die and coin

Probabilities... Hmm... Strange question, but interesting nonetheless.

From 1 to 6 you get two multiples of 3: 3 and 6. You will need to get 3 and 6 for your first toss. Then it gets a bit more complicated (or at least that's what I understand from if any other number comes). If you got a 6 the first time, now you need a 3 and vice versa. The coin toss will be simpler, since there's a fixed 50% chance of getting tails.

So two cases, each with the same probability. Getting a 3 the first time is 1/6, while getting a 6 the second time is again 1/6. Getting the tails will be 1/2. The probability for this event will be 1/6 * 1/6 * 1/2 = 1/72. Use the same reasoning to conclude that getting a 6 the first time and a 3 the second time yields the same probability.

Add them up to get 1/72 + 1/72 = 2/72 = 1/36.

s it 1/12??

DanaJ wrote:

Probabilities... Hmm... Strange question, but interesting nonetheless.

From 1 to 6 you get two multiples of 3: 3 and 6. You will need to get 3 and 6 for your first toss. Then it gets a bit more complicated (or at least that's what I understand from if any other number comes). If you got a 6 the first time, now you need a 3 and vice versa. The coin toss will be simpler, since there's a fixed 50% chance of getting tails.

So two cases, each with the same probability. Getting a 3 the first time is 1/6, while getting a 6 the second time is again 1/6. Getting the tails will be 1/2. The probability for this event will be 1/6 * 1/6 * 1/2 = 1/72. Use the same reasoning to conclude that getting a 6 the first time and a 3 the second time yields the same probability.

Add them up to get 1/72 + 1/72 = 2/72 = 1/36.

Answer is 0.

I'd like to see the official explanation, if the answer is 0. Maybe I misunderstood the problem.

Is this a GMAT kind of question?