MHB What is the Probability of Winning a Dice Game?

[pretzel1998]

Hi There!

This is my first time posting on this forums, so I hope I am following the right etiquette.

Im having some severe difficulties with the problems under Question 3 in this past exam paper. http://www.nzqa.govt.nz/nqfdocs/ncea-resource/exams/2014/91267-exm-2014.pdf
Can someone please check my answers and explain the questions where I faceplanted? Thanks

3. (a), I got 1/9 which I am pretty sure is good.
3. (b), I got 2/3 which I am pretty sure is good.
3. (c,i). I got 1/81, I am not very confident about this answer because the wording of the question is incredibly bad. I did 4/36 X 4/36 as that's the probability he gets 5 on the first roll, and then 5 on the second roll to win?
3. (c,ii). I got 13/1458, I don't think I got this right because it seems like such a crazy probability. I did the probability he got the sum of 5 on the first roll (4/36), multiplied by the probability that he did not get a sum of 5 on the second roll (13/18 I think? I might be wrong), the multiplied again by the probability that he will get a sum of 5. so in total (4/36 X 13/18 X 4/36 = 13/1458)? Really confused about this question.
3. (d) This question I had absolutely no idea what to do. I some how got 7/18, but I think I am wrong.

Thanks so much!

 

[Opalg]

Hi There!

This is my first time posting on this forums, so I hope I am following the right etiquette.

Im having some severe difficulties with the problems under Question 3 in this past exam paper. http://www.nzqa.govt.nz/nqfdocs/ncea-resource/exams/2014/91267-exm-2014.pdf
Can someone please check my answers and explain the questions where I faceplanted? Thanks

3. (a), I got 1/9 which I am pretty sure is good. I agree.
3. (b), I got 2/3 which I am pretty sure is good. I agree.
3. (c,i). I got 1/81, I am not very confident about this answer because the wording of the question is incredibly bad. I did 4/36 X 4/36 as that's the probability he gets 5 on the first roll, and then 5 on the second roll to win? I think that for part (c) you are meant to assume that he got a 5 on the first roll, so you only need to find the probability of scoring 5 on the second roll. If so, then the answer would be 1/9 rather than 1/81.
3. (c,ii). I got 13/1458, I don't think I got this right because it seems like such a crazy probability. I did the probability he got the sum of 5 on the first roll (4/36), multiplied by the probability that he did not get a sum of 5 on the second roll (13/18 I think? I might be wrong), the multiplied again by the probability that he will get a sum of 5. so in total (4/36 X 13/18 X 4/36 = 13/1458)? Really confused about this question. Here again, if you ignore the 1/9 that came from the first roll, you would get 13/162 rather than 13/1458, and I think that would be the answer they are looking for.
3. (d) This question I had absolutely no idea what to do. I some how got 7/18, but I think I am wrong.
On the first roll, M has a 2/9 probability of winning and W has a 1/9 probability. To make the game fair, the second roll must be arranged so that M's and W's overall probability of winning are both 1/2. The probability of a second roll being necessary is 2/3. If the second roll is arranged in such a way that M's probability of winning it is 15/36 (and W's probability of winning it is 21/36), then M's chances of winning the game are 2/9 (on the first roll) plus (2/3)x(15/36) (on the second roll). That comes to $\color{red}{\frac29 + \frac5{18} = \frac12}$, as required. So I think the answer should be 15/36.

Thanks so much!

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