Computing the probability example
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I have been doing the following example, and there is something I do not understand. It goes like this:
From an urn with four white balls and six black balls, one ball is drawn repeatedly with replacement until a white ball is obtained. Let X be the number of drawings resulting in a black ball.
Compute the probability $P(Xgeq 3)$.
Now, according to the solutions, the way to solve this is to calculate the following: $sum_{k=3}^∞ ((frac{6}{10})^kcdotfrac{4}{10}) $
My question: Why cannot I solve it like this: $P(Xgeq 3)=1-(P(X=0)+P(X=1)+P(X=2))$?
Thank you
probability probability-theory probability-distributions
asked Nov 18 at 21:37
Relax295
1099
add a comment |
up vote
1
down vote
favorite
I have been doing the following example, and there is something I do not understand. It goes like this:
From an urn with four white balls and six black balls, one ball is drawn repeatedly with replacement until a white ball is obtained. Let X be the number of drawings resulting in a black ball.
Compute the probability $P(Xgeq 3)$.
Now, according to the solutions, the way to solve this is to calculate the following: $sum_{k=3}^∞ ((frac{6}{10})^kcdotfrac{4}{10}) $
My question: Why cannot I solve it like this: $P(Xgeq 3)=1-(P(X=0)+P(X=1)+P(X=2))$?
Thank you
probability probability-theory probability-distributions
asked Nov 18 at 21:37
Relax295
1099
3
I much prefer your solution! I'd even go a bit further, and say that the answer is the probability of getting $BBB$ as your first three. Makes no difference what happens after that.
– lulu
Nov 18 at 21:38
3
As an editorial note: people often like the Geometric Series method because it's something you can just memorize and work through. Always better to think about the problem and then try to get a simpler solution.
– lulu
Nov 18 at 21:39
They are equivalent. It helps to know the general form though. For example, if you had to calculate $P(X geq 10)$, your method could feel a bit tedious.
– Aditya Dua
Nov 20 at 7:18
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have been doing the following example, and there is something I do not understand. It goes like this:
From an urn with four white balls and six black balls, one ball is drawn repeatedly with replacement until a white ball is obtained. Let X be the number of drawings resulting in a black ball.
Compute the probability $P(Xgeq 3)$.
Now, according to the solutions, the way to solve this is to calculate the following: $sum_{k=3}^∞ ((frac{6}{10})^kcdotfrac{4}{10}) $
My question: Why cannot I solve it like this: $P(Xgeq 3)=1-(P(X=0)+P(X=1)+P(X=2))$?
Thank you
probability probability-theory probability-distributions
asked Nov 18 at 21:37
Relax295
1099
I have been doing the following example, and there is something I do not understand. It goes like this:
From an urn with four white balls and six black balls, one ball is drawn repeatedly with replacement until a white ball is obtained. Let X be the number of drawings resulting in a black ball.
Compute the probability $P(Xgeq 3)$.
Now, according to the solutions, the way to solve this is to calculate the following: $sum_{k=3}^∞ ((frac{6}{10})^kcdotfrac{4}{10}) $
My question: Why cannot I solve it like this: $P(Xgeq 3)=1-(P(X=0)+P(X=1)+P(X=2))$?
Thank you
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Nov 18 at 21:37
Relax295
1099
asked Nov 18 at 21:37
Relax295
1099
asked Nov 18 at 21:37
Relax295
1099
asked Nov 18 at 21:37
Relax295
1099
asked Nov 18 at 21:37
Relax295
1099
1099
3
I much prefer your solution! I'd even go a bit further, and say that the answer is the probability of getting $BBB$ as your first three. Makes no difference what happens after that.
– lulu
Nov 18 at 21:38
3
As an editorial note: people often like the Geometric Series method because it's something you can just memorize and work through. Always better to think about the problem and then try to get a simpler solution.
– lulu
Nov 18 at 21:39
They are equivalent. It helps to know the general form though. For example, if you had to calculate $P(X geq 10)$, your method could feel a bit tedious.
– Aditya Dua
Nov 20 at 7:18
add a comment |
3
I much prefer your solution! I'd even go a bit further, and say that the answer is the probability of getting $BBB$ as your first three. Makes no difference what happens after that.
– lulu
Nov 18 at 21:38
3
As an editorial note: people often like the Geometric Series method because it's something you can just memorize and work through. Always better to think about the problem and then try to get a simpler solution.
– lulu
Nov 18 at 21:39
They are equivalent. It helps to know the general form though. For example, if you had to calculate $P(X geq 10)$, your method could feel a bit tedious.
– Aditya Dua
Nov 20 at 7:18
3
3
I much prefer your solution! I'd even go a bit further, and say that the answer is the probability of getting $BBB$ as your first three. Makes no difference what happens after that.
– lulu
Nov 18 at 21:38
I much prefer your solution! I'd even go a bit further, and say that the answer is the probability of getting $BBB$ as your first three. Makes no difference what happens after that.
– lulu
Nov 18 at 21:38
3
3
As an editorial note: people often like the Geometric Series method because it's something you can just memorize and work through. Always better to think about the problem and then try to get a simpler solution.
– lulu
Nov 18 at 21:39
As an editorial note: people often like the Geometric Series method because it's something you can just memorize and work through. Always better to think about the problem and then try to get a simpler solution.
– lulu
Nov 18 at 21:39
They are equivalent. It helps to know the general form though. For example, if you had to calculate $P(X geq 10)$, your method could feel a bit tedious.
– Aditya Dua
Nov 20 at 7:18
They are equivalent. It helps to know the general form though. For example, if you had to calculate $P(X geq 10)$, your method could feel a bit tedious.
– Aditya Dua
Nov 20 at 7:18
add a comment |
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I much prefer your solution! I'd even go a bit further, and say that the answer is the probability of getting $BBB$ as your first three. Makes no difference what happens after that.
– lulu
Nov 18 at 21:38
3
As an editorial note: people often like the Geometric Series method because it's something you can just memorize and work through. Always better to think about the problem and then try to get a simpler solution.
– lulu
Nov 18 at 21:39
They are equivalent. It helps to know the general form though. For example, if you had to calculate $P(X geq 10)$, your method could feel a bit tedious.
– Aditya Dua
Nov 20 at 7:18