Probability- rolling 10 and 11 before 7

[Roni1985]

Homework Statement

Two dice are rolled: what is the probability of rolling a 10 and an 11 before rolling a 7?

Homework Equations

The Attempt at a Solution

Is conditional probability the way to go here?

Let,
A= rolling 10
B= rolling 11
C= not rolling 7

P( A and B| C)=P(A and B and C)/P(C)

I'm getting 6/36 for the answer, but something is telling me it can't be this easy. I think I'm a little rusty :\

 

[Roni1985]

I think I have a lead, its an infinite sum... solving it right now...

 

[Roni1985]

All right,

This is the sum I'm getting:
inf
\sum nC2* \frac{3}{36}*\frac{2}{36}(\frac{30}{36})^(^n^-^2^)
n=2

Is this sum correct? Is it possible to calculate this sum without a calculator?

 

[Roni1985]

I just tried with a calculator and I'm getting 1 so I made a mistake somewhere :(

 

[Roni1985]

Sorry for another post, I think I found my mistake, it should be:
inf
<br /> \sum nC2* \frac{3}{36}*\frac{2}{36}*(\frac{30}{36})^(^n^-^3^)*(\frac{6}{36})<br />
n=3

inf
=<br /> \sum (n-1)*n/2* \frac{3}{36}*\frac{2}{36}*(\frac{30}{36})^(^n^-^3^)*(\frac{6}{36})<br />
n=3
It sums up to 43/216. But, I still can't find the answer without a calculator :\
And, I'm not sure if it's nC2 or nP2. I think the order does matter so it needs to be nP2... ?

 

[Oxymoron]

Can I just see if I understand the problem correctly? Are you asking for the probability that you roll one of these 3 pairs {(4,6),(5,5),(6,4)} and one of these 2 pairs {(5,6),(6,5)} before you roll any of these pairs {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}? Or is it or. The or case is easy which is why I was wondering.

Then if you roll any of the other 25 possible pairs (there are obviously 36 in total and the ones listed above are all mutually exclusive) the experiment continues?

 

[Oxymoron]

Because if you want the probability of a sum of 10 or a sum of 11 (as opposed to and) before a sum of 7 then you can work it out easily with infinite series:

First, define your random variable S to be the sum of the two dice. Then at each trial of the experiment you have

P(S=10) = \frac{3}{36}

P(S=11) = \frac{2}{36}

P(S=7) = \frac{6}{36}

P(\mbox{Experiment Continues}) = \frac{25}{36}

Then let p be the probability that we will roll a total of 10 or 11 before we roll a total of 7.

If we roll two dice the probability of success is 5/36, the probability of failure is 6/36 and else we begin experiment again with probability 25/36. And so

p = \frac{5}{36} + \frac{25}{36}\cdot\frac{5}{36} + \left(\frac{25}{36}\right)^2\frac{5}{36} + ...
\quad = \frac{5}{36}\left[1+\frac{25}{36}+\left(\frac{25}{36}\right)^2+...\right]
\quad = \frac{5}{36}\left(\frac{1}{1-\frac{25}{36}}\right)
\quad = \frac{5}{36}\frac{36}{25}
\quad = \frac{5}{11}

 

[Roni1985]

Can I just see if I understand the problem correctly? Are you asking for the probability that you roll one of these 3 pairs {(4,6),(5,5),(6,4)} and one of these 2 pairs {(5,6),(6,5)} before you roll any of these pairs {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}? Or is it or. The or case is easy which is why I was wondering.

Then if you roll any of the other 25 possible pairs (there are obviously 36 in total and the ones listed above are all mutually exclusive) the experiment continues?

Yes, its the "and"
Here are some scenarios:

10 11 7
11 10 10 5 4 11 7

meaning, you need to get 10 and 11, at least 1 time each, anywhere before you get 7.And, I don't think your answer is correct. I was able to solve it a different way and I got the answer, which is 17/132.

Thanks.

 

[Oxymoron]

Yeah okay, so it is 'and'. That makes it more difficult. I am also learning probability at the moment so I should have made that clear.

Do you think it might be possible to use a probability distrbution? I had a look at your infinite sum but it is not clear to me how you came to that expression. Could you perhaps explain it a little more?

 

[Robert1986]

What if you wanted compute the probability that a 10 is rolled before a 7? Then you would do something like this:
Let A be the event that a 10 is rolled.
Let B be the event that a 10 OR 7 is rolled.

Then, the probability that a 10 is rolled before a 7 is just the probability that a 10 has been rolled, given a sum of 7 or 10 has been rolled. That is, P(A|B).

It seems, and I might be wrong, that, here, you could do something like this:

Let A be the event that a sum of 10 AND a sum of 11 are rolled.
Let B be the event that (a sum of 10 AND a sum of 11 are rolled) or a sum of 7 is rolled. That is, you need to compute P(A|B).

 

[Oxymoron]

Let A be the event that a sum of 10 AND a sum of 11 are rolled.

You can't get a sum of 10 and a sum of 11 in a single event. So you would need at least 2 trials before you can get this. For example, if n is the number of trials then when n=1, P(A)=0. So you need to consider a sequence of trials which I think is where Roni1985 is getting his infinite series from.

 

[Robert1986]

You can't get a sum of 10 and a sum of 11 in a single event. So you would need at least 2 trials before you can get this. For example, if n is the number of trials then when n=1, P(A)=0. So you need to consider a sequence of trials which I think is where Roni1985 is getting his infinite series from.

Yes, that would be rather difficult to get a 10 and 11 on the same roll, wouldn't it? Ahh, pesky details :).

What about this,
A = event 10 is rolled
B = event 7 or 10 is rolled
C = event 11 is rolled
D = event 11 or 7 is rolled

Could you compute P(A|B)*P(C|D)? That is, the probability that a 10 has been rolled, given a 7 or 10 has been rolled, AND a 11 has been rolled, given an 11 or 7 has been rolled?