Consider a game in which 6 true dice are rolled. Find the probability of obtaining: a)exactly 1 ace.?

Question

b) at least 1 ace. c) exactly 2 aces.

Answer 1

We can use binomial distribution because we only have two cases for each dice:1) (1/6) of chance to get ace; 2) (5/6) of chances to get no ace.

[(1/6) + (5/6)]^6

a) C(6,1)(1/6)(5/6)^5 = 0.40188

b) 1 - C(6,6)(5/6)^6 = 0.66510

c) C(6,2)(1/6)^2(5/6)^4 = 0.20094

Answer 2

You have 6 dice with 6 sides each, so the total number of possible rolls is 6 to the 6th power. There are 6 possible ways to get only one ace, so the probability is 6/6^6, or 1/6^5.

Answer 3

a)
P(exactly1 Ace) = 6(1/6)(5/6)^5 = 0.4

b)
P(at least 1 Ace) = (1 - (5/6)^6) = 0.665

c)
P(exactly 2 Aces) = 2*4(1/6)^2(5/6)^4 = 0.107

Answer 4

"0" as u don't have aces in True Dice..if u define any one of the sequence of the sides as the ace then its
a) 1/6
b) 1/6
c) 1/3

Answer 5

a) 6 * (1/6)(5/6)^6 = (5/6)^6 <- Helmut and sahsjing agree

b) 1 - (5/6)^6 <- Helmut and sahsjing agree

c) (6 * 5 / 2) * (1/6)^2(5/6)^4 <- sahsjing agrees

Answer 6

I think it sounds like the game YAHTZEE..... not sure about the "probability" part though... no... wait, yahtzee only has 5 die... sorry....

Answer 7

what is an ace in this context |?