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First success timing

ProbabilityEasy

Each attempt of a test succeeds with probability $\frac{1}{4}$ , independently of the others. What is the probability that the first success occurs exactly on attempt number $3$ ?

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Hint

The first $2$ attempts must fail and the next one must succeed.

Solution

The failure probability of one attempt is $1 - \frac{1}{4} = \frac{3}{4}$ .
Both of the first $2$ attempts fail with probability $(\frac{3}{4}) \times (\frac{3}{4}) = \frac{9}{16}$ .
Attempt number $3$ then succeeds with probability $\frac{1}{4}$ .
The product is $\frac{9}{16} \times \frac{1}{4} = \frac{9}{64}$ .

Answer: $\frac{9}{64}$

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