Given Company ABC launching a product. If the product is successful, the company earns \$100 million in profit. In case of failure, the company loses \$200 million (i.e. negative profit). Probability of failure is $\frac{1}{10}$. The company can buy insurance for this product. Cost of insurance is \$30 million (paid before the launch). In case of failure the insurer will pay Company ABC \$200 million (thus compensating all the damages). Consider two cases: 1. Company ABC decided not to buy insurance. 2. Company ABC decided to buy insurance. Denote its profit in the first case by $X$ and in the second case by $Y$. (In the second case profit includes payments to/from the insurer taken with appropriate sign.) Find expected values and variances of $X$ and $Y$. Describe how buying of insurance affects the profit, its expected value and variance? Does buying insurance is cost-efficient in the long run? In which case Company ABC can decide to buy insurance and why?
Probability of success is $\frac{9}{10}$. Then computing expected value and variance for first case, I got the following:
$E(X)=\frac{9}{10}*\$100 - \frac{1}{10}*\$200=\$70$
$V(X)=E(X^{2})-E(X)^{2}=\$8100$
For Second Case, I computed expected value and variance as the following:
$E(Y)=1*-\$30 + \frac{9}{10}*\$100+\frac{1}{10}*(-\$200 + \$200)=\$60$
$V(Y)=E(Y^{2})-E(Y)^{2}=\$9900$
Comparing $X$ and $Y$, I am getting expected value of $X$ more than that of $Y$ and also variance of it is less than that of $Y$. Based on the above calculations, the conclusion is the first case $X$ is more profitable for Company ABC. But am not sure if my approach to the problem is correct.
Can you please let me know if the above approach to solve the given problem formulation is correct?