Three trucks carrying a total of 180 boxes of rubble arrive at a factory. The trucks carry, respectively, 51, 69, and 60 boxes of rubble.
(a) One of the 180 boxes selected uniformly at random. Let X denote the number of boxes of rubble that were on the truck carrying the randomly selected box. Find Var(X) for X as given in the problem
(b) One of the 3 truck drivers is also selected uniformly at random. Let Y denote the number of boxes of rubble on the chosen driver's truck. Find Var(Y) for Y as given in the problem
I know that the equation for variance is E(X)^2 - E(X)^2, and then the probability of picking in each trucks is 51/180, 69/180, and 60/180; for trucks, I would say there is 1/3 chance to pick 1 driver.
I don't know how to proceed. Please help!
Outline: You are on the right track with probabilities in the distributions of $X$ and $Y.$
(a) Random variable $X$ takes values $x_1=51, x_2=69,$ and $x_3=60$ with probabilities $p(x_1)=51/180,\;$ $p(x_2)=69/180,$ and $p(x_3)= 60/180,$ respectively.
(b) Random variable $Y$ takes values $51, 69,$ and $60$ with probabilities $1/3, 1/3,$ and $1/3,$ respectively.
Means and variances of random variables: If random variable $W$ takes values $w_1, w_2,$ and $w_3$ with probabilities $p(w_1),\; p(w_2),$ and $P(w_3),$ respectively, then $$\mu_{{}_W} = E(W) = \sum_{i=1}^3 w_ip(w_i),$$ and $$\sigma^2_W = Var(W) = \sum_{i=1}^3 (w_i -\mu_{{}_W})^2p(w_i)\\ = E(W^2) -\mu_{{}_W}^2 = \sum_{i=1}^3 w_i^2p(w_i) - \mu_{{}_W}^2.$$
You can use either the first or the second line in the last displayed equation for $Var(W),$ whichever you find most convenient. Simply plug in the values.