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From a past paper and mark scheme:

Q: Malik is playing a game in which he has to throw a 6 on a fair six-sided die to start the game. Find the probability that Malik needs at most ten attempts to throw a 6.

A: $1-(5/6)^{10}$

I do not understand the answer, where did it come from. I believe it is an application of the geometric distribution but I'm not sure. My guess would have been

$$\sum_{x=1}^{10}\text{Geo}(1/6)=\sum_{x=1}^{10}(1/6)(5/6)^{x-1}$$

but apparently the answer is a lot simpler. How would I obtain it?

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There's no need to invoke the geometric distribution (though you could); one can do it directly from simple probability calculations.

You should progress by answering these simpler questions:

  • What is the chance that Malik fails on the first throw?

  • Now what is the chance Malik fails for all ten throws in a row?

  • What is the chance that doesn't happen?

This "work out the probability of the complementary event" is a common strategy, since it can often be simpler to calculate than the event we're interested in.

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  • $\begingroup$ I'll mark this as accepted seeing as it follows the self-study guidelines and is a good answer, thanks :) $\endgroup$ – Tom Apr 4 '16 at 12:04
  • $\begingroup$ @Tom thanks. However, you should feel free to accept as you see fit; if you found the other answer more useful, you should not worry about un-accepting my answer (by clicking on the green "" mark) and accepting the other. (I will be happy either way.) $\endgroup$ – Glen_b Apr 4 '16 at 17:29
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So, if the Event is $E$ = "Getting a $6$ in at most $10$ attempts" = "Getting at least a $6$ in $10$ attempts".

The complementary event is $E^c$ = "Getting no $6$ in the first $10$ attempts".

And you well know that $P(E)= 1- P(E^c)$ .

$P(E^c) = \frac{5}{6} * \frac{5}{6} * ...* \frac{5}{6} $ 10 times.

$P(E^c) = (\frac{5}{6})^{10}$

$P(E) = 1 - (\frac{5}{6})^{10}$

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  • $\begingroup$ +1 It's a good answer, but please see the guidelines for answering self-study questions. You didn't really leave any part of their assigned work for the student to do themselves. $\endgroup$ – Glen_b Apr 4 '16 at 9:36
  • $\begingroup$ uh, you got me, sorry $\endgroup$ – Tommaso Guerrini Apr 4 '16 at 9:37
  • $\begingroup$ Thank you, the rewording of the question was insightful to me $\endgroup$ – Tom Apr 4 '16 at 12:02
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I always think of the geometric distribution when answering the question "how many trials until I see the first X?" For example, "how many coin tosses until I see the first Head, etc." So, for this problem, the intuition to reach for the geometric distribution would have made sense if the question was asking about how many rolls you needed until you got your first six, etc. If you want to know how many trials until you see N instances of X, you reach for the negative binomial distribution, which has the distinction of being neither negative nor binomial, ha.

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We are interested in the cumulative distribution function of a discrete geometric random variable. From the pmf of a geometric r.v, we have

$$ \text{Pr}(X=x) = p(1-p)^{x-1}$$ for $x \in \{1,2,3 \dots\}$

Therefore, consider

$$Pr(X\leq x) = p(1-p)^{x-1} + p(1-p)^{x-2} + \dots + p$$ $$Pr(X\leq x) = p(1+(1-p)+(1-p)^2 +\dots + (1-p)^{x-1})$$ By the geometric series, we also have $$Pr(X\leq x) = p\frac{1-(1-p)^x}{p} = 1-(1-p)^x$$

Now substitute $x=10$ into the result.

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