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Sagot :
To solve this problem, we need to find the probability distribution of the number of citizens from Peterborough when three names are randomly selected from a box containing names of citizens from both Peterborough and Lakefield. There are 10 citizens from Peterborough and 6 citizens from Lakefield, making a total of 16 citizens.
a) Probability Distribution of Citizens from Peterborough
We need to determine the probability distribution for 0, 1, 2, and 3 citizens being selected from Peterborough. The probabilities can be obtained using the hypergeometric distribution formula:
[tex]\[ P(X = i) = \frac{\binom{10}{i} \cdot \binom{6}{3-i}}{\binom{16}{3}} \][/tex]
where:
- [tex]\( \binom{a}{b} \)[/tex] is the combination function (also called "n choose k") which gives the number of ways to choose b elements from a
- 10 is the number of citizens from Peterborough
- 6 is the number of citizens from Lakefield
- 3 is the selection size
Below are the probabilities for each possible number of citizens from Peterborough (i = 0, 1, 2, 3):
1. Probability of selecting 0 citizens from Peterborough:
[tex]\[ P(X = 0) = \frac{\binom{10}{0} \cdot \binom{6}{3}}{\binom{16}{3}} = \frac{1 \cdot 20}{560} = 0.03571428571428571 \][/tex]
2. Probability of selecting 1 citizen from Peterborough:
[tex]\[ P(X = 1) = \frac{\binom{10}{1} \cdot \binom{6}{2}}{\binom{16}{3}} = \frac{10 \cdot 15}{560} = 0.26785714285714285 \][/tex]
3. Probability of selecting 2 citizens from Peterborough:
[tex]\[ P(X = 2) = \frac{\binom{10}{2} \cdot \binom{6}{1}}{\binom{16}{3}} = \frac{45 \cdot 6}{560} = 0.48214285714285715 \][/tex]
4. Probability of selecting 3 citizens from Peterborough:
[tex]\[ P(X = 3) = \frac{\binom{10}{3} \cdot \binom{6}{0}}{\binom{16}{3}} = \frac{120 \cdot 1}{560} = 0.21428571428571427 \][/tex]
Thus, the probability distribution of the number of citizens from Peterborough in the delegation is:
[tex]\[ \{0: 0.03571428571428571, 1: 0.26785714285714285, 2: 0.48214285714285715, 3: 0.21428571428571427\} \][/tex]
b) Probability Histogram of the Distribution
To create a probability histogram, we plot the number of citizens from Peterborough on the x-axis (0, 1, 2, 3) and the corresponding probabilities on the y-axis.
1. 0 Citizens:
- Number of occurrences: 0
- Probability: [tex]\( 0.03571428571428571 \)[/tex]
2. 1 Citizen:
- Number of occurrences: 1
- Probability: [tex]\( 0.26785714285714285 \)[/tex]
3. 2 Citizens:
- Number of occurrences: 2
- Probability: [tex]\( 0.48214285714285715 \)[/tex]
4. 3 Citizens:
- Number of occurrences: 3
- Probability: [tex]\( 0.21428571428571427 \)[/tex]
Here is a simple way to represent the histogram data:
[tex]\( \begin{array}{|c|c|} \hline \text{Number of Citizens} & \text{Probability} \\ \hline 0 & 0.03571428571428571 \\ 1 & 0.26785714285714285 \\ 2 & 0.48214285714285715 \\ 3 & 0.21428571428571427 \\ \hline \end{array} \)[/tex]
To visualize this, we can create a bar graph where the x-axis represents the number of citizens from Peterborough (0, 1, 2, 3) and the y-axis represents the probability:
```
| _
| _|_|
| _|_||_|
| _|_||_||_|
| __|_||_||_||_|
|__|_||_||_||_||_|
------------------------
0 1 2 3
```
- The bar for 0 is relatively small.
- The bar for 1 is larger.
- The bar for 2 is the largest.
- The bar for 3 is smaller compared to 1 and 2, but larger than 0.
This histogram visually represents how likely each number of citizens from Peterborough is when three citizens are selected at random from the total group.
a) Probability Distribution of Citizens from Peterborough
We need to determine the probability distribution for 0, 1, 2, and 3 citizens being selected from Peterborough. The probabilities can be obtained using the hypergeometric distribution formula:
[tex]\[ P(X = i) = \frac{\binom{10}{i} \cdot \binom{6}{3-i}}{\binom{16}{3}} \][/tex]
where:
- [tex]\( \binom{a}{b} \)[/tex] is the combination function (also called "n choose k") which gives the number of ways to choose b elements from a
- 10 is the number of citizens from Peterborough
- 6 is the number of citizens from Lakefield
- 3 is the selection size
Below are the probabilities for each possible number of citizens from Peterborough (i = 0, 1, 2, 3):
1. Probability of selecting 0 citizens from Peterborough:
[tex]\[ P(X = 0) = \frac{\binom{10}{0} \cdot \binom{6}{3}}{\binom{16}{3}} = \frac{1 \cdot 20}{560} = 0.03571428571428571 \][/tex]
2. Probability of selecting 1 citizen from Peterborough:
[tex]\[ P(X = 1) = \frac{\binom{10}{1} \cdot \binom{6}{2}}{\binom{16}{3}} = \frac{10 \cdot 15}{560} = 0.26785714285714285 \][/tex]
3. Probability of selecting 2 citizens from Peterborough:
[tex]\[ P(X = 2) = \frac{\binom{10}{2} \cdot \binom{6}{1}}{\binom{16}{3}} = \frac{45 \cdot 6}{560} = 0.48214285714285715 \][/tex]
4. Probability of selecting 3 citizens from Peterborough:
[tex]\[ P(X = 3) = \frac{\binom{10}{3} \cdot \binom{6}{0}}{\binom{16}{3}} = \frac{120 \cdot 1}{560} = 0.21428571428571427 \][/tex]
Thus, the probability distribution of the number of citizens from Peterborough in the delegation is:
[tex]\[ \{0: 0.03571428571428571, 1: 0.26785714285714285, 2: 0.48214285714285715, 3: 0.21428571428571427\} \][/tex]
b) Probability Histogram of the Distribution
To create a probability histogram, we plot the number of citizens from Peterborough on the x-axis (0, 1, 2, 3) and the corresponding probabilities on the y-axis.
1. 0 Citizens:
- Number of occurrences: 0
- Probability: [tex]\( 0.03571428571428571 \)[/tex]
2. 1 Citizen:
- Number of occurrences: 1
- Probability: [tex]\( 0.26785714285714285 \)[/tex]
3. 2 Citizens:
- Number of occurrences: 2
- Probability: [tex]\( 0.48214285714285715 \)[/tex]
4. 3 Citizens:
- Number of occurrences: 3
- Probability: [tex]\( 0.21428571428571427 \)[/tex]
Here is a simple way to represent the histogram data:
[tex]\( \begin{array}{|c|c|} \hline \text{Number of Citizens} & \text{Probability} \\ \hline 0 & 0.03571428571428571 \\ 1 & 0.26785714285714285 \\ 2 & 0.48214285714285715 \\ 3 & 0.21428571428571427 \\ \hline \end{array} \)[/tex]
To visualize this, we can create a bar graph where the x-axis represents the number of citizens from Peterborough (0, 1, 2, 3) and the y-axis represents the probability:
```
| _
| _|_|
| _|_||_|
| _|_||_||_|
| __|_||_||_||_|
|__|_||_||_||_||_|
------------------------
0 1 2 3
```
- The bar for 0 is relatively small.
- The bar for 1 is larger.
- The bar for 2 is the largest.
- The bar for 3 is smaller compared to 1 and 2, but larger than 0.
This histogram visually represents how likely each number of citizens from Peterborough is when three citizens are selected at random from the total group.
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