Three friends go for coffee. They decide who will pay the bill by each tossing a coin and then letting the "odd person" pay. There is no odd person if all three tosses produce the same result. If there is no odd person in the first round, they make a second round of tosses and they continue to do so until there is an odd person. What is the probability that exactly three rounds of tosses are made?

$\frac{3}{64}$

The correct answer is Option (2) → $\frac{3}{64}$ ##

$P(\text{not obtaining an odd person in a single round})$

$= P(\text{All three of them throw tails or All three of them throw heads})$

$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times 2 = \frac{1}{4}$

$P(\text{obtaining an odd person in a single round})$

$= 1 - P(\text{not obtaining an odd person in a single round})$

$= 1 - \frac{1}{4} = \frac{3}{4}$

The required probability

$= P(\text{"In first round, there is no odd person" and "In second round, there is no odd person" and "In third round, there is an odd person"})$

$= \frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{64}$