Probability of choosing the basket-i
up vote
2
down vote
favorite
Here is the question:
There are N+1 baskets 0,1,2,3,...,N. Where each basket i has i white balls and N-i black balls.
We choose randomly a basket and take out a ball after another with returning.
a)If we know that the first two withdraws we got a black and a white ball (not necessarily by order) what is the probability that we have chosen the basket i?for each i?
b)If we know that the first two withdraws were black balls what is the probability that the third withdraw will be black?
Ok,this is an a question in my homework, here is my approach:
a)for i equal 0/N the probability is zero (there is no white/black balls), for any other i the probability is 1/(N-1).
I am not sure about the answer,should we consider that we have picked two different balls, does that effect the probability of which basket we choose?
I know the sum of probabilities for the i-es should equal 1. Which my answer does satisfy.every other answer did not.
b)for b it is the same,will the the first two withdraws effect the third one? I think it does only tell that we have not picked the N-th basket no more.
my answer for b is 1/2.
I would like some hints...
Thank you guys, and sorry english is not my first language.
probability
asked Nov 22 at 18:03
Adddison
826
add a comment |
up vote
2
down vote
favorite
Here is the question:
There are N+1 baskets 0,1,2,3,...,N. Where each basket i has i white balls and N-i black balls.
We choose randomly a basket and take out a ball after another with returning.
a)If we know that the first two withdraws we got a black and a white ball (not necessarily by order) what is the probability that we have chosen the basket i?for each i?
b)If we know that the first two withdraws were black balls what is the probability that the third withdraw will be black?
Ok,this is an a question in my homework, here is my approach:
a)for i equal 0/N the probability is zero (there is no white/black balls), for any other i the probability is 1/(N-1).
I am not sure about the answer,should we consider that we have picked two different balls, does that effect the probability of which basket we choose?
I know the sum of probabilities for the i-es should equal 1. Which my answer does satisfy.every other answer did not.
b)for b it is the same,will the the first two withdraws effect the third one? I think it does only tell that we have not picked the N-th basket no more.
my answer for b is 1/2.
I would like some hints...
Thank you guys, and sorry english is not my first language.
probability
asked Nov 22 at 18:03
Adddison
826
Take a look at Bayes'rule
– saulspatz
Nov 22 at 18:08
I added another approach using Bayes's rule. @saulspatz
– Adddison
Nov 22 at 18:38
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Here is the question:
There are N+1 baskets 0,1,2,3,...,N. Where each basket i has i white balls and N-i black balls.
We choose randomly a basket and take out a ball after another with returning.
a)If we know that the first two withdraws we got a black and a white ball (not necessarily by order) what is the probability that we have chosen the basket i?for each i?
b)If we know that the first two withdraws were black balls what is the probability that the third withdraw will be black?
Ok,this is an a question in my homework, here is my approach:
a)for i equal 0/N the probability is zero (there is no white/black balls), for any other i the probability is 1/(N-1).
I am not sure about the answer,should we consider that we have picked two different balls, does that effect the probability of which basket we choose?
I know the sum of probabilities for the i-es should equal 1. Which my answer does satisfy.every other answer did not.
b)for b it is the same,will the the first two withdraws effect the third one? I think it does only tell that we have not picked the N-th basket no more.
my answer for b is 1/2.
I would like some hints...
Thank you guys, and sorry english is not my first language.
probability
asked Nov 22 at 18:03
Adddison
826
Here is the question:
There are N+1 baskets 0,1,2,3,...,N. Where each basket i has i white balls and N-i black balls.
We choose randomly a basket and take out a ball after another with returning.
a)If we know that the first two withdraws we got a black and a white ball (not necessarily by order) what is the probability that we have chosen the basket i?for each i?
b)If we know that the first two withdraws were black balls what is the probability that the third withdraw will be black?
Ok,this is an a question in my homework, here is my approach:
a)for i equal 0/N the probability is zero (there is no white/black balls), for any other i the probability is 1/(N-1).
I am not sure about the answer,should we consider that we have picked two different balls, does that effect the probability of which basket we choose?
I know the sum of probabilities for the i-es should equal 1. Which my answer does satisfy.every other answer did not.
b)for b it is the same,will the the first two withdraws effect the third one? I think it does only tell that we have not picked the N-th basket no more.
my answer for b is 1/2.
I would like some hints...
Thank you guys, and sorry english is not my first language.
probability
probability
asked Nov 22 at 18:03
Adddison
826
asked Nov 22 at 18:03
Adddison
826
asked Nov 22 at 18:03
Adddison
826
asked Nov 22 at 18:03
Adddison
826
asked Nov 22 at 18:03
Adddison
826
826
Take a look at Bayes'rule
– saulspatz
Nov 22 at 18:08
I added another approach using Bayes's rule. @saulspatz
– Adddison
Nov 22 at 18:38
add a comment |
Take a look at Bayes'rule
– saulspatz
Nov 22 at 18:08
I added another approach using Bayes's rule. @saulspatz
– Adddison
Nov 22 at 18:38
Take a look at Bayes'rule
– saulspatz
Nov 22 at 18:08
Take a look at Bayes'rule
– saulspatz
Nov 22 at 18:08
I added another approach using Bayes's rule. @saulspatz
– Adddison
Nov 22 at 18:38
I added another approach using Bayes's rule. @saulspatz
– Adddison
Nov 22 at 18:38
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
another approach:(using bayes`s equation)
let:
A=choosing basket i
B=two two ball withdrawn are blak and white
P(A|B)=$ frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{frac{1}{2}} $ =4 * $frac{i*(N-i)}{N^3+N^2}$
But the sum of i-es does not equal 1.I think I am wrong somewhere.
@saulspatz
answered Nov 22 at 18:36
Adddison
826
P(B) is wrong. Basically, if you are in a basket that has a very uneven distribution of black/white balls, you are much more likely to draw 2 balls of the same color than one black/one white. You have to add your enumerators to find the real $P(B)$.
– Ingix
Nov 22 at 18:45
could P(B) be ((i/N) *((N-i)/N))? @Ingix
– Adddison
Nov 22 at 18:58
It it is the sum over all those values (or rather 2 times that, as you correctly wrote in the solution), when $i$ ranges from 0 to N. Note that event B does not mention any basket number, so $P(B)$ cannot contain any $i$.
– Ingix
Nov 22 at 21:37
ok,so it's the sum from i=0 to N of the following: 2*(i/N)*((N-i)/N)*(1/N). I think it does equal N/6. but what is the mathematic proof for that?or how do I know the sum?
– Adddison
Nov 25 at 12:49
add a comment |
up vote
1
down vote
ok,here is my second approach:
a)$frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{2*sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})*frac{1}{N+1}} $ =$ frac{ frac{i}{N} * frac{N-i}{N}}{sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})} $
I don't know how to get rid of the sigma
b)P(third withdraw is black|first two withdraws are black)=$ frac{P(the three withdraws are black)}{P(the first two withdraws are black)} $=
$frac{sum_{i=0}^N frac{i^3}{N^3}}{sum_{i=0}^N frac{i^2}{N^2}}$=$sum_{i=0}^N frac{frac{i^3}{N}}{sum_{i=0}^N i^2}$
here too I could'nt get rid of i from the whole equation.any help please?
answered Nov 25 at 13:28
Adddison
826
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
1
down vote
another approach:(using bayes`s equation)
let:
A=choosing basket i
B=two two ball withdrawn are blak and white
P(A|B)=$ frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{frac{1}{2}} $ =4 * $frac{i*(N-i)}{N^3+N^2}$
But the sum of i-es does not equal 1.I think I am wrong somewhere.
@saulspatz
answered Nov 22 at 18:36
Adddison
826
P(B) is wrong. Basically, if you are in a basket that has a very uneven distribution of black/white balls, you are much more likely to draw 2 balls of the same color than one black/one white. You have to add your enumerators to find the real $P(B)$.
– Ingix
Nov 22 at 18:45
could P(B) be ((i/N) *((N-i)/N))? @Ingix
– Adddison
Nov 22 at 18:58
It it is the sum over all those values (or rather 2 times that, as you correctly wrote in the solution), when $i$ ranges from 0 to N. Note that event B does not mention any basket number, so $P(B)$ cannot contain any $i$.
– Ingix
Nov 22 at 21:37
ok,so it's the sum from i=0 to N of the following: 2*(i/N)*((N-i)/N)*(1/N). I think it does equal N/6. but what is the mathematic proof for that?or how do I know the sum?
– Adddison
Nov 25 at 12:49
add a comment |
up vote
1
down vote
another approach:(using bayes`s equation)
let:
A=choosing basket i
B=two two ball withdrawn are blak and white
P(A|B)=$ frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{frac{1}{2}} $ =4 * $frac{i*(N-i)}{N^3+N^2}$
But the sum of i-es does not equal 1.I think I am wrong somewhere.
@saulspatz
answered Nov 22 at 18:36
Adddison
826
P(B) is wrong. Basically, if you are in a basket that has a very uneven distribution of black/white balls, you are much more likely to draw 2 balls of the same color than one black/one white. You have to add your enumerators to find the real $P(B)$.
– Ingix
Nov 22 at 18:45
could P(B) be ((i/N) *((N-i)/N))? @Ingix
– Adddison
Nov 22 at 18:58
It it is the sum over all those values (or rather 2 times that, as you correctly wrote in the solution), when $i$ ranges from 0 to N. Note that event B does not mention any basket number, so $P(B)$ cannot contain any $i$.
– Ingix
Nov 22 at 21:37
ok,so it's the sum from i=0 to N of the following: 2*(i/N)*((N-i)/N)*(1/N). I think it does equal N/6. but what is the mathematic proof for that?or how do I know the sum?
– Adddison
Nov 25 at 12:49
add a comment |
up vote
1
down vote
up vote
1
down vote
another approach:(using bayes`s equation)
let:
A=choosing basket i
B=two two ball withdrawn are blak and white
P(A|B)=$ frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{frac{1}{2}} $ =4 * $frac{i*(N-i)}{N^3+N^2}$
But the sum of i-es does not equal 1.I think I am wrong somewhere.
@saulspatz
answered Nov 22 at 18:36
Adddison
826
another approach:(using bayes`s equation)
let:
A=choosing basket i
B=two two ball withdrawn are blak and white
P(A|B)=$ frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{frac{1}{2}} $ =4 * $frac{i*(N-i)}{N^3+N^2}$
But the sum of i-es does not equal 1.I think I am wrong somewhere.
@saulspatz
answered Nov 22 at 18:36
Adddison
826
answered Nov 22 at 18:36
Adddison
826
answered Nov 22 at 18:36
Adddison
826
answered Nov 22 at 18:36
Adddison
826
826
P(B) is wrong. Basically, if you are in a basket that has a very uneven distribution of black/white balls, you are much more likely to draw 2 balls of the same color than one black/one white. You have to add your enumerators to find the real $P(B)$.
– Ingix
Nov 22 at 18:45
could P(B) be ((i/N) *((N-i)/N))? @Ingix
– Adddison
Nov 22 at 18:58
It it is the sum over all those values (or rather 2 times that, as you correctly wrote in the solution), when $i$ ranges from 0 to N. Note that event B does not mention any basket number, so $P(B)$ cannot contain any $i$.
– Ingix
Nov 22 at 21:37
ok,so it's the sum from i=0 to N of the following: 2*(i/N)*((N-i)/N)*(1/N). I think it does equal N/6. but what is the mathematic proof for that?or how do I know the sum?
– Adddison
Nov 25 at 12:49
add a comment |
P(B) is wrong. Basically, if you are in a basket that has a very uneven distribution of black/white balls, you are much more likely to draw 2 balls of the same color than one black/one white. You have to add your enumerators to find the real $P(B)$.
– Ingix
Nov 22 at 18:45
could P(B) be ((i/N) *((N-i)/N))? @Ingix
– Adddison
Nov 22 at 18:58
It it is the sum over all those values (or rather 2 times that, as you correctly wrote in the solution), when $i$ ranges from 0 to N. Note that event B does not mention any basket number, so $P(B)$ cannot contain any $i$.
– Ingix
Nov 22 at 21:37
ok,so it's the sum from i=0 to N of the following: 2*(i/N)*((N-i)/N)*(1/N). I think it does equal N/6. but what is the mathematic proof for that?or how do I know the sum?
– Adddison
Nov 25 at 12:49
P(B) is wrong. Basically, if you are in a basket that has a very uneven distribution of black/white balls, you are much more likely to draw 2 balls of the same color than one black/one white. You have to add your enumerators to find the real $P(B)$.
– Ingix
Nov 22 at 18:45
P(B) is wrong. Basically, if you are in a basket that has a very uneven distribution of black/white balls, you are much more likely to draw 2 balls of the same color than one black/one white. You have to add your enumerators to find the real $P(B)$.
– Ingix
Nov 22 at 18:45
could P(B) be ((i/N) *((N-i)/N))? @Ingix
– Adddison
Nov 22 at 18:58
could P(B) be ((i/N) *((N-i)/N))? @Ingix
– Adddison
Nov 22 at 18:58
It it is the sum over all those values (or rather 2 times that, as you correctly wrote in the solution), when $i$ ranges from 0 to N. Note that event B does not mention any basket number, so $P(B)$ cannot contain any $i$.
– Ingix
Nov 22 at 21:37
It it is the sum over all those values (or rather 2 times that, as you correctly wrote in the solution), when $i$ ranges from 0 to N. Note that event B does not mention any basket number, so $P(B)$ cannot contain any $i$.
– Ingix
Nov 22 at 21:37
ok,so it's the sum from i=0 to N of the following: 2*(i/N)*((N-i)/N)*(1/N). I think it does equal N/6. but what is the mathematic proof for that?or how do I know the sum?
– Adddison
Nov 25 at 12:49
ok,so it's the sum from i=0 to N of the following: 2*(i/N)*((N-i)/N)*(1/N). I think it does equal N/6. but what is the mathematic proof for that?or how do I know the sum?
– Adddison
Nov 25 at 12:49
add a comment |
up vote
1
down vote
ok,here is my second approach:
a)$frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{2*sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})*frac{1}{N+1}} $ =$ frac{ frac{i}{N} * frac{N-i}{N}}{sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})} $
I don't know how to get rid of the sigma
b)P(third withdraw is black|first two withdraws are black)=$ frac{P(the three withdraws are black)}{P(the first two withdraws are black)} $=
$frac{sum_{i=0}^N frac{i^3}{N^3}}{sum_{i=0}^N frac{i^2}{N^2}}$=$sum_{i=0}^N frac{frac{i^3}{N}}{sum_{i=0}^N i^2}$
here too I could'nt get rid of i from the whole equation.any help please?
answered Nov 25 at 13:28
Adddison
826
add a comment |
up vote
1
down vote
ok,here is my second approach:
a)$frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{2*sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})*frac{1}{N+1}} $ =$ frac{ frac{i}{N} * frac{N-i}{N}}{sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})} $
I don't know how to get rid of the sigma
b)P(third withdraw is black|first two withdraws are black)=$ frac{P(the three withdraws are black)}{P(the first two withdraws are black)} $=
$frac{sum_{i=0}^N frac{i^3}{N^3}}{sum_{i=0}^N frac{i^2}{N^2}}$=$sum_{i=0}^N frac{frac{i^3}{N}}{sum_{i=0}^N i^2}$
here too I could'nt get rid of i from the whole equation.any help please?
answered Nov 25 at 13:28
Adddison
826
add a comment |
up vote
1
down vote
up vote
1
down vote
ok,here is my second approach:
a)$frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{2*sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})*frac{1}{N+1}} $ =$ frac{ frac{i}{N} * frac{N-i}{N}}{sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})} $
I don't know how to get rid of the sigma
b)P(third withdraw is black|first two withdraws are black)=$ frac{P(the three withdraws are black)}{P(the first two withdraws are black)} $=
$frac{sum_{i=0}^N frac{i^3}{N^3}}{sum_{i=0}^N frac{i^2}{N^2}}$=$sum_{i=0}^N frac{frac{i^3}{N}}{sum_{i=0}^N i^2}$
here too I could'nt get rid of i from the whole equation.any help please?
answered Nov 25 at 13:28
Adddison
826
ok,here is my second approach:
a)$frac{P(B|A)*P(A)}{P(B)} $= $ frac{2*( frac{i}{N} * frac{N-i}{N})*( frac{1}{N+1}) }{2*sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})*frac{1}{N+1}} $ =$ frac{ frac{i}{N} * frac{N-i}{N}}{sum_{i=1}^n a_n(frac{k}{N}*frac{N-k}{N})} $
I don't know how to get rid of the sigma
b)P(third withdraw is black|first two withdraws are black)=$ frac{P(the three withdraws are black)}{P(the first two withdraws are black)} $=
$frac{sum_{i=0}^N frac{i^3}{N^3}}{sum_{i=0}^N frac{i^2}{N^2}}$=$sum_{i=0}^N frac{frac{i^3}{N}}{sum_{i=0}^N i^2}$
here too I could'nt get rid of i from the whole equation.any help please?
answered Nov 25 at 13:28
Adddison
826
answered Nov 25 at 13:28
Adddison
826
answered Nov 25 at 13:28
Adddison
826
answered Nov 25 at 13:28
Adddison
826
826
add a comment |
add a comment |
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Take a look at Bayes'rule
– saulspatz
Nov 22 at 18:08
I added another approach using Bayes's rule. @saulspatz
– Adddison
Nov 22 at 18:38