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4. Probability Distributions of Discrete Variables

24. A citizen's delegation to Queen's Park is to be formed by selecting three names at random from a box. The box contains the names of 10 citizens from Peterborough and 6 citizens from Lakefield.

a) Show the probability distribution of the number of citizens that came from Peterborough. (3 marks)

b) Make a probability histogram of the distribution. (3 marks)


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.