An insurance company examines its pool of auto insurance customers and gathers the following information:
All customers insure at least one car
$64\%$ of the customers insure more than one car.
$20\%$ of the customers insure a sports car.
Of those customers who insure more than one car, $15\%$ insure a sports car.
What is the probability that a randomly selected customer insures exactly one car and that car is not a sports car?
My attempt: Let $S$ denote the event that a policyholder insures a sports car. Let $X$ denote the number of cars insured. We need to find $P(\{X = 1\}\cap S')$. Note that $$P(\{X = 1\}\cap S') = P(\{X = 1\}| S')\cdot P(S') = [1-P(\{X = 1\}| S)]\cdot (1-0.2) = [1-P(\{X = 1\}| S)]\cdot 0.8$$
Also, $0.2 = [P(S|X=1) \cdot 0.36] + [0.64 \cdot 0.15] \implies P(S|X=1) \approx 0.289$
Using Bayes' Theorem, $P(\{X = 1\}| S) = \dfrac{P(S|X=1) \cdot P(X=1)}{P(S|X=1) \cdot P(X=1) + P(S|X>1) \cdot P(X>1)} = \dfrac{0.289 \cdot 0.36}{(0.289 \cdot 0.36) + (0.15 \cdot 0.64)} \approx 0.52$.
Therefore, $P(\{X = 1\}\cap S') = [1-0.52] \cdot 0.8 \approx 0.384$, which is not correct. Can someone please let me know where I went wrong? Thanks!
$A=$"insures more than a car"
$B=$"insures a sport car"
We are asked to get $$P(\bar A \cap \bar B)=P(\overline{A\cup B})=1-P(A\cup B)$$
We obtain $$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0,64+0,20-P(A\cap B)$$
We need $$P(B/A)=\dfrac{P(A\cap B)}{P(A)}\implies P(A\cap B)=P(B/A)P(A)=0,15\cdot 0,64 =0,096$$
So, finally
$$P(\bar A \cap \bar B)=1-P(A\cup B)=1-(0,64+0,20-P(A\cap B))=1-(0,64+0,20-0,096).$$
That is
$$P(\bar A \cap \bar B)=1-(0,64+0,20-0,096)=0,256.$$