Question

The scheduling officer for a local police department is trying to schedule additional patrol units in each of two neighbourhoods – southern and northern. She knows that on any given day, the probabilities of major crimes and minor crimes being committed in the northern neighbourhood were 0.418 and 0.612, respectively, and that the corresponding probabilities in the southern neighbourhood were 0.355 and 0.520. Assuming that all crime occur independent of each other and likewise that crime in the two neighbourhoods are independent of each other, what is the probability that no crime of either type is committed in either neighbourhood on any given day?

• A. 0.069
• B. 0.225
• C. 0.69
• D. 0.775
• E. None of the above

Hint:

When multiple independent events are involved, always compute the “no occurrence” probability separately for each case and then multiply them together. This approach prevents mistakes in probability problems.

Answer 1

The Correct Option is A

Step 1: Probability of no crime in northern neighbourhood.
- Probability of no major crime in northern area = \(1 - 0.418 = 0.582\).
- Probability of no minor crime in northern area = \(1 - 0.612 = 0.388\). Since major and minor crimes are independent: \[ P(\text{no crime in northern}) = 0.582 \times 0.388 = 0.2256 \]

Step 2: Probability of no crime in southern neighbourhood.
- Probability of no major crime in southern area = \(1 - 0.355 = 0.645\).
- Probability of no minor crime in southern area = \(1 - 0.520 = 0.480\). So, \[ P(\text{no crime in southern}) = 0.645 \times 0.480 = 0.3096 \]

Step 3: Combine both neighbourhoods.
Since crimes in northern and southern neighbourhoods are independent: \[ P(\text{no crime in either neighbourhood}) = 0.2256 \times 0.3096 \] \[ = 0.0698 \approx 0.069 \]

Final Answer:
\[ \boxed{0.069} \]