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I have seen a question from someone's old exam. Two dice, one has 6 sides and the other 4 sides. Question:

  • a) What is the probability that the possible outcome(sum of two dice) would be at least 9?

  • b) How many times you have to roll those two dice so that the outcome of getting 9 would be more than 95% probability?

I have learned quantile, possibility, normal distribution and binormal distribution and so on, still I feel it's hard to recall from my memory the knowledge I need to solve the question.

For question a, I think it might have three possible outcomes to get at least 9:

six sides: 5 and 6
four sides: 3 and 4
so 5+4; 6+3; 6+4  so  3 outcomes / 24 alternatives (4x6)==>1/8=12.5% ?

For question b, do I need z-score or t-score, etc?

BTW, the original question was not in English and I am not sure if you could fully understand my question. If you believe I did not write in a correct way, please feel free to modify it. Thank you!

BruceET
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almo
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    A general answer to this kind of question (a) about dice appears at https://stats.stackexchange.com/questions/116792/dungeons-dragons-attack-hit-probability-success-percentage/116913#116913. For (b), a good search term is "hypergeometric distribution." – whuber Feb 14 '21 at 12:30
  • Thank you @whuber, I will check it out – almo Feb 14 '21 at 13:00
  • :D Thanks @BruceET, I correct it now – almo Feb 14 '21 at 15:28
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    Enumeration seems OK for this simple case. Alternatively, simulation in R, code `sum = replicate(10^6, sample(1:6, 1)+sample(1:4, 1)); mean(sum>=9)` returns $0.124988,$ which should be good to three places, so your answer $1/8$ to (a) seems fine. Then maybe geometric dist'n for (b) // BTW in English `mouse : mice :: die : dice`. "Dices" is verb for chopping things into small pieces. – BruceET Feb 14 '21 at 15:33
  • Thanks @BruceET for your help:) – almo Feb 14 '21 at 15:41

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