If the probability that an officer makes an arrest on a traffic stop is 0.12, construct a probability distribution for the probability of an officer arresting 0 to 5 drivers when 5 traffic stops are conducted by the officer… (Each space is worth 0.5 points for a total of 12 points.) For this table, you will round to the nearest thousandths. Review Table 6.3 on p. 146 in your textbook for help.
r	             (n)                 pr                  q n-r                  p®
0				
1				
2				
3				
4				
5

In this problem, we're dealing with **binomial probability distribution** because the scenario is composed of independent trials (traffic stops), and each trial has two possible outcomes: an arrest (success) or no arrest (failure). The probability of success (arrest) in each trial is given as \( p = 0.12 \), and the probability of failure \( q = 1 - p = 0.88 \). 

We need to calculate the probability of making 0 to 5 arrests in 5 traffic stops. The binomial probability formula is:

\[
P(r) = \binom{n}{r} p^r q^{n-r}
\]
Where:
- \( P(r) \) is the probability of exactly \( r \) arrests,
- \( n \) is the number of trials (5 traffic stops),
- \( r \) is the number of arrests (0 to 5),
- \( p \) is the probability of success (arrest) in each trial,
- \( q \) is the probability of failure (no arrest) in each trial,
- \( \binom{n}{r} \) is the binomial coefficient.

We'll calculate the probabilities for each possible value of \( r \) from 0 to 5 and round to the nearest thousandth.

### Table structure:
| \( r \) | \( n \) (Total trials) | \( p^r \) | \( q^{n-r} \) | \( P(r) \) |
|--------|-----------------------|-----------|---------------|------------|
|   0    |          5            |           |               |            |
|   1    |          5            |           |               |            |
|   2    |          5            |           |               |            |
|   3    |          5            |           |               |            |
|   4    |          5            |           |               |            |
|   5    |          5            |           |               |            |

Now, let’s calculate the binomial probabilities for each case.Here is the completed probability distribution table for the number of arrests \( r \) when the officer conducts 5 traffic stops:

| \( r \) (Number of arrests) | \( n \) (Total trials) | \( p^r \) | \( q^{n-r} \) | \( P(r) \) (Probability) |
|----------------------------|------------------------|-----------|---------------|--------------------------|
| 0                          | 5                      | \( (0.12)^0 \) | \( (0.88)^5 \)  | 0.528                    |
| 1                          | 5                      | \( (0.12)^1 \) | \( (0.88)^4 \)  | 0.360                    |
| 2                          | 5                      | \( (0.12)^2 \) | \( (0.88)^3 \)  | 0.098                    |
| 3                          | 5                      | \( (0.12)^3 \) | \( (0.88)^2 \)  | 0.013                    |
| 4                          | 5                      | \( (0.12)^4 \) | \( (0.88)^1 \)  | 0.001                    |
| 5                          | 5                      | \( (0.12)^5 \) | \( (0.88)^0 \)  | 0.000                    |

The probabilities are rounded to the nearest thousandth.

Would you like a detailed explanation of any of these calculations?

### 5 Related Questions:
1. How do we compute the binomial coefficient \( \binom{n}{r} \)?
2. How does the probability of success \( p \) affect the distribution?
3. What does the value \( P(0) = 0.528 \) represent in this context?
4. How would the distribution change if the probability of arrest were higher, say \( p = 0.2 \)?
5. Can this method be applied to situations with more than two outcomes?

### Tip:
The sum of all probabilities in a binomial distribution should always equal 1.

If the probability that an officer makes an arrest on a traffic stop is 0.12, construct a probability distribution for the probability of an officer arresting 0 to 5 drivers when 5 traffic stops are conducted by the officer. Round to the nearest thousandth.

Explore how to calculate the binomial probability distribution for an officer making arrests on traffic stops. Given that the probability of an arrest is 0.12, this problem demonstrates the application of binomial distribution to predict arrests for 0 to 5 drivers out of 5 traffic stops. Includes detailed steps and a table of probabilities rounded to the nearest thousandth.

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