I have a problem with exercise. Basically, I am not sure if I am right and how should it be. That's why if you can solve it so I will understand. I am reading this but I have difficulties and I am trying alone. The exercise says:

We throw continually one dice until we have succeeded for the first time a result below than 3.What is the possibility, to be more than 3 tries, until succeed the given result(the number to be smaller than 3)?

What I did I take this type from possibilities of Geometry: $f(x)=p(1-p)^{x-1}$ but I am not sure until now if this is the type or another. I believe this is it from what I have read

  • $\begingroup$ If I understand you, you are asking for the expended number of tosses you have to make until you get a result less than $3$, yes? Are you including that last toss or not? That is, if you toss, in succession, $5, 6, 5, 2$ is the answer (in that case) four or three? Either way, this is an example of a Geometric Distribution. $\endgroup$ – lulu Sep 11 at 10:24
  • $\begingroup$ Sentences start with capital letters. There is a space between each sentence. Sigh. $\endgroup$ – Klangen Sep 11 at 10:26
  • $\begingroup$ i have been confused little bit with your comments and answers $\endgroup$ – s.m Sep 11 at 10:34
  • $\begingroup$ What exactly don't you understand? If you have a question about a specific answer, please leave a comment below that answer. $\endgroup$ – Toby Mak Sep 11 at 10:36
  • $\begingroup$ @TobyMak i did it :) $\endgroup$ – s.m Sep 11 at 10:41

You require the score on each of the first 3 tries to be in {3,4,5,6}.

The probability is therefore $(\frac {4}{6})^3$.

  • $\begingroup$ The question says "What is the possibility, to be more than 3 tries, until succeed the given result(the number to be smaller than 3)" .Example it can be 8 times the dice i throw it i am not saying only the 3 times .I am not looking for the first 3 tries.I am looking for only the first try i got a number less than 3.When i got 1 the number or 2 the number then i will have success.It maybe be 8 trial .I hope you got me now.I think you counted the numbers of the dice 3,4,5,6 . $\endgroup$ – s.m Sep 11 at 10:41
  • $\begingroup$ No, you only care about the first 3 tries. Whatever happens after those 3 tries don't matter. For example, it would not matter if you roll a die 4 times or 8 times, since both of them have taken over 3 tries. $\endgroup$ – Kyky Sep 11 at 10:55
  • $\begingroup$ thanks a lot man ,so it isnt Geometric Distribution as i thought? $\endgroup$ – s.m Sep 11 at 10:56
  • $\begingroup$ Thanks @Kyky, I was trying to think of a way to explain that which wouldn't cause any confusion! $\endgroup$ – S. Dolan Sep 11 at 11:00
  • $\begingroup$ @S.Dolan man sorry for asking again but what you thought to solve this ?(i mean type ) etc .It isn't Geometric Distribution what it is i mean $\endgroup$ – s.m Sep 11 at 11:08

You want the probability that you don't succeed (in getting a number below $3$) in the first three trials. This means you want the first three outcomes to be $3,4,5$ or $6$. The answer is $(\frac 4 6)(\frac 4 6)(\frac 4 6)=(\frac 2 3)^{3}$

  • $\begingroup$ why should i want 3,4,5,6? and not until 10 for example? $\endgroup$ – s.m Sep 11 at 10:45

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