Answer :
To determine the value of [tex]\( n \)[/tex] such that the probability of Marie picking a green ball on the [tex]\( n \)[/tex]-th pick is [tex]\(\frac{21}{220}\)[/tex], we need to carefully consider the setup of the problem and then solve for [tex]\( n \)[/tex] accordingly.
First, let's define the given situation:
1. Marie picks balls at random from a total of 12 balls.
2. She stops picking once she picks a green ball.
3. The probability that she picks a green ball on the [tex]\( n \)[/tex]-th pick is [tex]\(\frac{21}{220}\)[/tex].
Given that the provided probability for picking a green ball specifically on the [tex]\( n \)[/tex]-th pick, we need to understand the nature of this probability:
The way this is usually interpreted is that Marie will continue to pick non-green balls for the first [tex]\( n - 1 \)[/tex] picks and then pick a green ball on the [tex]\( n \)[/tex]-th pick.
To compute the exact probability of this sequence of events, we typically use a hypergeometric distribution. However, if we simplify the scenario and assume that we have this exact probability for this question, our goal is then to find [tex]\( n \)[/tex].
From the given probability equation for [tex]\( n \)[/tex]:
[tex]\[ P(\text{pick green ball on } n\text{-th pick}) = \frac{21}{220} \][/tex]
Solving for [tex]\( n \)[/tex]:
There is no straightforward algebraic formulation provided directly connecting the equation to [tex]\( n \)[/tex].
From the direct calculation, since the required statement results in [tex]\( \text{No solution} ([]) \)[/tex]:
This suggests that under the given constraints, there's no valid pick number [tex]\( n \)[/tex] that would result in the probability of [tex]\(\frac{21}{220}\)[/tex] for Marie to pick a green ball on the [tex]\( n \)[/tex]-th pick.
Therefore, based on the provided information, we conclude:
[tex]\[ \boxed{n \text{ does not exist for which the probability is } \frac{21}{220}} \][/tex]
This suggests that either the problem constraints or the probability provided is not possible with the designed picking method and total ball count.
First, let's define the given situation:
1. Marie picks balls at random from a total of 12 balls.
2. She stops picking once she picks a green ball.
3. The probability that she picks a green ball on the [tex]\( n \)[/tex]-th pick is [tex]\(\frac{21}{220}\)[/tex].
Given that the provided probability for picking a green ball specifically on the [tex]\( n \)[/tex]-th pick, we need to understand the nature of this probability:
The way this is usually interpreted is that Marie will continue to pick non-green balls for the first [tex]\( n - 1 \)[/tex] picks and then pick a green ball on the [tex]\( n \)[/tex]-th pick.
To compute the exact probability of this sequence of events, we typically use a hypergeometric distribution. However, if we simplify the scenario and assume that we have this exact probability for this question, our goal is then to find [tex]\( n \)[/tex].
From the given probability equation for [tex]\( n \)[/tex]:
[tex]\[ P(\text{pick green ball on } n\text{-th pick}) = \frac{21}{220} \][/tex]
Solving for [tex]\( n \)[/tex]:
There is no straightforward algebraic formulation provided directly connecting the equation to [tex]\( n \)[/tex].
From the direct calculation, since the required statement results in [tex]\( \text{No solution} ([]) \)[/tex]:
This suggests that under the given constraints, there's no valid pick number [tex]\( n \)[/tex] that would result in the probability of [tex]\(\frac{21}{220}\)[/tex] for Marie to pick a green ball on the [tex]\( n \)[/tex]-th pick.
Therefore, based on the provided information, we conclude:
[tex]\[ \boxed{n \text{ does not exist for which the probability is } \frac{21}{220}} \][/tex]
This suggests that either the problem constraints or the probability provided is not possible with the designed picking method and total ball count.
