Suppose we roll a fair six-sided die 20 times and draw ten cards from a standard 52-card deck. Let X be the number of "6"s rolled plus the number of Jack, Queen, King, or Aces drawn (There are 16 such cards in the 52).
(a) Calculate the Expected value, Variance, and Standard deviation of X.
Hint: Let X1 be the number of "6"s rolled and X2 be the number of Jacks or better drawn. Then, X = X1 +X2, and X1 and X2 are independent.
(b) What is the probability that we roll at least five "6"'s and, at the same time, draw at least 4 Jacks, Queens, Kings, or Aces?

a) Expected value = 6.406

Variance = 4.905

Standard deviation = 2.45

b) The probability is 0.08547

Step-by-step explanation:

a) Let's suppose that:

X₁ = number of 6´s

X₂ = number of Jack, Queen, King or Aces

The mean of X₁ is:

MeanX₁ = n * p = 20 * (1/6) = 3.33

The variance of X₁ is:

The mean of X₂ is:

MeanX₂ = 10 * (16/52) = 3.076

The variance of X₂ is:

The expect value of X is:

Xexp = MeanX₁ + MeanX₂ = 3.33 + 3.076 = 6.406

The variance of X is:

VarX = VarX₁ + VarX₂ = 2.775 + 2.13 = 4.905

The standard deviation is:

Xdevi = 4.905/2 = 2.45

b) The probability of drawing at least five six out of 20 rolls is equal to:

∑(1/6)ˣ(5/6)²⁰⁻ˣ = 0.231 with x = 5

The probability of at least 4 Jack, Queen, Kings or Aces is:

∑(16/52)ˣ(1-(16/52))¹⁰⁻ˣ = 0.37 with x = 4

The probability of given event is equal to:

P = 0.231 * 0.37 = 0.08547

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1

4x and 6y

im not sure though