A six-sided number cube is weighted so that the probabilities of throwing 2, 3, 4, 5, or 6 are equal, and the probability of throwing a 1 is twice the probability of throwing a 2. If the number cube is thrown twice, what is the probability that the sum of the numbers thrown will be 4?
I understand why the individual probabilities are 2/7 for 1, and 1/7 for 2 through 6, and I understand that the combinations are (1, 3), (3, 1) and (2, 2). My questions is why is (2, 2) only counted once? If (1, 3) and the reverse (3, 1) are considered, why is (2, 2) not considered twice, since the first and second dice could be (2, 2) and they could also switch? I'm guessing my misunderstanding is on a fundamental level, but I thought the answer would have been 6/49, but the correct answer is 5/49.
Thanks in advance.
I understand why the individual probabilities are 2/7 for 1, and 1/7 for 2 through 6, and I understand that the combinations are (1, 3), (3, 1) and (2, 2). My questions is why is (2, 2) only counted once? If (1, 3) and the reverse (3, 1) are considered, why is (2, 2) not considered twice, since the first and second dice could be (2, 2) and they could also switch? I'm guessing my misunderstanding is on a fundamental level, but I thought the answer would have been 6/49, but the correct answer is 5/49.
Thanks in advance.
