Answer: E(Y) = 1.6 and Var(Y)=1.12
Step-by-step explanation:
Since we have given that
X 0 1 2
P(X) 0.4 0.4 0.2
Here, number of games = 2
So, 
Since 
are independent variables.
so, ![E[Y]=2E[X]\\\\Var[Y]=2Var[X]](https://tex.z-dn.net/?f=E%5BY%5D%3D2E%5BX%5D%5C%5C%5C%5CVar%5BY%5D%3D2Var%5BX%5D)
So, we get that
![E(X)=0.4\times 0+0.4\times 1+0.2\times 2=0.8\\\\and Var[x]=E[x^2]-(E[x])^2\\\\E[x^2]=0\times 0.4+1\times 0.4+4\times 0.2=1.2\\\\So, Var[x]=1.2-(0.8)^2\\\\Var[x]=1.2-0.64=0.56](https://tex.z-dn.net/?f=E%28X%29%3D0.4%5Ctimes%200%2B0.4%5Ctimes%201%2B0.2%5Ctimes%202%3D0.8%5C%5C%5C%5Cand%20Var%5Bx%5D%3DE%5Bx%5E2%5D-%28E%5Bx%5D%29%5E2%5C%5C%5C%5CE%5Bx%5E2%5D%3D0%5Ctimes%200.4%2B1%5Ctimes%200.4%2B4%5Ctimes%200.2%3D1.2%5C%5C%5C%5CSo%2C%20Var%5Bx%5D%3D1.2-%280.8%29%5E2%5C%5C%5C%5CVar%5Bx%5D%3D1.2-0.64%3D0.56)
So, E[y]=2×0.8=1.6
and Var[y]=2×0.56=1.12
Hence, E(Y) = 1.6 and Var(Y)=1.12
The mean is the average of the numbers, so if x = all 11 numbers added together, then you have the equation:
x / 11 = 7, so x = 77 (this is all of the numbers added together)
if the number 13 is removed, the equation becomes:
(77 - 13) / 10 = M (M is the new mean)
so the new mean is 6.4
n(A-B) denotes elements which are in A but not in B
n(Au B) denotes elements in A and B
n(AnB) denotes elements that are common in A and B
Now I will add one more set
n(B-A) which denotes elements in B but not in A
So, n(AuB) = n(A-B) + n( B-A) +n(AnB)
70 = 18 +n(B-A) + 25
70 = 43 + n(B-A)
n(B-A) = 70-43
n(B-A) = 27
So, n(B) = n( B-A) + n( AnB)
= 27+25
= 52
