Probability remains same Why?
$begingroup$
A box has $m$ black and $n$ white balls. A ball is drawn at random and put back with $k$ additional balls of the same colour as that of the drawn ball. Now a ball is drawn again. Find the probability that it is a white ball.
I got the answer $frac{n}{m+n}$.
I just want to understand why it's not dependent on $k$.
probability probability-theory conditional-probability elementary-probability
edited Dec 19 '18 at 17:01
jgon
15.7k32143
asked Dec 19 '18 at 13:45
Maths_RocksMaths_Rocks
30417
$endgroup$
|
show 2 more comments
$begingroup$
A box has $m$ black and $n$ white balls. A ball is drawn at random and put back with $k$ additional balls of the same colour as that of the drawn ball. Now a ball is drawn again. Find the probability that it is a white ball.
I got the answer $frac{n}{m+n}$.
I just want to understand why it's not dependent on $k$.
probability probability-theory conditional-probability elementary-probability
edited Dec 19 '18 at 17:01
jgon
15.7k32143
asked Dec 19 '18 at 13:45
Maths_RocksMaths_Rocks
30417
$endgroup$
$begingroup$
This problem with different numbers was posted a few minutes ago and has already some answers. It is important that you attempt to solve this though. From your tag I assume you already know that is solved using conditional probability.
$endgroup$
– Test123
Dec 19 '18 at 13:49
1
$begingroup$
Possible duplicate of probability - getting a red ball in thenext draw is
$endgroup$
– Test123
Dec 19 '18 at 13:50
$begingroup$
Does the bucket now contain $n+m+K$ balls in total?
$endgroup$
– user334732
Dec 19 '18 at 16:53
1
$begingroup$
Strongly disagree with those voting to close as duplicates. The OP here knows how to do the problem, and is instead asking about why there's no dependence on $k$, which isn't addressed by any of the answers in the supposed duplicate.
$endgroup$
– jgon
Dec 19 '18 at 16:58
$begingroup$
I edited your question for formatting. You were inconsistent with the capitalization of $k$, so I chose lower case $k$ to make it consistent. Feel free to change that if you intended $k$ to be capital.
$endgroup$
– jgon
Dec 19 '18 at 17:02
|
show 2 more comments
$begingroup$
A box has $m$ black and $n$ white balls. A ball is drawn at random and put back with $k$ additional balls of the same colour as that of the drawn ball. Now a ball is drawn again. Find the probability that it is a white ball.
I got the answer $frac{n}{m+n}$.
I just want to understand why it's not dependent on $k$.
probability probability-theory conditional-probability elementary-probability
edited Dec 19 '18 at 17:01
jgon
15.7k32143
asked Dec 19 '18 at 13:45
Maths_RocksMaths_Rocks
30417
$endgroup$
A box has $m$ black and $n$ white balls. A ball is drawn at random and put back with $k$ additional balls of the same colour as that of the drawn ball. Now a ball is drawn again. Find the probability that it is a white ball.
I got the answer $frac{n}{m+n}$.
I just want to understand why it's not dependent on $k$.
probability probability-theory conditional-probability elementary-probability
probability probability-theory conditional-probability elementary-probability
edited Dec 19 '18 at 17:01
jgon
15.7k32143
asked Dec 19 '18 at 13:45
Maths_RocksMaths_Rocks
30417
edited Dec 19 '18 at 17:01
jgon
15.7k32143
asked Dec 19 '18 at 13:45
Maths_RocksMaths_Rocks
30417
edited Dec 19 '18 at 17:01
jgon
15.7k32143
edited Dec 19 '18 at 17:01
jgon
15.7k32143
edited Dec 19 '18 at 17:01
jgon
15.7k32143
15.7k32143
asked Dec 19 '18 at 13:45
Maths_RocksMaths_Rocks
30417
asked Dec 19 '18 at 13:45
Maths_RocksMaths_Rocks
30417
asked Dec 19 '18 at 13:45
Maths_RocksMaths_Rocks
30417
30417
$begingroup$
This problem with different numbers was posted a few minutes ago and has already some answers. It is important that you attempt to solve this though. From your tag I assume you already know that is solved using conditional probability.
$endgroup$
– Test123
Dec 19 '18 at 13:49
1
$begingroup$
Possible duplicate of probability - getting a red ball in thenext draw is
$endgroup$
– Test123
Dec 19 '18 at 13:50
$begingroup$
Does the bucket now contain $n+m+K$ balls in total?
$endgroup$
– user334732
Dec 19 '18 at 16:53
1
$begingroup$
Strongly disagree with those voting to close as duplicates. The OP here knows how to do the problem, and is instead asking about why there's no dependence on $k$, which isn't addressed by any of the answers in the supposed duplicate.
$endgroup$
– jgon
Dec 19 '18 at 16:58
$begingroup$
I edited your question for formatting. You were inconsistent with the capitalization of $k$, so I chose lower case $k$ to make it consistent. Feel free to change that if you intended $k$ to be capital.
$endgroup$
– jgon
Dec 19 '18 at 17:02
|
show 2 more comments
$begingroup$
This problem with different numbers was posted a few minutes ago and has already some answers. It is important that you attempt to solve this though. From your tag I assume you already know that is solved using conditional probability.
$endgroup$
– Test123
Dec 19 '18 at 13:49
1
$begingroup$
Possible duplicate of probability - getting a red ball in thenext draw is
$endgroup$
– Test123
Dec 19 '18 at 13:50
$begingroup$
Does the bucket now contain $n+m+K$ balls in total?
$endgroup$
– user334732
Dec 19 '18 at 16:53
1
$begingroup$
Strongly disagree with those voting to close as duplicates. The OP here knows how to do the problem, and is instead asking about why there's no dependence on $k$, which isn't addressed by any of the answers in the supposed duplicate.
$endgroup$
– jgon
Dec 19 '18 at 16:58
$begingroup$
I edited your question for formatting. You were inconsistent with the capitalization of $k$, so I chose lower case $k$ to make it consistent. Feel free to change that if you intended $k$ to be capital.
$endgroup$
– jgon
Dec 19 '18 at 17:02
$begingroup$
This problem with different numbers was posted a few minutes ago and has already some answers. It is important that you attempt to solve this though. From your tag I assume you already know that is solved using conditional probability.
$endgroup$
– Test123
Dec 19 '18 at 13:49
$begingroup$
This problem with different numbers was posted a few minutes ago and has already some answers. It is important that you attempt to solve this though. From your tag I assume you already know that is solved using conditional probability.
$endgroup$
– Test123
Dec 19 '18 at 13:49
1
1
$begingroup$
Possible duplicate of probability - getting a red ball in thenext draw is
$endgroup$
– Test123
Dec 19 '18 at 13:50
$begingroup$
Possible duplicate of probability - getting a red ball in thenext draw is
$endgroup$
– Test123
Dec 19 '18 at 13:50
$begingroup$
Does the bucket now contain $n+m+K$ balls in total?
$endgroup$
– user334732
Dec 19 '18 at 16:53
$begingroup$
Does the bucket now contain $n+m+K$ balls in total?
$endgroup$
– user334732
Dec 19 '18 at 16:53
1
1
$begingroup$
Strongly disagree with those voting to close as duplicates. The OP here knows how to do the problem, and is instead asking about why there's no dependence on $k$, which isn't addressed by any of the answers in the supposed duplicate.
$endgroup$
– jgon
Dec 19 '18 at 16:58
$begingroup$
Strongly disagree with those voting to close as duplicates. The OP here knows how to do the problem, and is instead asking about why there's no dependence on $k$, which isn't addressed by any of the answers in the supposed duplicate.
$endgroup$
– jgon
Dec 19 '18 at 16:58
$begingroup$
I edited your question for formatting. You were inconsistent with the capitalization of $k$, so I chose lower case $k$ to make it consistent. Feel free to change that if you intended $k$ to be capital.
$endgroup$
– jgon
Dec 19 '18 at 17:02
$begingroup$
I edited your question for formatting. You were inconsistent with the capitalization of $k$, so I chose lower case $k$ to make it consistent. Feel free to change that if you intended $k$ to be capital.
$endgroup$
– jgon
Dec 19 '18 at 17:02
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Look at this probability from the other side. Consider two disjoint events.
Event $B_1$ means that the second ball was in the urn from the very beginning. Event $B_2$ means that the second ball is taken out of those balls that were added to the urn. Event $A$ means that the second ball is white. Consider conditional probabilities $mathbb P(Amid B_1)$ and $mathbb P(Amid B_2)$:
$$mathbb P(Amid B_1) = frac{n}{n+m}$$
since no ball added is taken into account when $B_1$ happens.
And also
$$mathbb P(Amid B_2) = frac{n}{n+m}.$$
This is the probability of the second ball to be white if it is selected from $k$ balls added. In other words, this is the probability that $k$ white balls are added into the urn. This is exactly the probability that the first ball was white.
Since conditional probabilities of $A$ under both $B_1$ and $B_2$ are the same, the probabilities of $B_1$ and $B_2$ are irrelevant and
$$mathbb P(A) = mathbb P(B_1)frac{n}{n+m}+mathbb P(B_2)frac{n}{n+m}=frac{n}{n+m}.$$
Note that only $mathbb P(B_1)$ and $mathbb P(B_2)$ depend on $k$.
answered Dec 19 '18 at 16:43
NChNCh
6,8663825
$endgroup$
$begingroup$
Very nice! (+1)
$endgroup$
– jgon
Dec 19 '18 at 16:59
$begingroup$
Thanx Nch for the help
$endgroup$
– Maths_Rocks
Dec 19 '18 at 17:05
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Look at this probability from the other side. Consider two disjoint events.
Event $B_1$ means that the second ball was in the urn from the very beginning. Event $B_2$ means that the second ball is taken out of those balls that were added to the urn. Event $A$ means that the second ball is white. Consider conditional probabilities $mathbb P(Amid B_1)$ and $mathbb P(Amid B_2)$:
$$mathbb P(Amid B_1) = frac{n}{n+m}$$
since no ball added is taken into account when $B_1$ happens.
And also
$$mathbb P(Amid B_2) = frac{n}{n+m}.$$
This is the probability of the second ball to be white if it is selected from $k$ balls added. In other words, this is the probability that $k$ white balls are added into the urn. This is exactly the probability that the first ball was white.
Since conditional probabilities of $A$ under both $B_1$ and $B_2$ are the same, the probabilities of $B_1$ and $B_2$ are irrelevant and
$$mathbb P(A) = mathbb P(B_1)frac{n}{n+m}+mathbb P(B_2)frac{n}{n+m}=frac{n}{n+m}.$$
Note that only $mathbb P(B_1)$ and $mathbb P(B_2)$ depend on $k$.
answered Dec 19 '18 at 16:43
NChNCh
6,8663825
$endgroup$
$begingroup$
Very nice! (+1)
$endgroup$
– jgon
Dec 19 '18 at 16:59
$begingroup$
Thanx Nch for the help
$endgroup$
– Maths_Rocks
Dec 19 '18 at 17:05
add a comment |
$begingroup$
Look at this probability from the other side. Consider two disjoint events.
Event $B_1$ means that the second ball was in the urn from the very beginning. Event $B_2$ means that the second ball is taken out of those balls that were added to the urn. Event $A$ means that the second ball is white. Consider conditional probabilities $mathbb P(Amid B_1)$ and $mathbb P(Amid B_2)$:
$$mathbb P(Amid B_1) = frac{n}{n+m}$$
since no ball added is taken into account when $B_1$ happens.
And also
$$mathbb P(Amid B_2) = frac{n}{n+m}.$$
This is the probability of the second ball to be white if it is selected from $k$ balls added. In other words, this is the probability that $k$ white balls are added into the urn. This is exactly the probability that the first ball was white.
Since conditional probabilities of $A$ under both $B_1$ and $B_2$ are the same, the probabilities of $B_1$ and $B_2$ are irrelevant and
$$mathbb P(A) = mathbb P(B_1)frac{n}{n+m}+mathbb P(B_2)frac{n}{n+m}=frac{n}{n+m}.$$
Note that only $mathbb P(B_1)$ and $mathbb P(B_2)$ depend on $k$.
answered Dec 19 '18 at 16:43
NChNCh
6,8663825
$endgroup$
$begingroup$
Very nice! (+1)
$endgroup$
– jgon
Dec 19 '18 at 16:59
$begingroup$
Thanx Nch for the help
$endgroup$
– Maths_Rocks
Dec 19 '18 at 17:05
add a comment |
$begingroup$
Look at this probability from the other side. Consider two disjoint events.
Event $B_1$ means that the second ball was in the urn from the very beginning. Event $B_2$ means that the second ball is taken out of those balls that were added to the urn. Event $A$ means that the second ball is white. Consider conditional probabilities $mathbb P(Amid B_1)$ and $mathbb P(Amid B_2)$:
$$mathbb P(Amid B_1) = frac{n}{n+m}$$
since no ball added is taken into account when $B_1$ happens.
And also
$$mathbb P(Amid B_2) = frac{n}{n+m}.$$
This is the probability of the second ball to be white if it is selected from $k$ balls added. In other words, this is the probability that $k$ white balls are added into the urn. This is exactly the probability that the first ball was white.
Since conditional probabilities of $A$ under both $B_1$ and $B_2$ are the same, the probabilities of $B_1$ and $B_2$ are irrelevant and
$$mathbb P(A) = mathbb P(B_1)frac{n}{n+m}+mathbb P(B_2)frac{n}{n+m}=frac{n}{n+m}.$$
Note that only $mathbb P(B_1)$ and $mathbb P(B_2)$ depend on $k$.
answered Dec 19 '18 at 16:43
NChNCh
6,8663825
$endgroup$
Look at this probability from the other side. Consider two disjoint events.
Event $B_1$ means that the second ball was in the urn from the very beginning. Event $B_2$ means that the second ball is taken out of those balls that were added to the urn. Event $A$ means that the second ball is white. Consider conditional probabilities $mathbb P(Amid B_1)$ and $mathbb P(Amid B_2)$:
$$mathbb P(Amid B_1) = frac{n}{n+m}$$
since no ball added is taken into account when $B_1$ happens.
And also
$$mathbb P(Amid B_2) = frac{n}{n+m}.$$
This is the probability of the second ball to be white if it is selected from $k$ balls added. In other words, this is the probability that $k$ white balls are added into the urn. This is exactly the probability that the first ball was white.
Since conditional probabilities of $A$ under both $B_1$ and $B_2$ are the same, the probabilities of $B_1$ and $B_2$ are irrelevant and
$$mathbb P(A) = mathbb P(B_1)frac{n}{n+m}+mathbb P(B_2)frac{n}{n+m}=frac{n}{n+m}.$$
Note that only $mathbb P(B_1)$ and $mathbb P(B_2)$ depend on $k$.
answered Dec 19 '18 at 16:43
NChNCh
6,8663825
edited Dec 20 '18 at 0:34
answered Dec 19 '18 at 16:43
NChNCh
6,8663825
answered Dec 19 '18 at 16:43
NChNCh
6,8663825
answered Dec 19 '18 at 16:43
NChNCh
6,8663825
6,8663825
$begingroup$
Very nice! (+1)
$endgroup$
– jgon
Dec 19 '18 at 16:59
$begingroup$
Thanx Nch for the help
$endgroup$
– Maths_Rocks
Dec 19 '18 at 17:05
add a comment |
$begingroup$
Very nice! (+1)
$endgroup$
– jgon
Dec 19 '18 at 16:59
$begingroup$
Thanx Nch for the help
$endgroup$
– Maths_Rocks
Dec 19 '18 at 17:05
$begingroup$
Very nice! (+1)
$endgroup$
– jgon
Dec 19 '18 at 16:59
$begingroup$
Very nice! (+1)
$endgroup$
– jgon
Dec 19 '18 at 16:59
$begingroup$
Thanx Nch for the help
$endgroup$
– Maths_Rocks
Dec 19 '18 at 17:05
$begingroup$
Thanx Nch for the help
$endgroup$
– Maths_Rocks
Dec 19 '18 at 17:05
add a comment |
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$begingroup$
This problem with different numbers was posted a few minutes ago and has already some answers. It is important that you attempt to solve this though. From your tag I assume you already know that is solved using conditional probability.
$endgroup$
– Test123
Dec 19 '18 at 13:49
1
$begingroup$
Possible duplicate of probability - getting a red ball in thenext draw is
$endgroup$
– Test123
Dec 19 '18 at 13:50
$begingroup$
Does the bucket now contain $n+m+K$ balls in total?
$endgroup$
– user334732
Dec 19 '18 at 16:53
1
$begingroup$
Strongly disagree with those voting to close as duplicates. The OP here knows how to do the problem, and is instead asking about why there's no dependence on $k$, which isn't addressed by any of the answers in the supposed duplicate.
$endgroup$
– jgon
Dec 19 '18 at 16:58
$begingroup$
I edited your question for formatting. You were inconsistent with the capitalization of $k$, so I chose lower case $k$ to make it consistent. Feel free to change that if you intended $k$ to be capital.
$endgroup$
– jgon
Dec 19 '18 at 17:02