution, which is just $\frac{1}{T}$.

If someone could offer a hint or solution, I would appreciate it.

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For Geometric distribution, the mean is $\frac{1}{p}$ so as you said yourself $\frac{1}{T}=p$. Now we have to suppose that $T$ is integer.

The event of getting $k$ tails out on $n$ tosses is then modeled using Binomial distribution. Assume the random variable corresponding to this event is $X$. Then to find the probability that you get at least 2 tails in $3T$ tosses, we calculate first the inverse probability, i.e. the probability that we get at most 1 tails in $3T$ tosses which is: $$ \Pr(X\leq 1)=(1-p)^{3T}+{3T \choose 1}(1-p)^{3T-1}p $$ Hence $$ \Pr(X\geq 2)=1-(1-p)^{3T}-{3T \choose 1}(1-p)^{3T-1}p $$

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