Someone plays darts that has a join density function based on the surface z = 3 – r where z is the likelihood of the dart landing at $(r,\theta)$. How can I find the value of C so that f is a joint density function and then find the probability that the thrown dart lands in S?
$$g(r,\theta)=
\begin{cases}
C_{1}(3-r), & \text{if} (r,\theta)∈ R\\
0, & \text{otherwise}
\end{cases}$$
R is the region inside of r = 1 + tan $(\theta /4)$ (which includes the little loop) and S is just the region inside the inner loop.
So far this is how I am solving for C but how can I actually get the probability? Also is this correct so far?
Best Answer
Limits of your integral do not seem correct.
The outer loop of the region $R$ forms for $\theta \in \displaystyle \small \bigg(\big (0, \frac{4\pi}{3} \big), \big(\frac{7\pi}{3}, \frac{8\pi}{3} \big) , \big(\frac{11\pi}{3}, 4\pi\big)\bigg)$
And the inner loop $S$ forms for $ \ \theta \in \displaystyle \small \big(\frac{8\pi}{3}, \frac{11\pi}{3}\big)$.
So the integral to find $\small C$ should be,
$\displaystyle \small \int_{0}^{4 \pi/3} \int_{0}^{|1 + \tan (\theta/4)|} C(3 - r) \ r \ dr \ d\theta + \int_{7 \pi /3}^{8 \pi/3} \int_{0}^{|1 + \tan (\theta/4)|} C(3 - r) \ r \ dr \ d\theta$
$+ \displaystyle \small \int_{11 \pi /3}^{4 \pi} \int_{0}^{|1 + \tan (\theta/4)|} C(3 - r) \ r \ dr \ d\theta$
$ \small \approx 14.043 C \implies \small C \approx 0.0712 $
Then the desired probability is,
$\displaystyle \small \int_{8 \pi /3}^{11 \pi / 3} \int_{0}^{|1 + \tan (\theta/4)|} C(3 - r) \ r \ dr \ d\theta \approx 0.0551 $