Conditional Probability of a Uniform Random Subset.
up vote
2
down vote
favorite
Question:
Let X = $({1,2,3,..,10})$ Let $Y$ be a uniformly random subset of $X$. Define the events:
A = “$Y$ contains at least $4$ elements",
B = “all elements of $Y$ are even".
What is $Pr(A|B)$?
Answer: 0.1875
Attempt:
I know that P(A $bigcap$ B) / P(B) is what I have to ultimately find.
For $P(B)$ = $frac{5}{10}$ = $frac{1}{2}$
For P(A) => Must have exactly 4, 5, 6, 7, 8, 9, 10 elements
So, I used the binomial for this by doing:
P(A) =
$10choose4$$.$ $frac{1}{10}$$^4$$.$$(1-frac{1}{10}$)$^6$ +
$10choose5$$.$ $frac{1}{10}$$^5$$.$$(1-frac{1}{10}$)$^5$ +
$10choose6$$.$ $frac{1}{10}$$^6$$.$$(1-frac{1}{10}$)$^4$ +
$10choose7$$.$ $frac{1}{10}$$^7$$.$$(1-frac{1}{10}$)$^3$ +
$10choose8$$.$ $frac{1}{10}$$^8$$.$$(1-frac{1}{10}$)$^2$ +
$10choose9$$.$ $frac{1}{10}$$^9$$.$$(1-frac{1}{10}$)$^1$ +
$10choose10$$.$ $frac{1}{10}$$^{10}$$.$$(1-frac{1}{10}$)$^0$ +
= $0.012795$
P(A $bigcap$B) = $frac{1}{2}*0.012795$
Pr(A|B) = $frac{frac{1}{2}*0.012795}{2}$
Pr(A|B) = $0.012795$
Where did I go wrong with this approach? I find finding P(A) very time consuming, I feel like there has to be a much more simpler approach.
probability probability-theory discrete-mathematics random-variables conditional-probability
add a comment |
up vote
2
down vote
favorite
Question:
Let X = $({1,2,3,..,10})$ Let $Y$ be a uniformly random subset of $X$. Define the events:
A = “$Y$ contains at least $4$ elements",
B = “all elements of $Y$ are even".
What is $Pr(A|B)$?
Answer: 0.1875
Attempt:
I know that P(A $bigcap$ B) / P(B) is what I have to ultimately find.
For $P(B)$ = $frac{5}{10}$ = $frac{1}{2}$
For P(A) => Must have exactly 4, 5, 6, 7, 8, 9, 10 elements
So, I used the binomial for this by doing:
P(A) =
$10choose4$$.$ $frac{1}{10}$$^4$$.$$(1-frac{1}{10}$)$^6$ +
$10choose5$$.$ $frac{1}{10}$$^5$$.$$(1-frac{1}{10}$)$^5$ +
$10choose6$$.$ $frac{1}{10}$$^6$$.$$(1-frac{1}{10}$)$^4$ +
$10choose7$$.$ $frac{1}{10}$$^7$$.$$(1-frac{1}{10}$)$^3$ +
$10choose8$$.$ $frac{1}{10}$$^8$$.$$(1-frac{1}{10}$)$^2$ +
$10choose9$$.$ $frac{1}{10}$$^9$$.$$(1-frac{1}{10}$)$^1$ +
$10choose10$$.$ $frac{1}{10}$$^{10}$$.$$(1-frac{1}{10}$)$^0$ +
= $0.012795$
P(A $bigcap$B) = $frac{1}{2}*0.012795$
Pr(A|B) = $frac{frac{1}{2}*0.012795}{2}$
Pr(A|B) = $0.012795$
Where did I go wrong with this approach? I find finding P(A) very time consuming, I feel like there has to be a much more simpler approach.
probability probability-theory discrete-mathematics random-variables conditional-probability
3
Your set contains $100$ elements but you are working with $10$. Is there a typo?
– Anurag A
Nov 30 at 4:20
@Anurag A yes it was a typo, set has 10 elements not 100
– Toby
Nov 30 at 4:37
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Question:
Let X = $({1,2,3,..,10})$ Let $Y$ be a uniformly random subset of $X$. Define the events:
A = “$Y$ contains at least $4$ elements",
B = “all elements of $Y$ are even".
What is $Pr(A|B)$?
Answer: 0.1875
Attempt:
I know that P(A $bigcap$ B) / P(B) is what I have to ultimately find.
For $P(B)$ = $frac{5}{10}$ = $frac{1}{2}$
For P(A) => Must have exactly 4, 5, 6, 7, 8, 9, 10 elements
So, I used the binomial for this by doing:
P(A) =
$10choose4$$.$ $frac{1}{10}$$^4$$.$$(1-frac{1}{10}$)$^6$ +
$10choose5$$.$ $frac{1}{10}$$^5$$.$$(1-frac{1}{10}$)$^5$ +
$10choose6$$.$ $frac{1}{10}$$^6$$.$$(1-frac{1}{10}$)$^4$ +
$10choose7$$.$ $frac{1}{10}$$^7$$.$$(1-frac{1}{10}$)$^3$ +
$10choose8$$.$ $frac{1}{10}$$^8$$.$$(1-frac{1}{10}$)$^2$ +
$10choose9$$.$ $frac{1}{10}$$^9$$.$$(1-frac{1}{10}$)$^1$ +
$10choose10$$.$ $frac{1}{10}$$^{10}$$.$$(1-frac{1}{10}$)$^0$ +
= $0.012795$
P(A $bigcap$B) = $frac{1}{2}*0.012795$
Pr(A|B) = $frac{frac{1}{2}*0.012795}{2}$
Pr(A|B) = $0.012795$
Where did I go wrong with this approach? I find finding P(A) very time consuming, I feel like there has to be a much more simpler approach.
probability probability-theory discrete-mathematics random-variables conditional-probability
Question:
Let X = $({1,2,3,..,10})$ Let $Y$ be a uniformly random subset of $X$. Define the events:
A = “$Y$ contains at least $4$ elements",
B = “all elements of $Y$ are even".
What is $Pr(A|B)$?
Answer: 0.1875
Attempt:
I know that P(A $bigcap$ B) / P(B) is what I have to ultimately find.
For $P(B)$ = $frac{5}{10}$ = $frac{1}{2}$
For P(A) => Must have exactly 4, 5, 6, 7, 8, 9, 10 elements
So, I used the binomial for this by doing:
P(A) =
$10choose4$$.$ $frac{1}{10}$$^4$$.$$(1-frac{1}{10}$)$^6$ +
$10choose5$$.$ $frac{1}{10}$$^5$$.$$(1-frac{1}{10}$)$^5$ +
$10choose6$$.$ $frac{1}{10}$$^6$$.$$(1-frac{1}{10}$)$^4$ +
$10choose7$$.$ $frac{1}{10}$$^7$$.$$(1-frac{1}{10}$)$^3$ +
$10choose8$$.$ $frac{1}{10}$$^8$$.$$(1-frac{1}{10}$)$^2$ +
$10choose9$$.$ $frac{1}{10}$$^9$$.$$(1-frac{1}{10}$)$^1$ +
$10choose10$$.$ $frac{1}{10}$$^{10}$$.$$(1-frac{1}{10}$)$^0$ +
= $0.012795$
P(A $bigcap$B) = $frac{1}{2}*0.012795$
Pr(A|B) = $frac{frac{1}{2}*0.012795}{2}$
Pr(A|B) = $0.012795$
Where did I go wrong with this approach? I find finding P(A) very time consuming, I feel like there has to be a much more simpler approach.
probability probability-theory discrete-mathematics random-variables conditional-probability
probability probability-theory discrete-mathematics random-variables conditional-probability
edited Nov 30 at 4:37
asked Nov 30 at 4:18
Toby
866
866
3
Your set contains $100$ elements but you are working with $10$. Is there a typo?
– Anurag A
Nov 30 at 4:20
@Anurag A yes it was a typo, set has 10 elements not 100
– Toby
Nov 30 at 4:37
add a comment |
3
Your set contains $100$ elements but you are working with $10$. Is there a typo?
– Anurag A
Nov 30 at 4:20
@Anurag A yes it was a typo, set has 10 elements not 100
– Toby
Nov 30 at 4:37
3
3
Your set contains $100$ elements but you are working with $10$. Is there a typo?
– Anurag A
Nov 30 at 4:20
Your set contains $100$ elements but you are working with $10$. Is there a typo?
– Anurag A
Nov 30 at 4:20
@Anurag A yes it was a typo, set has 10 elements not 100
– Toby
Nov 30 at 4:37
@Anurag A yes it was a typo, set has 10 elements not 100
– Toby
Nov 30 at 4:37
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
$5$ elements of $X$ are even.
To have at least $4$ elements and all elements are even, you have either $4$ elements ($5$ ways, choose one to leave out) or $5$ elements ($1$ way).
$$frac{1+5}{2^{5}}=frac{6}{32}$$
Remark:
We are choosing uniformly over all subsets. $P(B)= frac{2^5}{2^{10}} ne frac{5}{10}$.
We don't have to find $P(A)$, what is of interest is $P(A cap B)$.
We do not have $P(A cap B)=P(A)P(B)$ in general, we need independence.
How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
– Toby
Nov 30 at 4:55
$P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
– Siong Thye Goh
Nov 30 at 4:57
if the events were A= 1 is an element of Y and B= 7 is an element of Y and subset Y has size =5, would I just take the 2^1/2^10 as the probability of A and B? And for P(A int B) = 2^2/2^5. So P(A|B) = 2^2/2^5 / 2/2^10?
– Toby
Nov 30 at 21:38
not really, if $A$ is $1$ must be an element of $Y$, then there are $2^9$ of them, because you get to choose for the ohters whether they should be included. You can use the similar reasoning for the rest.
– Siong Thye Goh
2 days ago
add a comment |
up vote
2
down vote
Why are you using binomial distribution to evaluate the probabilities for event $A$, when $Y$ is a uniformly random subset of $X$? That means that every element (ie subset of $X$) of the powerset of $X$ (the sample space) has an equal probability of being selected. Just count the favoured and total spaces.
$mathsf P(Acap B)=mathsf P(B)~mathsf P(A)$ is only true for independent events, of which these are not. Evaluate $mathsf P(Acap B)$ using a similar method as you should have evaluated $mathsf P(A)$.
$mathsf P(B) neq tfrac 12$. The event of "all the elements of $Y$ are even" is not exactly half of the sample space, which may be partitioned by "all the elements of $Y$ are even", "all of the elements of $Y$ are odd", and "some of the elements of $Y$ are even, and some are odd." Rethink.
I'm not sure on how to calculate $mathsf P(Acap B)$? Is the binomial method applicable here? For uniformly random subsets, I take it that the binomial can't be used?
– Toby
Nov 30 at 4:54
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$5$ elements of $X$ are even.
To have at least $4$ elements and all elements are even, you have either $4$ elements ($5$ ways, choose one to leave out) or $5$ elements ($1$ way).
$$frac{1+5}{2^{5}}=frac{6}{32}$$
Remark:
We are choosing uniformly over all subsets. $P(B)= frac{2^5}{2^{10}} ne frac{5}{10}$.
We don't have to find $P(A)$, what is of interest is $P(A cap B)$.
We do not have $P(A cap B)=P(A)P(B)$ in general, we need independence.
How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
– Toby
Nov 30 at 4:55
$P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
– Siong Thye Goh
Nov 30 at 4:57
if the events were A= 1 is an element of Y and B= 7 is an element of Y and subset Y has size =5, would I just take the 2^1/2^10 as the probability of A and B? And for P(A int B) = 2^2/2^5. So P(A|B) = 2^2/2^5 / 2/2^10?
– Toby
Nov 30 at 21:38
not really, if $A$ is $1$ must be an element of $Y$, then there are $2^9$ of them, because you get to choose for the ohters whether they should be included. You can use the similar reasoning for the rest.
– Siong Thye Goh
2 days ago
add a comment |
up vote
3
down vote
accepted
$5$ elements of $X$ are even.
To have at least $4$ elements and all elements are even, you have either $4$ elements ($5$ ways, choose one to leave out) or $5$ elements ($1$ way).
$$frac{1+5}{2^{5}}=frac{6}{32}$$
Remark:
We are choosing uniformly over all subsets. $P(B)= frac{2^5}{2^{10}} ne frac{5}{10}$.
We don't have to find $P(A)$, what is of interest is $P(A cap B)$.
We do not have $P(A cap B)=P(A)P(B)$ in general, we need independence.
How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
– Toby
Nov 30 at 4:55
$P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
– Siong Thye Goh
Nov 30 at 4:57
if the events were A= 1 is an element of Y and B= 7 is an element of Y and subset Y has size =5, would I just take the 2^1/2^10 as the probability of A and B? And for P(A int B) = 2^2/2^5. So P(A|B) = 2^2/2^5 / 2/2^10?
– Toby
Nov 30 at 21:38
not really, if $A$ is $1$ must be an element of $Y$, then there are $2^9$ of them, because you get to choose for the ohters whether they should be included. You can use the similar reasoning for the rest.
– Siong Thye Goh
2 days ago
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$5$ elements of $X$ are even.
To have at least $4$ elements and all elements are even, you have either $4$ elements ($5$ ways, choose one to leave out) or $5$ elements ($1$ way).
$$frac{1+5}{2^{5}}=frac{6}{32}$$
Remark:
We are choosing uniformly over all subsets. $P(B)= frac{2^5}{2^{10}} ne frac{5}{10}$.
We don't have to find $P(A)$, what is of interest is $P(A cap B)$.
We do not have $P(A cap B)=P(A)P(B)$ in general, we need independence.
$5$ elements of $X$ are even.
To have at least $4$ elements and all elements are even, you have either $4$ elements ($5$ ways, choose one to leave out) or $5$ elements ($1$ way).
$$frac{1+5}{2^{5}}=frac{6}{32}$$
Remark:
We are choosing uniformly over all subsets. $P(B)= frac{2^5}{2^{10}} ne frac{5}{10}$.
We don't have to find $P(A)$, what is of interest is $P(A cap B)$.
We do not have $P(A cap B)=P(A)P(B)$ in general, we need independence.
answered Nov 30 at 4:40
Siong Thye Goh
95.7k1462116
95.7k1462116
How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
– Toby
Nov 30 at 4:55
$P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
– Siong Thye Goh
Nov 30 at 4:57
if the events were A= 1 is an element of Y and B= 7 is an element of Y and subset Y has size =5, would I just take the 2^1/2^10 as the probability of A and B? And for P(A int B) = 2^2/2^5. So P(A|B) = 2^2/2^5 / 2/2^10?
– Toby
Nov 30 at 21:38
not really, if $A$ is $1$ must be an element of $Y$, then there are $2^9$ of them, because you get to choose for the ohters whether they should be included. You can use the similar reasoning for the rest.
– Siong Thye Goh
2 days ago
add a comment |
How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
– Toby
Nov 30 at 4:55
$P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
– Siong Thye Goh
Nov 30 at 4:57
if the events were A= 1 is an element of Y and B= 7 is an element of Y and subset Y has size =5, would I just take the 2^1/2^10 as the probability of A and B? And for P(A int B) = 2^2/2^5. So P(A|B) = 2^2/2^5 / 2/2^10?
– Toby
Nov 30 at 21:38
not really, if $A$ is $1$ must be an element of $Y$, then there are $2^9$ of them, because you get to choose for the ohters whether they should be included. You can use the similar reasoning for the rest.
– Siong Thye Goh
2 days ago
How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
– Toby
Nov 30 at 4:55
How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
– Toby
Nov 30 at 4:55
$P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
– Siong Thye Goh
Nov 30 at 4:57
$P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
– Siong Thye Goh
Nov 30 at 4:57
if the events were A= 1 is an element of Y and B= 7 is an element of Y and subset Y has size =5, would I just take the 2^1/2^10 as the probability of A and B? And for P(A int B) = 2^2/2^5. So P(A|B) = 2^2/2^5 / 2/2^10?
– Toby
Nov 30 at 21:38
if the events were A= 1 is an element of Y and B= 7 is an element of Y and subset Y has size =5, would I just take the 2^1/2^10 as the probability of A and B? And for P(A int B) = 2^2/2^5. So P(A|B) = 2^2/2^5 / 2/2^10?
– Toby
Nov 30 at 21:38
not really, if $A$ is $1$ must be an element of $Y$, then there are $2^9$ of them, because you get to choose for the ohters whether they should be included. You can use the similar reasoning for the rest.
– Siong Thye Goh
2 days ago
not really, if $A$ is $1$ must be an element of $Y$, then there are $2^9$ of them, because you get to choose for the ohters whether they should be included. You can use the similar reasoning for the rest.
– Siong Thye Goh
2 days ago
add a comment |
up vote
2
down vote
Why are you using binomial distribution to evaluate the probabilities for event $A$, when $Y$ is a uniformly random subset of $X$? That means that every element (ie subset of $X$) of the powerset of $X$ (the sample space) has an equal probability of being selected. Just count the favoured and total spaces.
$mathsf P(Acap B)=mathsf P(B)~mathsf P(A)$ is only true for independent events, of which these are not. Evaluate $mathsf P(Acap B)$ using a similar method as you should have evaluated $mathsf P(A)$.
$mathsf P(B) neq tfrac 12$. The event of "all the elements of $Y$ are even" is not exactly half of the sample space, which may be partitioned by "all the elements of $Y$ are even", "all of the elements of $Y$ are odd", and "some of the elements of $Y$ are even, and some are odd." Rethink.
I'm not sure on how to calculate $mathsf P(Acap B)$? Is the binomial method applicable here? For uniformly random subsets, I take it that the binomial can't be used?
– Toby
Nov 30 at 4:54
add a comment |
up vote
2
down vote
Why are you using binomial distribution to evaluate the probabilities for event $A$, when $Y$ is a uniformly random subset of $X$? That means that every element (ie subset of $X$) of the powerset of $X$ (the sample space) has an equal probability of being selected. Just count the favoured and total spaces.
$mathsf P(Acap B)=mathsf P(B)~mathsf P(A)$ is only true for independent events, of which these are not. Evaluate $mathsf P(Acap B)$ using a similar method as you should have evaluated $mathsf P(A)$.
$mathsf P(B) neq tfrac 12$. The event of "all the elements of $Y$ are even" is not exactly half of the sample space, which may be partitioned by "all the elements of $Y$ are even", "all of the elements of $Y$ are odd", and "some of the elements of $Y$ are even, and some are odd." Rethink.
I'm not sure on how to calculate $mathsf P(Acap B)$? Is the binomial method applicable here? For uniformly random subsets, I take it that the binomial can't be used?
– Toby
Nov 30 at 4:54
add a comment |
up vote
2
down vote
up vote
2
down vote
Why are you using binomial distribution to evaluate the probabilities for event $A$, when $Y$ is a uniformly random subset of $X$? That means that every element (ie subset of $X$) of the powerset of $X$ (the sample space) has an equal probability of being selected. Just count the favoured and total spaces.
$mathsf P(Acap B)=mathsf P(B)~mathsf P(A)$ is only true for independent events, of which these are not. Evaluate $mathsf P(Acap B)$ using a similar method as you should have evaluated $mathsf P(A)$.
$mathsf P(B) neq tfrac 12$. The event of "all the elements of $Y$ are even" is not exactly half of the sample space, which may be partitioned by "all the elements of $Y$ are even", "all of the elements of $Y$ are odd", and "some of the elements of $Y$ are even, and some are odd." Rethink.
Why are you using binomial distribution to evaluate the probabilities for event $A$, when $Y$ is a uniformly random subset of $X$? That means that every element (ie subset of $X$) of the powerset of $X$ (the sample space) has an equal probability of being selected. Just count the favoured and total spaces.
$mathsf P(Acap B)=mathsf P(B)~mathsf P(A)$ is only true for independent events, of which these are not. Evaluate $mathsf P(Acap B)$ using a similar method as you should have evaluated $mathsf P(A)$.
$mathsf P(B) neq tfrac 12$. The event of "all the elements of $Y$ are even" is not exactly half of the sample space, which may be partitioned by "all the elements of $Y$ are even", "all of the elements of $Y$ are odd", and "some of the elements of $Y$ are even, and some are odd." Rethink.
edited Nov 30 at 4:44
answered Nov 30 at 4:37
Graham Kemp
84.5k43378
84.5k43378
I'm not sure on how to calculate $mathsf P(Acap B)$? Is the binomial method applicable here? For uniformly random subsets, I take it that the binomial can't be used?
– Toby
Nov 30 at 4:54
add a comment |
I'm not sure on how to calculate $mathsf P(Acap B)$? Is the binomial method applicable here? For uniformly random subsets, I take it that the binomial can't be used?
– Toby
Nov 30 at 4:54
I'm not sure on how to calculate $mathsf P(Acap B)$? Is the binomial method applicable here? For uniformly random subsets, I take it that the binomial can't be used?
– Toby
Nov 30 at 4:54
I'm not sure on how to calculate $mathsf P(Acap B)$? Is the binomial method applicable here? For uniformly random subsets, I take it that the binomial can't be used?
– Toby
Nov 30 at 4:54
add a comment |
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3
Your set contains $100$ elements but you are working with $10$. Is there a typo?
– Anurag A
Nov 30 at 4:20
@Anurag A yes it was a typo, set has 10 elements not 100
– Toby
Nov 30 at 4:37