Variance when changing from one random discrete variable to another

Question

There 16 people divided to 8 pairs, each pair does the same task and has $p=0.8$ chance of completing it successfully, independent of the results of the other pairs

1. find Var(X), where X is the number of pairs which completed the task successfully
2. find Var(Y), where Y is the number of people which completed the task successfully

For (1) this is a classic binomial random variable so $$Var(X)= np(1-p) = 8 * (0.8)(1-0.8) = 0.128$$

As for (2), instinctively, I want to say that $Var(Y) = 2 * Var(X)$ because the $n=16$ instead of $8$ and the probability that a person completes the task remains $0.8$. But on the other hand I thought that I can look at $Y$ as twice the number of pairs that completed the task so $$Var(Y)=Var(2\cdot X)=4\cdot Var(X) = 4 * 1.28 = 5.12$$ but I'm not sure which way is correct

Answer 1

Part of your problem is that you have not precisely defined the following:

1. For a given pair of people, if their task is completed successfully, are both people considered to have completed the task?
2. Is it possible for exactly one person in a given pair to complete the task successfully?

To illustrate, I will describe a scenario consistent with my interpretation of your question. Suppose the task is for each person to roll a fair ten-sided die numbered from $1$ to $10$ inclusive, and the task is successful if the sum of the numbers they roll is not divisible by $5$. For example, if the numbers rolled are $4$ and $2$, then they succeeded, but if they roll $10$ and $5$ they failed. You can check that the probability this occurs is exactly $p = 0.8$. Then they can only both succeed or both fail.

In this case, the number of people succeeding can only be among the set $Y \in \{0, 2, 4, 6, \ldots, 16\}$, whereas the number of pairs succeeding can be among the set $X \in \{0, 1, 2, \ldots, 8\}$. Moreover, $Y = 2X$ for all outcomes of $X$.

It is not true that $\operatorname{Var}[Y] = 2 \operatorname{Var}[X]$. This is wrong. For a scalar constant $c$, we have $$\operatorname{Var}[cX] = c^2 \operatorname{Var}[X],$$ and for $c = 2$, this means the variance of $Y$ is four times the variance of $X$. The flaw in your first line of reasoning is that $Y$ is not itself binomial on $\{0, 1, 2, \ldots, 16\}$, because it is not possible to observe an odd number of people succeeding. Therefore, $\operatorname{Var}[Y] \ne 16 p(1-p)$.