A company offers a group insurance with $b$ payable in the event of the travel accident. The question states that $q$ is the probability of death by travel accident for an individual. In a certain group of $100$ lives, the independence assumption fails because three specific individuals always travel together. If one dies in an accident, all three are assumed to die. Determine the Variance of this insurance.
For each insured，
$$V(X)=b^2q(1-q)\; .$$
My answer is to segregate the $100$ into two parts. For the $97$, the variance is $97 V(X)=97 b^2 q(1-q)$; For the specific three, this is a Bernoulli distribution with $\text{Pr}(Y=3b)=1-(1-q)^3$. By Bernoulli shortcut, $$V(Y)= (3b)^2 (1-q)^3 (1-(1-q)^3)\; .$$
Then
$$97 V(X)+V(Y)=97 b^2 q(1-q)+(3b)^2 (1-q)^3 (1-(1-q)^3) \; .$$
But the correct answer states that the variance of the $3$ is $V(3X)=3^2 b^2 q(1-q)$. That means the specific $3$ is a Bernoulli distribution with $\text{Pr}(Y=3b)=q$.
I don't know why the death probability of this $3$ is $q$, but not $1-(1-q)^3$.