# Probability of a specific event at a given trial

Recently I have founded some problems about probabilities, that ask to find the probability of an event at a given trial.

A dollar coin is toss several times until ones get "one dollar" face up. What is the probability to toss the coin at least $3$ times?

I thought to apply for the binomial law.But the binomial law gives the probability for a number of favorable trials, and the question ask for a specific trial.How can I solve this kind of problem?

Is there any methodology that one can apply for this kind of problems?

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Hint: It is the same as the probability of two tails in a row, because you need to toss at least $3$ times precisely if the first two tosses are tails. And the probability of two tails in a row is $\frac{1}{4}$.
Remark: For fun, let's also do this problem the hard way. We need at least $6$ tosses if we toss $2$ tails then a head, or if we toss $3$ tails then a head, or we toss $4$ tails then a head, or we toss $5$ tails then a head, or $\dots$.
The probability of $2$ tails then a head is $\left(\frac{1}{2}\right)^3$. The probability of $3$ tails then a head is $\left(\frac{1}{2}\right)^4$. The probability of $4$ tails then a head is $\left(\frac{1}{2}\right)^5$. And so on. Add up. The required probability is $$\left(\frac{1}{2}\right)^3 +\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^5+\cdots.$$ This is an infinite geometric series, which can be summed in the usual way. We get $\frac{1}{4}$.
The probability of two tails in a row is $\frac{1}{2}\times \frac{1}{2}$. That ($1/4$) is the answer to the question you asked. No complementary event needed. You need at least $3$ tosses exactly when the first two tosses are tails. –  André Nicolas Apr 17 '12 at 17:37