Hi @cc4300 Welcome! Please note that if we copied (transferred) the three given joint probabilities to a 3* 3 matrix (G * B) then only the diagonals would have values; the other six cells would contain zeros; e.g. P[G=3, B=7] = 0%. I say that in order to emphasize that the unconditional (aka, marginal) probabilities are:
P(G=3) = 20%, P(G=9) = 60%, P(G=12) = 20%, and:
P(B=5) = 20%, P(B=7) = 60%, P(B=9) = 20%.
Variances are univariate: there is one for G and one for B. The exhibit utilizes the most helpful version: variance(X) = E(X^2) - [E(X)]^2; see https://en.wikipedia.org/wiki/Variance
E(G) = 3.0*20% + 9.0*60% + 12.0*20% = 8.40; see first column (G) under Expected Value
E(B) = 5.0*20% + 7.0*60% + 9.0*20% = 7.00; see second column (B) under Expected Value.
E(G^2) = (3.0^2) *20% + (9.0^2) *60% + (12.0^2)*20% = 79.20; see fourth column (G^2) under Expected Value
E(B^2) = (5.0^2)*20% + (7.0^2)*60% + (9.0^2)*20% = 50.60; see fifth column (B) under Expected Value.
Now we can calculate each random variable's variance:
σ^2(G) = E(G^2) - [E(G)]^2 = 79.20 - 8.40^2 = 8.64
σ^2(B) = E(B^2) - [E(B)]^2 = 50.60 - 7.0^2 = 1.60
Although many would probably prefer to get to the same place with:
σ^2(G) = 20%*(3.0 - 8.40)^2 + 60%*(9.0 - 8.40)^2 + 20%*(12.0-8.4)^2 = 8.64
σ^2(B) = 20%*(5.0 - 7.0)^2 + 60%*(7.0 - 7.0)^2 + 20%*(9.0-7.0)^2 = 1.60
... these are calculating variance as the expected squared difference of each observation from the mean.
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