Conditional Probability of a Uniform Random Subset.












2












$begingroup$



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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Your set contains $100$ elements but you are working with $10$. Is there a typo?
    $endgroup$
    – Anurag A
    Nov 30 '18 at 4:20










  • $begingroup$
    @Anurag A yes it was a typo, set has 10 elements not 100
    $endgroup$
    – Toby
    Nov 30 '18 at 4:37
















2












$begingroup$



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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Your set contains $100$ elements but you are working with $10$. Is there a typo?
    $endgroup$
    – Anurag A
    Nov 30 '18 at 4:20










  • $begingroup$
    @Anurag A yes it was a typo, set has 10 elements not 100
    $endgroup$
    – Toby
    Nov 30 '18 at 4:37














2












2








2


1



$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 4:37







Toby

















asked Nov 30 '18 at 4:18









TobyToby

1577




1577








  • 3




    $begingroup$
    Your set contains $100$ elements but you are working with $10$. Is there a typo?
    $endgroup$
    – Anurag A
    Nov 30 '18 at 4:20










  • $begingroup$
    @Anurag A yes it was a typo, set has 10 elements not 100
    $endgroup$
    – Toby
    Nov 30 '18 at 4:37














  • 3




    $begingroup$
    Your set contains $100$ elements but you are working with $10$. Is there a typo?
    $endgroup$
    – Anurag A
    Nov 30 '18 at 4:20










  • $begingroup$
    @Anurag A yes it was a typo, set has 10 elements not 100
    $endgroup$
    – Toby
    Nov 30 '18 at 4:37








3




3




$begingroup$
Your set contains $100$ elements but you are working with $10$. Is there a typo?
$endgroup$
– Anurag A
Nov 30 '18 at 4:20




$begingroup$
Your set contains $100$ elements but you are working with $10$. Is there a typo?
$endgroup$
– Anurag A
Nov 30 '18 at 4:20












$begingroup$
@Anurag A yes it was a typo, set has 10 elements not 100
$endgroup$
– Toby
Nov 30 '18 at 4:37




$begingroup$
@Anurag A yes it was a typo, set has 10 elements not 100
$endgroup$
– Toby
Nov 30 '18 at 4:37










2 Answers
2






active

oldest

votes


















3












$begingroup$

$5$ 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.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
    $endgroup$
    – Toby
    Nov 30 '18 at 4:55












  • $begingroup$
    $P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
    $endgroup$
    – Siong Thye Goh
    Nov 30 '18 at 4:57










  • $begingroup$
    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?
    $endgroup$
    – Toby
    Nov 30 '18 at 21:38










  • $begingroup$
    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.
    $endgroup$
    – Siong Thye Goh
    Dec 1 '18 at 4:33



















2












$begingroup$


  1. 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.


  2. $mathsf P(Acap B)=mathsf P(B)~mathsf P(A)$ 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 $mathsf P(A)$.


  3. $mathsf P(B) neq tfrac 12$.   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.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – Toby
    Nov 30 '18 at 4:54











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

$5$ 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.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
    $endgroup$
    – Toby
    Nov 30 '18 at 4:55












  • $begingroup$
    $P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
    $endgroup$
    – Siong Thye Goh
    Nov 30 '18 at 4:57










  • $begingroup$
    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?
    $endgroup$
    – Toby
    Nov 30 '18 at 21:38










  • $begingroup$
    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.
    $endgroup$
    – Siong Thye Goh
    Dec 1 '18 at 4:33
















3












$begingroup$

$5$ 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.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
    $endgroup$
    – Toby
    Nov 30 '18 at 4:55












  • $begingroup$
    $P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
    $endgroup$
    – Siong Thye Goh
    Nov 30 '18 at 4:57










  • $begingroup$
    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?
    $endgroup$
    – Toby
    Nov 30 '18 at 21:38










  • $begingroup$
    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.
    $endgroup$
    – Siong Thye Goh
    Dec 1 '18 at 4:33














3












3








3





$begingroup$

$5$ 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.







share|cite|improve this answer









$endgroup$



$5$ 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.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 '18 at 4:40









Siong Thye GohSiong Thye Goh

100k1465117




100k1465117












  • $begingroup$
    How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
    $endgroup$
    – Toby
    Nov 30 '18 at 4:55












  • $begingroup$
    $P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
    $endgroup$
    – Siong Thye Goh
    Nov 30 '18 at 4:57










  • $begingroup$
    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?
    $endgroup$
    – Toby
    Nov 30 '18 at 21:38










  • $begingroup$
    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.
    $endgroup$
    – Siong Thye Goh
    Dec 1 '18 at 4:33


















  • $begingroup$
    How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
    $endgroup$
    – Toby
    Nov 30 '18 at 4:55












  • $begingroup$
    $P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
    $endgroup$
    – Siong Thye Goh
    Nov 30 '18 at 4:57










  • $begingroup$
    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?
    $endgroup$
    – Toby
    Nov 30 '18 at 21:38










  • $begingroup$
    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.
    $endgroup$
    – Siong Thye Goh
    Dec 1 '18 at 4:33
















$begingroup$
How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
$endgroup$
– Toby
Nov 30 '18 at 4:55






$begingroup$
How do I calculate $mathsf P(Acap B)$? I thought I can use the binomial to calculate P(A)
$endgroup$
– Toby
Nov 30 '18 at 4:55














$begingroup$
$P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
$endgroup$
– Siong Thye Goh
Nov 30 '18 at 4:57




$begingroup$
$P(A cap B) = frac{1+5}{2^{10}}$ only those $6$ sets that I described satisfies the condition.
$endgroup$
– Siong Thye Goh
Nov 30 '18 at 4:57












$begingroup$
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?
$endgroup$
– Toby
Nov 30 '18 at 21:38




$begingroup$
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?
$endgroup$
– Toby
Nov 30 '18 at 21:38












$begingroup$
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.
$endgroup$
– Siong Thye Goh
Dec 1 '18 at 4:33




$begingroup$
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.
$endgroup$
– Siong Thye Goh
Dec 1 '18 at 4:33











2












$begingroup$


  1. 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.


  2. $mathsf P(Acap B)=mathsf P(B)~mathsf P(A)$ 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 $mathsf P(A)$.


  3. $mathsf P(B) neq tfrac 12$.   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.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – Toby
    Nov 30 '18 at 4:54
















2












$begingroup$


  1. 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.


  2. $mathsf P(Acap B)=mathsf P(B)~mathsf P(A)$ 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 $mathsf P(A)$.


  3. $mathsf P(B) neq tfrac 12$.   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.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – Toby
    Nov 30 '18 at 4:54














2












2








2





$begingroup$


  1. 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.


  2. $mathsf P(Acap B)=mathsf P(B)~mathsf P(A)$ 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 $mathsf P(A)$.


  3. $mathsf P(B) neq tfrac 12$.   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.







share|cite|improve this answer











$endgroup$




  1. 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.


  2. $mathsf P(Acap B)=mathsf P(B)~mathsf P(A)$ 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 $mathsf P(A)$.


  3. $mathsf P(B) neq tfrac 12$.   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.








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 4:44

























answered Nov 30 '18 at 4:37









Graham KempGraham Kemp

84.9k43378




84.9k43378












  • $begingroup$
    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?
    $endgroup$
    – Toby
    Nov 30 '18 at 4:54


















  • $begingroup$
    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?
    $endgroup$
    – Toby
    Nov 30 '18 at 4:54
















$begingroup$
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?
$endgroup$
– Toby
Nov 30 '18 at 4:54




$begingroup$
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?
$endgroup$
– Toby
Nov 30 '18 at 4:54


















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