To find the probability of exactly 2 traffic accidents requiring medical assistance during the specified time slot, we can use the Poisson probability formula. The Poisson distribution is suitable here because it models the number of events happening in a fixed interval of time, given the average number of events.
The formula for the Poisson probability is:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
Where:
- \( P(X = k) \) is the probability of observing \( k \) events in an interval.
- \( \lambda \) is the average number of events in the interval.
- \( k \) is the number of events we want to find the probability for.
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
In this case:
- \( \lambda = 1 \) (the average number of accidents).
- \( k = 2 \) (the number of accidents we want to find the probability for).
Substituting the values into the formula:
\[ P(X = 2) = \frac{e^{-1} \cdot 1^2}{2!} \]
Calculating \( 2! \):
\[ 2! = 2 \]
Now plugging everything in:
\[ P(X = 2) = \frac{e^{-1} \cdot 1^2}{2} = \frac{e^{-1}}{2} \]
Next, we need to find \( e^{-1} \):
\[ e^{-1} \approx 0.367879 \]
Now substitute this into the probability:
\[ P(X = 2) = \frac{0.367879}{2} \approx 0.1839395 \]
Finally, we round this to the nearest thousandth:
\[ \boxed{0.184} \]
So, the chance that there will be a need for exactly 2 ambulances on the Hollywood Freeway during that time slot is approximately \( 0.184 \).