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I'm learning probability, specifically discrete probability distributions, and need help with the following problem :

A bag contains ten balls numbered $1$ to $10$. We randomly select them until we find the ball with the number $7$.

$(1)$ What is the probability that the ball with the number $7$ is found for the first time in the $n$-th attempt, if every time we select a ball we replace it in the bag?

$(2)$ What is the probability that the ball with the number $7$ is found for the first time in the $n$-th attempt, if we draw without replacement?

My thoughts : the fact that we are looking first the first success makes me think that this problem is related to the geometric distribution, i.e. the distribution of the number of trials needed to get the first success in repeated Bernoulli trials.

So if we let $X$ represents the number of trials need to get the first $7$ then, in both cases, we are looking for $P(X = n)$. For $(1)$, the probability of success (drawing a $7$) is $p = 1/10$ and the probability of failure is $1-p = 9/10$. These probabilities stays constant from trial to trial since we are drawing with replacement. Therefore

$$P(X = n) = (1-p)^{n-1}p = \big(\frac{9}{10}\big)^{n-1}\big(\frac{1}{10}\big)$$

Is my work correct for $(1)$? For $(2)$ I have no idea. Does $(2)$ still follows the geometric distribution? Any help will be greatly appreciated.

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  • $\begingroup$ Yes (1) is correct; (2) is much easier though! $\endgroup$ – Lord Shark the Unknown Jun 18 '17 at 10:09
  • $\begingroup$ For $2$. note that every order of the $7$ is equally likely. $\endgroup$ – lulu Jun 18 '17 at 10:10
  • $\begingroup$ Your approach to (1) makes sense. For (2) the number of balls changes with each step. The number of balls when making the $k^{th}$ draw is $11-k$, you should be able to figure out the probability of a success or failure? Again you will want fails for$k=1\ldots n-1$ and a success for $k=n$. $\endgroup$ – Paul Aljabar Jun 18 '17 at 10:20
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Hint for (2): Let´s have a look on a example. You draw a 7 at the fourth time. The probability for that is

$$P(X=4)=\frac{9}{10}\cdot \frac{8}{9}\cdot \frac{7}{8}\cdot \frac{1}{7}$$

Now you can cancel $9,8,7$. What is the result ? This result can be generalized.

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  • $\begingroup$ Great hint, thanks! $\endgroup$ – user347616 Jun 18 '17 at 12:02
  • $\begingroup$ @Elix Your´re welcome. $\endgroup$ – callculus Jun 18 '17 at 12:08

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