Answer:
Assume that all the coins involved here are fair coins.
a) Probability of finding the "odd" person in one round: 
.
b) Probability of finding the "odd" person in the 
th round: 
.
c) Expected number of rounds: 
.
Step-by-step explanation:
<h3>a)</h3>
To decide the "odd" person, either of the following must happen:
- There are
heads and 
tail, or - There are head and tails.
Assume that the coins here all are all fair. In other words, each has a 
chance of landing on the head and a
The binomial distribution can model the outcome of 
coin-tosses. The chance of getting 
heads out of
- The chance of getting heads (and consequently, tail) would be
. - The chance of getting heads (and consequently, tails) would be
.
These two events are mutually-exclusive. 
would be the chance that either of them will occur. That's the same as the chance of determining the "odd" person in one round.
<h3>b)</h3>
Since the coins here are all fair, the chance of determining the "odd" person would be in all rounds.
When the chance 
of getting a success in each round is the same, the geometric distribution would give the probability of getting the first success (that is, to find the "odd" person) in the th round: 
. That's the same as the probability of getting one success after 
unsuccessful attempts.
In this case, 
. Therefore, the probability of succeeding on round round would be
.
<h3>c)</h3>
Let is the chance of success on each round in a geometric distribution. The expected value of that distribution would be 
.
In this case, since , the expected value would be 
.
