The question clearly states that one is six, which eliminates it from consideration. By declaring its outcome, its probability has resolved to 1/1. We are now left to calculate the probability of the OTHER die, which is a simple 1/6 chance.
It's the exact same problem as saying "You are wearing a blue shirt and rolled a die. What are the odds that die comes up six?"
Also, for the question the Livejournaler THOUGHT he was asking, I would reply that the correct answer is NONE of the above. ODDS are the ratio of correct to incorrect, whereas PROBABILITY is the ratio of correct to all possible outcomes. Therefore, if there are 11 ways to have at least one six out of two die, but only one outcome that has both be sixes, the ODDS are 1/10. (In my math classes, we'd different the two with odds like 1:10 vs probability like 1/11, though that's apparently not a hard rule.)
(no subject)
Date: 2011-08-18 01:20 pm (UTC)The question clearly states that one is six, which eliminates it from consideration. By declaring its outcome, its probability has resolved to 1/1. We are now left to calculate the probability of the OTHER die, which is a simple 1/6 chance.
It's the exact same problem as saying "You are wearing a blue shirt and rolled a die. What are the odds that die comes up six?"
Also, for the question the Livejournaler THOUGHT he was asking, I would reply that the correct answer is NONE of the above. ODDS are the ratio of correct to incorrect, whereas PROBABILITY is the ratio of correct to all possible outcomes. Therefore, if there are 11 ways to have at least one six out of two die, but only one outcome that has both be sixes, the ODDS are 1/10. (In my math classes, we'd different the two with odds like 1:10 vs probability like 1/11, though that's apparently not a hard rule.)