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I have a probability problem which i cannot solve yet. I want to the manuals says that the correct answers is:

$${9 \choose3}\times \left(\frac 56\right)^6 \times\left(\frac16\right)^4$$

I understand the $\left(\frac 56\right)^6 \times\left(\frac16\right)^4$ part. Assume that the probability of success is $\frac 16$ and the probability of failure is $\frac 56$.

What I don't understand is the use of the term $9\choose3$. The first term is to take all the possible variations in sequence in account.

$$\frac{n!}{k!*(n-k)!}$$

But you throw $10$ times, so shouldn't $n$ be $10$? The times of success is $4$, so why isn't $k = 4$?

In my opinion the solutions should be:

$${10\choose 4}\times \left(\frac 56\right)^6 \times\left(\frac16\right)^4$$

But this is not correct.

Could somebody give me feedback on what is wrong with my thinking?

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    $\begingroup$ The last throw is assumed to be a 4. Hence, you pick the remaining 3 throws out of the preceding 9 throws. $\endgroup$ Mar 6, 2020 at 12:32
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    $\begingroup$ It is the answer on the question: "what is the probability that exactly 10 throws are needed if it is your aim to throw 4 times a six?" I wouldn't say that your thinking is wrong but rather that the question is not well posed. $\endgroup$
    – drhab
    Mar 6, 2020 at 12:41
  • $\begingroup$ It is clear now! Thank you for your effort :) $\endgroup$
    – Tim
    Mar 6, 2020 at 12:43
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    $\begingroup$ Can you give type up the exact wording of the problem from wherever you got it? Based on the wording that you currently have, your thought process is right. $\endgroup$ Mar 6, 2020 at 14:09
  • $\begingroup$ Hi, the exact problem statement is: Imagine throwing a die until you have thrown 6 for the fourth time. calculate the chance that you will succeed in 10 times. I think the problem was my in the understanding of the problem. Thanks for your effort! Ter $\endgroup$
    – Tim
    Mar 7, 2020 at 15:17

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