When three friends go for coffee, they decide who will pay the check by each flipping a coin and then letting the “odd person” pay. If all three flips produce the same result (so that there is no odd person), then they make a second round of flips, and they continue to do so until there is an odd person. What is the probability that

1. exactly $3$rounds of flips are made?
2. more than $4$rounds are needed?

Three friends decide to flip coin and pay the bill.

If all three flips produce the same result hen they make a second round of flips, and they continue to do so until there is an odd person.

We have to find that what is the probability exactly 3 rounds of flips are made.

It is easily seen that probability to have a successful round is $p=0.75$.

Let $X~G\left(0.75\right)$

Therefore,

$P(x=3)={0.25}^2\times 0.75$

$=0.046875$

The probability that exactly $3$rounds of flips are made is $0.046875$

Three friends decide to flip coin and pay the bill.

If all three flips produce the same result hen they make a second round of flips, and they continue to do so until there is an odd person.

We have to find what is the probability that more than $4$rounds are needed?

More than $4$rounds

$p(x>4)=1-p(x\le 4)$

$=1-\left(1-{0.25}^4\right)$

$={0.25}^4$

Therefore, the probability that more than $4$rounds are needed:$P(x>4)=0.00390625$

The probability that more than $4$rounds are needed:$0.00390625$