Conditional Probability

[mattgm8]

I am not sure how to work the following problem:

Two standard dice are rolled. What is the probability that a sum of 7 or 11 is rolled, given that the first roll is a 3?

I have the probablity of rolling a 3 as 2/36. A sum of 7 as 6/36 and an 11 as 2/36.
I can only figure an answer as (0.5 x 0.17)/ 0.05 = 0.17

Is this the proper way to approach this problem ??

Thanks

 

[Loren]

I don't think the question is clear. I understand the question to be that you roll 2 dice. Are you rolling them simultaneously or one at a time. If you are rolling simultaneously and getting a 3 on the first roll, it doesn't make any difference when you roll the second time. The dice don't remember the first roll. So the probability of getting a 7 or 11 on the second roll is 6 ways to get a 7 plus 2 ways to get an 11 for a probability of 8/36.
If you are rolling them one at a time and you get a 3 on the first roll, you can't get either a 7 or an 11 on the second roll because a die only has up to 6 dots on a side.

Maybe you can copy the original problem exactly as it appears, word for word, then show us your thinking.

 

[wjm11]

Two standard dice are rolled. What is the probability that a sum of 7 or 11 is rolled, given that the first roll is a 3?

I have the probablity of rolling a 3 as 2/36. A sum of 7 as 6/36 and an 11 as 2/36.

I interpreted this to mean one more die needed to be rolled, then added to the first die roll of 3.

First of all, forget about the probability of rolling a 3; the 3 is already “given.” The question simplifies to “What rolls of the second die will result in a sum of 7 or 11?”.

Obviously we cannot roll an 8 to reach the target of 11, so the only solution that will work is to roll a 4 to reach 7.

The probability of rolling a 4 is 1/6.

Make sense?

 

[mattgm8]

Thank you both for your input !!

The original question is posted just as I have it. I also thought the question wasn’t clear.
I thought of it both ways as well:
1) I took it as if I roll a 3 on the 1st roll (using both die) then I cannot produce a “sum” of 11 with the second roll. So only the probability of reaching a 7 works.
2)OR that the 3 doesn’t come into play. So the probability that I need to find is that of rolling a 7 or an 11.

 

[Denis]

Can't you simply ask your teacher to clarify the darn thing ? :shock: