Given the information below $$ \begin{array}{|c|c|c|c|c|} \hline \textbf{Group} & \textbf{A} & \textbf{B} & \textbf{C}& \textbf{Total} \\ \hline \text{Age }26-35 & 60 & 40 & 50&150 \\ \text{Age }36-45 & 80 & 50 & 70&200 \\ \text{Age }46-55 & 40 & 50 & 60&150\\ \hline \textbf{Total}& 180 & 140 & 180&500\\ \hline \end{array} $$ where
A = Number of people who are having Health Insurance
B = Number of people who are having Life Insurance
C = Number of people who are having Critical Illness Insurance
Questions:
- If 2 people simultaneously and randomly selected, what is the chance that 1 person having Health Insurance and 1 person having Life Insurance from group age $36-45$?
- If 4 people simultaneously and randomly selected, what is the chance that 1 person having Health Insurance, 1 person having Life Insurance, and 2 people having Critical Illness Insurance from group age $26-35$?
For the first question, we obtain the probability of 1 person who is having Health Insurance $\frac{80}{500}=0.16$ and the probability of 1 person who is having Life Insurance $\frac{50}{500}=0.10$, so if 2 people simultaneously and randomly selected, the chance that 1 person having Health Insurance and 1 person having Life Insurance from group age $36-45$ is $0.16\times0.10=0.016$. But I'm not quite sure. Is this correct?
For the second question, I don't have any idea on how to approach the problem. My guts tell me it involves combination or permutation, but again I'm not so sure. Help is needed. Thanks.
Assuming independent lives, $$\begin{align}\mathbb P\left(\{A\cap(36-45)\}\cap\{B\cap (36-45)\}\right)\end{align}\\=\mathbb P\left(\{A\cap(36-45)\}\right)\cdot\mathbb P\left(\{B\cap(36-45)\}\right)$$
Both calculations are entirely analogous : $$\mathbb P\left(A,(36-45)\right)=\mathbb P\left(A\:|\:(36-45)\right)\cdot\mathbb P\left((36-45)\right)=\frac{80}{200}\cdot\frac{200}{500}=0.16\\\mathbb P\left(B,(36-45)\right)=\mathbb P\left(B\:|\:(36-45)\right)\cdot\mathbb P\left((36-45)\right)=\frac{50}{200}\cdot\frac{200}{500}=0.10$$
So your first answer is good.
For the second part, use the same process for indivual lives and for the two insured on a Critical Illness Insurance we have $$\mathbb P(C,(36-45))^2.$$