Calculating the Variance

[DoubtfulTurnip]

I don't think I've done this right, I used this equation [math]Variance = x∗σ^2(x) =n∗p(1−p)[/math]

 

[BigBeachBanana]

View attachment 31204

View attachment 31205
I don't think I've done this right, I used this equation [math]Variance = x∗σ2(x) =n∗p(1−p)[/math]

[math]Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y)[/math]Since rolling die are independent events, [imath]Cov(X,Y) =0[/imath]
This reduce to: [math]Var(X+Y)=Var(X) + Var(Y)[/math]You found [imath]Var(X)\medspace \& \medspace Var(Y)[/imath]. Your answer is correct.

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You can prove that if [imath]Z=X+Y[/imath], where [imath]X\sim Binom(n,p)[/imath] and [imath]Y\sim Binom(m,p)[/imath], then [imath]Z\sim Binom(n+m,p)[/imath]. It follows that
[math]Var(Z)=Var(X+Y)=(n+m)(p)(1-p)[/math]In your case:
[math]Var(X+Y)=(n+n)\left(\frac{1}{6}\right)\left(1-\frac{1}{6}\right)[/math]