		2.	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	




0				
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This problem involves finding the probability distribution for the number of drivers arrested during 5 traffic stops. Since the officer has a probability of 0.12 of making an arrest at each stop, we can model this situation using a **binomial distribution**, where:

- \( n = 5 \) (the number of trials or stops),
- \( p = 0.12 \) (the probability of an arrest on each stop),
- \( r \) is the number of arrests (which ranges from 0 to 5).

The probability mass function (PMF) for a binomial distribution is given by:

\[
P(r) = \binom{n}{r} p^r (1-p)^{n-r}
\]

Where:
- \( \binom{n}{r} \) is the binomial coefficient, calculated as \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \),
- \( p^r \) is the probability of having \( r \) arrests,
- \( (1-p)^{n-r} \) is the probability of having \( n-r \) non-arrests.

### Let's compute the probabilities for \( r = 0, 1, 2, 3, 4, 5 \):

1. **For \( r = 0 \) (no arrests):**
\[
P(0) = \binom{5}{0} (0.12)^0 (1-0.12)^{5-0} = 1 \times 1 \times (0.88)^5 = 0.527
\]

2. **For \( r = 1 \) (one arrest):**
\[
P(1) = \binom{5}{1} (0.12)^1 (1-0.12)^{5-1} = 5 \times 0.12 \times (0.88)^4 = 0.359
\]

3. **For \( r = 2 \) (two arrests):**
\[
P(2) = \binom{5}{2} (0.12)^2 (1-0.12)^{5-2} = 10 \times (0.12)^2 \times (0.88)^3 = 0.096
\]

4. **For \( r = 3 \) (three arrests):**
\[
P(3) = \binom{5}{3} (0.12)^3 (1-0.12)^{5-3} = 10 \times (0.12)^3 \times (0.88)^2 = 0.015
\]

5. **For \( r = 4 \) (four arrests):**
\[
P(4) = \binom{5}{4} (0.12)^4 (1-0.12)^{5-4} = 5 \times (0.12)^4 \times (0.88) = 0.001
\]

6. **For \( r = 5 \) (five arrests):**
\[
P(5) = \binom{5}{5} (0.12)^5 (1-0.12)^{5-5} = 1 \times (0.12)^5 \times 1 = 0.000
\]

### Final Probability Distribution Table:

| \( r \) (number of arrests) | Probability \( P(r) \) |
|-----------------------------|------------------------|
| 0                           | 0.527                  |
| 1                           | 0.359                  |
| 2                           | 0.096                  |
| 3                           | 0.015                  |
| 4                           | 0.001                  |
| 5                           | 0.000                  |

These probabilities are rounded to the nearest thousandths as required.

---

Would you like a detailed explanation of how these calculations are done or have any questions?

### Related Questions:
1. How would the distribution change if the probability of arrest increases to 0.20?
2. What is the expected number of arrests in this scenario?
3. How do you calculate the variance of a binomial distribution?
4. What is the cumulative probability of arresting at least 2 drivers?
5. How does the binomial distribution differ from the normal distribution?

**Tip**: The sum of the probabilities in a probability distribution must always equal 1, or close to it due to rounding errors.

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. For this table, you will round to the nearest thousandths.

Learn how to construct a probability distribution for the number of arrests made during 5 traffic stops, given a 0.12 probability of arrest per stop. This problem involves the use of binomial distribution formulas and rounding probabilities to the nearest thousandths.

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