Question

In a game of dice, you roll two standard six-sided dice and examine their sum.

a) What is the probability of rolling a 7? 1/6
b) If you roll the dice 20 times, what is the probability of rolling a total of 3 sums of 7?
c) How many times would you expect to roll the dice before you rolled a sum of 12? 36

Answer 1

a) 1/6

b) 0.2379

c) 36 times

Explanation:

a) Possible outcome of sum = 7 are: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) = 6/6^2 = 1/6

b) Let X be a binomial distribution representing the number of sums of 7 obtained

P(X = 3) = (20 nCr 3)(1/6)^3(5/6)^17 = 0.2379

c) The probability of rolling a 6 on a fair-sided die is 1/6. To get a sum of 12, we need two 6's. So, we multiply the probabilities: 1/6 x 1/6 = 1/36. Therefore, we should expect a sum of 12 about every 36 times we roll the dice.

Answer 2

Possible outcome of sum = 7 are: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) = 6/6^2 = 1/6

Let X be a binomial distribution representing the number of sums of 7 obtained
P(X = 3) = 20C3(1/6)^17(5/6)^3 = 3.9 x 10^-11

You will roll the dice 36 times to obtain a sum of 12.