The (hypothetical) game of Blorple is played with a set of 12 specialty dice. Th

Question

The (hypothetical) game of Blorple is played with a set of 12 specialty dice. The dice are fair, six-sided dice, but instead of numbered faces they have colored faces. Each die has a green side, a blue side, a red side, a purple side, and two white sides. Let G,B, R, P. W be the number of green, blue, red, purple, and white sides which come up when all 12 dice are rolled.

a) Compute P(G =2, B =3, R = 2, P = 1, W =4).

b) Compute P (G =3, B =2, R = 1, P =3, W = 4).

c) Compute E [W], the expected number of white dice rolled.

d) Compute P(G + B + R = 8).

e) Compute P (G = 4W = 0) ..

Explanation / Answer

(a) P(G =2, B =3, R = 2, P = 1, W =4)

= 12_C_2 * (1/6)^2 * 10_C_3 * (1/6)^3 * 7_C_2 * (1/6)^2 * 5_C_1 * (1/6)^1 * 4_C_4 * (2/6)^4

16 * (1/6)^12 * 66 * 120 * 21 * 5 * 1

= .0061125

(b) P (G =3, B =2, R = 1, P =3, W = 4) = 0

since we only have 12 dice and these numbers add to 13

(c) Note that we can treat the number of White dice that comes up as having a binomial distribution with n = 12 and p = 2/6

So the mean of this distribution is

E[W] = np = 12 (2/6) = 4.

(d) Note that if G + B + R = 8, then P + W = 4.

So P(G + B + R = 8) = P (P + W = 4)

= P(P = 4, W = 0) + P(P=3, W = 1) + P(P=2, W=2) + P(P=1, W = 4) + P(P=0, W=1)

= 12_C_4 (1/6)^4 * (3/6)^8 + 12_C_3 * 9_C_1 (1/6)^3 * (2/6)^1 * (3/6)^8 + 12_C_2 * 9_C_2 (1/6)^2 * (2/6)^2 * (3/6)^8 +

12_C_1 * 9_C_3 (1/6)^1 * (2/6)^3 * (3/6)^8 + 9_C_4 * (2/6)^4 * (3/6)^8

= .07246

(e) If we are given that W = 0, then it is like we're rolling 4 sided dice. We can treat the number of green dice as binomial with n = 12 and p = 1/4.

P(G = 4) = 12_C_4 * (1/4)^4 * (3/4)^8 = .1936

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