Question 1174461
Let $S$ be the event that a bike is a sports bike, and $R$ be the event that a bike is a regular bike.
Let $T$ be the event that a bike is stolen.
We are given:
$P(T|S) = 0.09$ (probability a sports bike is stolen)
$P(T|R) = 0.005$ (probability a regular bike is stolen)
$P(S) = 0.19$ (proportion of sports bikes)
$P(R) = 1 - P(S) = 1 - 0.19 = 0.81$ (proportion of regular bikes)

We want to find $P(S|T)$, the probability that a stolen bike is a sports bike, and $P(T)$, the probability that a bike is stolen.

Using Bayes' theorem:
$P(S|T) = \frac{P(T|S)P(S)}{P(T)}$

First, we find $P(T)$, the probability a bike is stolen:
$P(T) = P(T|S)P(S) + P(T|R)P(R)$
$P(T) = (0.09)(0.19) + (0.005)(0.81)$
$P(T) = 0.0171 + 0.00405$
$P(T) = 0.02115$

Now we can find $P(S|T)$:
$P(S|T) = \frac{P(T|S)P(S)}{P(T)}$
$P(S|T) = \frac{(0.09)(0.19)}{0.02115}$
$P(S|T) = \frac{0.0171}{0.02115}$
$P(S|T) \approx 0.8085$

Therefore:
The probability that a stolen bike is a sports bike is approximately 0.8085.
The probability that a bike will be stolen is 0.02115.

Final Answer: The final answer is $\boxed{0.8085, 0.02115}$