Example 6.23. The probability of placing an order in a tailor shop is $p<1.0$. A persistent customer visits all the tailor shops in the city until he succeeds. What is the probability that he will succeed no earlier than on the third attempt, if on average the number of attempts is 5.
Solution. The random variable is distributed according to the geometric law. Therefore, $M(X)=5=\frac{1}{p}$, from which $p=0.2$. Let's find $P(1)=p \cdot q^{1-1}=0.2$ and $P(2)=p \cdot q^{2-1}=0.2 \cdot 0.8=0.16$ — the probabilities that the client will achieve success on the first and second attempts, respectively.

The events that the client will achieve success no earlier than the third attempt or on the first or second attempt form a complete group of events, i.e., $P(m \geq 3)+[P(m=1)+P(m=2)]=1$. Therefore, $P(m \geq 3)=1-[P(1)+P(2)]=1-(0.2+0.16)=0.64$.