>>11211467For the first problem:
Var[X] = E[(X-E[X])^2] = E[X^2] - 2E[X]^2 + E[X]^2 = E[X^2] - E[X]^2
47 = E[(X-1)^2] = E[X^2] - 2E[X] + 1 = 61 - 2E[X] + 1
E[X] = 15/2
Var[X] = 61 - (15/2)^2 = 4.75
sqrt(Var[X]) = sqrt(4.75) = 2.17944947177
First answer is A.
For the second problem:
Ignore the spades (but don't relax).
There are 2 diamonds and 5 non-diamonds
There are 2*(2C2) = 2 ways to get 2 diamonds having probability 1/42 each.
There are 2*(2C1)*(5C1) = 20 ways to get 1 diamond with probability 1/42 each.
There are 2*(5C2) = 20 ways to get 0 diamonds with probability 1/42 each.
The average number of diamonds is 2*2/42 + 20*1/42 + 20*0/42 = 0.57142857142 = 4/7.
The variance is 2*((2-4/7)^2)/42 + 20*((1-4/7)^2)/42 + 20*((0-4/7)^2)/42 = 0.34013605442
The second answer is D
