Thread: Probability and dice

Refer to dice that are loaded so that the numbers 2, 4, and 6 are equally likely to appear. 1, 3, and 5 are also equally likely to appear, but 1 is three times as likely as 2 is to appear.

#1)Two dice are rolled. What is the probability of getting a sum of 6 or doubles given that at least one die shows 2?
To roll 6: 5/36
To roll doubles: 6/36
To have at least one 2: 11/36
(5/36)/(11/36) = 5/11
(6/36)/ (11/36) = 6/11

It this correct?

#2)Two dice are rolled. What is the probability of getting a sum of 6 or a sum of 8 given that at least one die shows 2?
To roll 6: 5/36
To roll 8: 5/36
To have at least one 2: 11/36
(5/36)/(11/36) = 5/11
(5/36)/ (11/36) = 5/11

It this correct?
Thank you.

You have not used the funny dice.

p(2) = 1/9
p(6 total) = 14/81
p(8 total) = 11/81
p(doubles) = 5/27

how to use the funny dice?

Refer to dice that are loaded so that the numbers 2, 4, and 6 are equally likely to appear. 1, 3, and 5 are also equally likely to appear, but 1 is three times as likely as 2 is to appear.

2, 4, and 6 are equally likely. Give them one point each.

1, 3, and 5 are equally likely and twice as likely as 2, 4, and 6. Give them two points each.

That's a total of 1+2+1+2+1+2 = 9 points.

p(2)=p(4)=p(6) = 1/9
p(1)=p(3)=p(5) = 2/9

That is quite a bit different from a fair die, which produces p(1)=p(2)=p(3)=p(4)=p(5)=p(6)=1/6

1, 3, and 5 are equally likely and twice as likely as 2, 4, and 6. Why twice as likely as 2, 4, and 6?

Thanks