Answer:
(10) Person B
(11) Person B
(12) ![P(5\ or\ 6) = 60\%](https://tex.z-dn.net/?f=P%285%5C%20or%5C%206%29%20%3D%2060%5C%25)
(13) Person B
Step-by-step explanation:
Given
Person A
5 coins (records the outcome of Heads)
Person
Rolls 2 dice (recorded the larger number)
Person A
First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)
Next, we list out all number of heads in each roll (sorted)
![Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}](https://tex.z-dn.net/?f=Head%20%3D%20%5C%7B5%2C4%2C4%2C4%2C4%2C4%2C3%2C3%2C3%2C3%2C3%2C3%2C3%2C3%2C3%2C3%2C2%2C2%2C2%2C2%2C2%2C2%2C2%2C2%2C2%2C2%2C1%2C1%2C1%2C1%2C1%2C0%5C%7D)
![n(Head) = 32](https://tex.z-dn.net/?f=n%28Head%29%20%3D%2032)
Person B
First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)
Next, we list out the highest in each toss (sorted)
![Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}](https://tex.z-dn.net/?f=Dice%20%3D%20%5C%7B2%2C2%2C3%2C3%2C3%2C3%2C4%2C4%2C4%2C4%2C4%2C4%2C5%2C5%2C5%2C5%2C5%2C5%2C5%2C5%2C6%2C6%2C6%2C6%2C6%2C6%2C6%2C6%2C6%2C6%5C%7D)
![n(Dice) = 30](https://tex.z-dn.net/?f=n%28Dice%29%20%3D%2030)
Question 10: Who is likely to get number 5
From person A list of outcomes, the proportion of 5 is:
![Pr(5) = \frac{n(5)}{n(Head)}](https://tex.z-dn.net/?f=Pr%285%29%20%3D%20%5Cfrac%7Bn%285%29%7D%7Bn%28Head%29%7D)
![Pr(5) = \frac{1}{32}](https://tex.z-dn.net/?f=Pr%285%29%20%3D%20%5Cfrac%7B1%7D%7B32%7D)
![Pr(5) = 0.03125](https://tex.z-dn.net/?f=Pr%285%29%20%3D%200.03125)
From person B list of outcomes, the proportion of 5 is:
![Pr(5) = \frac{n(5)}{n(Dice)}](https://tex.z-dn.net/?f=Pr%285%29%20%3D%20%5Cfrac%7Bn%285%29%7D%7Bn%28Dice%29%7D)
![Pr(5) = \frac{8}{30}](https://tex.z-dn.net/?f=Pr%285%29%20%3D%20%5Cfrac%7B8%7D%7B30%7D)
![Pr(5) = 0.267](https://tex.z-dn.net/?f=Pr%285%29%20%3D%200.267)
<em>From the above calculations: </em>
<em> Hence, person B is more likely to get 5</em>
Question 11: Person with Higher median
For person A
![Median = \frac{n(Head) + 1}{2}th](https://tex.z-dn.net/?f=Median%20%3D%20%5Cfrac%7Bn%28Head%29%20%2B%201%7D%7B2%7Dth)
![Median = \frac{32 + 1}{2}th](https://tex.z-dn.net/?f=Median%20%3D%20%5Cfrac%7B32%20%2B%201%7D%7B2%7Dth)
![Median = \frac{33}{2}th](https://tex.z-dn.net/?f=Median%20%3D%20%5Cfrac%7B33%7D%7B2%7Dth)
![Median = 16.5th](https://tex.z-dn.net/?f=Median%20%3D%2016.5th)
This means that the median is the mean of the 16th and the 17th item
So,
![Median = \frac{3+2}{2}](https://tex.z-dn.net/?f=Median%20%3D%20%5Cfrac%7B3%2B2%7D%7B2%7D)
![Median = \frac{5}{2}](https://tex.z-dn.net/?f=Median%20%3D%20%5Cfrac%7B5%7D%7B2%7D)
![Median = 2.5](https://tex.z-dn.net/?f=Median%20%3D%202.5)
For person B
![Median = \frac{n(Dice) + 1}{2}th](https://tex.z-dn.net/?f=Median%20%3D%20%5Cfrac%7Bn%28Dice%29%20%2B%201%7D%7B2%7Dth)
![Median = \frac{30 + 1}{2}th](https://tex.z-dn.net/?f=Median%20%3D%20%5Cfrac%7B30%20%2B%201%7D%7B2%7Dth)
![Median = \frac{31}{2}th](https://tex.z-dn.net/?f=Median%20%3D%20%5Cfrac%7B31%7D%7B2%7Dth)
![Median = 15.5th](https://tex.z-dn.net/?f=Median%20%3D%2015.5th)
This means that the median is the mean of the 15th and the 16th item. So,
![Median = \frac{5+5}{2}](https://tex.z-dn.net/?f=Median%20%3D%20%5Cfrac%7B5%2B5%7D%7B2%7D)
![Median = \frac{10}{2}](https://tex.z-dn.net/?f=Median%20%3D%20%5Cfrac%7B10%7D%7B2%7D)
![Median = 5](https://tex.z-dn.net/?f=Median%20%3D%205)
<em>Person B has a greater median of 5</em>
Question 12: Probability that B gets 5 or 6
This is calculated as:
![P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}](https://tex.z-dn.net/?f=P%285%5C%20or%5C%206%29%20%3D%20%5Cfrac%7Bn%285%5C%20or%5C%206%29%7D%7Bn%28Dice%29%7D)
From the sample space of person B, we have:
![n(5\ or\ 6) =n(5) + n(6)](https://tex.z-dn.net/?f=n%285%5C%20or%5C%206%29%20%3Dn%285%29%20%2B%20n%286%29)
![n(5\ or\ 6) =8+10](https://tex.z-dn.net/?f=n%285%5C%20or%5C%206%29%20%3D8%2B10)
![n(5\ or\ 6) = 18](https://tex.z-dn.net/?f=n%285%5C%20or%5C%206%29%20%3D%2018)
So, we have:
![P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}](https://tex.z-dn.net/?f=P%285%5C%20or%5C%206%29%20%3D%20%5Cfrac%7Bn%285%5C%20or%5C%206%29%7D%7Bn%28Dice%29%7D)
![P(5\ or\ 6) = \frac{18}{30}](https://tex.z-dn.net/?f=P%285%5C%20or%5C%206%29%20%3D%20%5Cfrac%7B18%7D%7B30%7D)
![P(5\ or\ 6) = 0.60](https://tex.z-dn.net/?f=P%285%5C%20or%5C%206%29%20%3D%200.60)
![P(5\ or\ 6) = 60\%](https://tex.z-dn.net/?f=P%285%5C%20or%5C%206%29%20%3D%2060%5C%25)
Question 13: Person with higher probability of 3 or more
Person A
![n(3\ or\ more) = 16](https://tex.z-dn.net/?f=n%283%5C%20or%5C%20more%29%20%3D%2016)
So:
![P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}](https://tex.z-dn.net/?f=P%283%5C%20or%5C%20more%29%20%3D%20%5Cfrac%7Bn%283%5C%20or%5C%20more%29%7D%7Bn%28Head%29%7D)
![P(3\ or\ more) = \frac{16}{32}](https://tex.z-dn.net/?f=P%283%5C%20or%5C%20more%29%20%3D%20%5Cfrac%7B16%7D%7B32%7D)
![P(3\ or\ more) = 0.50](https://tex.z-dn.net/?f=P%283%5C%20or%5C%20more%29%20%3D%200.50)
![P(3\ or\ more) = 50\%](https://tex.z-dn.net/?f=P%283%5C%20or%5C%20more%29%20%3D%2050%5C%25)
Person B
![n(3\ or\ more) = 28](https://tex.z-dn.net/?f=n%283%5C%20or%5C%20more%29%20%3D%2028)
So:
![P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}](https://tex.z-dn.net/?f=P%283%5C%20or%5C%20more%29%20%3D%20%5Cfrac%7Bn%283%5C%20or%5C%20more%29%7D%7Bn%28Dice%29%7D)
![P(3\ or\ more) = \frac{28}{30}](https://tex.z-dn.net/?f=P%283%5C%20or%5C%20more%29%20%3D%20%5Cfrac%7B28%7D%7B30%7D)
![P(3\ or\ more) = 0.933](https://tex.z-dn.net/?f=P%283%5C%20or%5C%20more%29%20%3D%200.933)
![P(3\ or\ more) = 93.3\%](https://tex.z-dn.net/?f=P%283%5C%20or%5C%20more%29%20%3D%2093.3%5C%25)
By comparison:
![93.3\% > 50\%](https://tex.z-dn.net/?f=93.3%5C%25%20%3E%2050%5C%25)
Hence, person B has a higher probability of 3 or more