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Eva8 [605]
3 years ago
8

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 rolle

d 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?
Mathematics
1 answer:
Lera25 [3.4K]3 years ago
5 0

Answer:

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:

Var-X_{1} =np(1-p)=3.33(1-(1/6))=2.775

The mean of X₂ is:

MeanX₂ = 10 * (16/52) = 3.076

The variance of X₂ is:

Var-X_{2} =3.076(1-(16/52))=2.13

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

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