OK I have answered these questions but unsure as to whether or not they are done correctly.
We roll two fair 6-sided dice. Each one of the 36 possible outcomes is assumed to be equally likely.
a) Find the probability that doubles were rolled
b) Given that the roll resulted in a sum of 3 or less, find the conditional probability that doubles were rolled
c) Find the probability that at least one die is 3.
d) Given that the two dice land on different numbers, find the conditional probability that at least one die is a three.
Right so for part (a) i draw a two way table to get the answer 1/6
part (b) i do
A = doubles = 1/6
B = (sum <= 3) = (4/36) = (1/9)
P(A | B) = (P(A∩B))/P(B)
part (c) i do (1/36 * 5/36) + (1/36 * 5/36) + (1/36 * 5/36) + (1/36 * 5/36) + (1/36 * 5/36)
= 25/1296 (seems a bit odd)
part (d) i'm not entirely sure about yet, gonna give it a think but need assurance part c is right which i doubt
I differ on some answers
part b: possibilities are: (1,1), (2,1) and (1,2) - 3 cases total. Of these, 1 is a double. So the conditional probability is 1/3.
part c: There are 6+6-1=11 cases where a 3 is rolled. So the probability is 11/36.
part d: Of the 30 cases where doubles aren't rolled, 10 have a 3 in them. So the conditional probability is 10/30.