Respuesta :
Answer:
Assume that all the coins involved here are fair coins.
a) Probability of finding the "odd" person in one round: [tex]\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}[/tex].
b) Probability of finding the "odd" person in the [tex]k[/tex]th round: [tex]\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left( 1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}[/tex].
c) Expected number of rounds: [tex]\displaystyle \frac{2^{n - 1}}{n}[/tex].
Step-by-step explanation:
a)
To decide the "odd" person, either of the following must happen:
- There are [tex](n - 1)[/tex] heads and [tex]1[/tex] tail, or
- There are [tex]1[/tex] head and [tex](n - 1)[/tex] tails.
Assume that the coins here all are all fair. In other words, each has a [tex]50\,\%[/tex] chance of landing on the head and a
The binomial distribution can model the outcome of [tex]n[/tex] coin-tosses. The chance of getting [tex]x[/tex] heads out of
- The chance of getting [tex](n - 1)[/tex] heads (and consequently, [tex]1[/tex] tail) would be [tex]\displaystyle {n \choose n - 1}\cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left(\frac{1}{2}\right)^{n - (n - 1)} = n\cdot \left(\frac{1}{2}\right)^n[/tex].
- The chance of getting [tex]1[/tex] heads (and consequently, [tex](n - 1)[/tex] tails) would be [tex]\displaystyle {n \choose 1}\cdot \left(\frac{1}{2}\right)^{1} \cdot \left(\frac{1}{2}\right)^{n - 1} = n\cdot \left(\frac{1}{2}\right)^n[/tex].
These two events are mutually-exclusive. [tex]\displaystyle n\cdot \left(\frac{1}{2}\right)^n + n\cdot \left(\frac{1}{2}\right)^n = 2\,n \cdot \left(\frac{1}{2}\right)^n = n \cdot \left(\frac{1}{2}\right)^{n - 1}[/tex] 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.
b)
Since the coins here are all fair, the chance of determining the "odd" person would be [tex]\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}[/tex] in all rounds.
When the chance [tex]p[/tex] 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 [tex]k[/tex]th round: [tex](1 - p)^{k - 1} \cdot p[/tex]. That's the same as the probability of getting one success after [tex](k - 1)[/tex] unsuccessful attempts.
In this case, [tex]\displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}[/tex]. Therefore, the probability of succeeding on round [tex]k[/tex] round would be
[tex]\displaystyle \underbrace{\left(1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}}_{(1 - p)^{k - 1}} \cdot \underbrace{n \cdot \left(\frac{1}{2}\right)^{n - 1}}_{p}[/tex].
c)
Let [tex]p[/tex] is the chance of success on each round in a geometric distribution. The expected value of that distribution would be [tex]\displaystyle \frac{1}{p}[/tex].
In this case, since [tex]\displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}[/tex], the expected value would be [tex]\displaystyle \frac{1}{p} = \frac{1}{\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}}= \frac{2^{n - 1}}{n}[/tex].
Answer:
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:
a)
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.
b)
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
.
c)
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
Step-by-step explanation:
