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 
.
Given that th<span>e coordinates of the vertices of △DEF are D(2, −1) , E(7, −1) , and F(2, −3) and the coordinates of the vertices of △D′E′F′ are D′(0, −1) , E′(−5, −1) , and F′(0, −3) .
Notice that the y-coordinates of the pre-image and that of the image are the same, which means that there is a reflection across the y-axis.
A refrection across the y-axis results in the change in sign of the x-coordinates of the pre-image and the image while the y-coordinate of the image remains the same as that of the pre-image.
A refrection across the y-axis of </span>△DEF with vertices D(2, −1) , E(7, −1) , and F(2, −3)
will result in and image with vertices (-2, -1), (-7, -1) and (-2, -3) respectively.
Notice that the x-coordinate of the final image △D′E′F′ with vertices <span>D′(0, −1) , E′(−5, −1) , and F′(0, −3) is 2 units greater than the vertices of the result of recting the pre-image across the y-axis.
This means that the result of refrecting the pre-image was shifted two places to the right.
Therefore, </span>the sequence of transformations that maps △DEF to △D′E′F′ are reflection across the y-axis and translation 2 units right.
