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				
1				
2				
3				
4				
5				





The situation described is a binomial probability distribution, where the officer has a fixed probability of making an arrest on each traffic stop, \( p = 0.12 \), and the number of trials (traffic stops) is 5. The probability distribution formula for a binomial distribution is:

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

Where:
- \( X \) is the number of arrests,
- \( n = 5 \) (number of traffic stops),
- \( r \) is the number of arrests (which ranges from 0 to 5),
- \( p = 0.12 \) is the probability of making an arrest,
- \( \binom{n}{r} \) is the binomial coefficient, which calculates how many ways \( r \) arrests can occur in 5 stops.

Let's calculate the probabilities for each \( r \) value from 0 to 5:

\[
P(X = 0) = \binom{5}{0} \cdot (0.12)^0 \cdot (1 - 0.12)^5
\]
\[
P(X = 1) = \binom{5}{1} \cdot (0.12)^1 \cdot (1 - 0.12)^4
\]
\[
P(X = 2) = \binom{5}{2} \cdot (0.12)^2 \cdot (1 - 0.12)^3
\]
\[
P(X = 3) = \binom{5}{3} \cdot (0.12)^3 \cdot (1 - 0.12)^2
\]
\[
P(X = 4) = \binom{5}{4} \cdot (0.12)^4 \cdot (1 - 0.12)^1
\]
\[
P(X = 5) = \binom{5}{5} \cdot (0.12)^5 \cdot (1 - 0.12)^0
\]

I'll calculate each of these probabilities and round them to the nearest thousandth.Here is the probability distribution for the officer arresting 0 to 5 drivers during 5 traffic stops:

\[
\begin{array}{|c|c|}
\hline
r & P(X = r) \\
\hline
0 & 0.528 \\
1 & 0.360 \\
2 & 0.098 \\
3 & 0.013 \\
4 & 0.001 \\
5 & 0.000 \\
\hline
\end{array}
\]

This table shows the probabilities of arresting 0, 1, 2, 3, 4, or 5 drivers.

Would you like further details or have any questions?

Here are 5 related questions:
1. How does the binomial distribution formula change if the number of traffic stops increases?
2. What happens to the distribution if the probability of arrest increases?
3. How do you calculate the expected number of arrests in a binomial distribution?
4. What is the cumulative probability of arresting at least 2 drivers?
5. Can we approximate this binomial distribution with a normal distribution?

**Tip:** The expected value in a binomial distribution is given by \( E(X) = n \cdot p \).

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.

Explore the probability distribution for an officer arresting 0 to 5 drivers during 5 traffic stops, with a 0.12 chance of making an arrest. This problem involves binomial distribution and rounds the probabilities to the nearest thousandths. Learn key probability concepts, including the binomial theorem, suitable for Grades 10-12.

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