Mixing
Probability of two aces when we get at least one ace
$begingroup$
"In drawing two cards from a deck (without returning the first card) what is the probability of two aces when you get at least one ace?"
I am aware there is exact same question answered here.
Probability of two aces when you get at least one ace
I know and understand then answers mentioned.
I would like to approach the problem in a different way.
Let us denote A = event that we get ace from first suite in case of success, i.e. when two aces are drawn., or alternately speaking, atleast one ace is from first suite
similarly B,C,D.
What we want to know is $P(Acup B cup Ccup D) $
$=sum_A ^D{P(X)} - {4 choose 2 }P(A cap B)$(due to symmetry) + {ignoring the other intersections since we can't have more than two cards}
$= 4*frac {3}{51} - {4 choose 2 } *frac{1}{{52 choose 2 }}$
$= 3/13$
which is not right answer, obviously I did some mistake, but where?
probability probability-theory conditional-probability
asked Jan 9 at 9:19
q126yq126y
239212
$endgroup$
|
show 1 more comment
$begingroup$
"In drawing two cards from a deck (without returning the first card) what is the probability of two aces when you get at least one ace?"
I am aware there is exact same question answered here.
Probability of two aces when you get at least one ace
I know and understand then answers mentioned.
I would like to approach the problem in a different way.
Let us denote A = event that we get ace from first suite in case of success, i.e. when two aces are drawn., or alternately speaking, atleast one ace is from first suite
similarly B,C,D.
What we want to know is $P(Acup B cup Ccup D) $
$=sum_A ^D{P(X)} - {4 choose 2 }P(A cap B)$(due to symmetry) + {ignoring the other intersections since we can't have more than two cards}
$= 4*frac {3}{51} - {4 choose 2 } *frac{1}{{52 choose 2 }}$
$= 3/13$
which is not right answer, obviously I did some mistake, but where?
probability probability-theory conditional-probability
asked Jan 9 at 9:19
q126yq126y
239212
$endgroup$
$begingroup$
Where does $3/51$ come from. Surely the probability of drawing the ace of spades is $1/26$?
$endgroup$
– Lord Shark the Unknown
Jan 9 at 9:23
$begingroup$
@LordSharktheUnknown Given we have ace of spades, we have 3/51 probability that we will have two aces.
$endgroup$
– q126y
Jan 9 at 9:24
$begingroup$
@Peter I know the answer comes out to be 1/33 using bayes' theorem. I want to find gaps in my understanding by approaching it differently
$endgroup$
– q126y
Jan 9 at 9:26
$begingroup$
@Peter I know that. I have mentioned that answers to the question are available here math.stackexchange.com/questions/1390830/… I want to get more grip on my understanding by approaching it in this manner. What am I doing wrong with this approach? How should I correct it? Is formulation of the problem wrong, or something else. Thanks Edit: or why this approach doesn't work.
$endgroup$
– q126y
Jan 9 at 9:30
$begingroup$
OK, so you only want to analyze your concrete approach - You used the conditional probability within the formula for the probability of $P(Acup Bcup Ccup D)$. This might be the issue.
$endgroup$
– Peter
Jan 9 at 9:37
|
show 1 more comment
$begingroup$
"In drawing two cards from a deck (without returning the first card) what is the probability of two aces when you get at least one ace?"
I am aware there is exact same question answered here.
Probability of two aces when you get at least one ace
I know and understand then answers mentioned.
I would like to approach the problem in a different way.
Let us denote A = event that we get ace from first suite in case of success, i.e. when two aces are drawn., or alternately speaking, atleast one ace is from first suite
similarly B,C,D.
What we want to know is $P(Acup B cup Ccup D) $
$=sum_A ^D{P(X)} - {4 choose 2 }P(A cap B)$(due to symmetry) + {ignoring the other intersections since we can't have more than two cards}
$= 4*frac {3}{51} - {4 choose 2 } *frac{1}{{52 choose 2 }}$
$= 3/13$
which is not right answer, obviously I did some mistake, but where?
probability probability-theory conditional-probability
asked Jan 9 at 9:19
q126yq126y
239212
$endgroup$
"In drawing two cards from a deck (without returning the first card) what is the probability of two aces when you get at least one ace?"
I am aware there is exact same question answered here.
Probability of two aces when you get at least one ace
I know and understand then answers mentioned.
I would like to approach the problem in a different way.
Let us denote A = event that we get ace from first suite in case of success, i.e. when two aces are drawn., or alternately speaking, atleast one ace is from first suite
similarly B,C,D.
What we want to know is $P(Acup B cup Ccup D) $
$=sum_A ^D{P(X)} - {4 choose 2 }P(A cap B)$(due to symmetry) + {ignoring the other intersections since we can't have more than two cards}
$= 4*frac {3}{51} - {4 choose 2 } *frac{1}{{52 choose 2 }}$
$= 3/13$
which is not right answer, obviously I did some mistake, but where?
probability probability-theory conditional-probability
probability probability-theory conditional-probability
asked Jan 9 at 9:19
q126yq126y
239212
asked Jan 9 at 9:19
q126yq126y
239212
asked Jan 9 at 9:19
q126yq126y
239212
asked Jan 9 at 9:19
q126yq126y
239212
asked Jan 9 at 9:19
q126yq126y
239212
239212
$begingroup$
Where does $3/51$ come from. Surely the probability of drawing the ace of spades is $1/26$?
$endgroup$
– Lord Shark the Unknown
Jan 9 at 9:23
$begingroup$
@LordSharktheUnknown Given we have ace of spades, we have 3/51 probability that we will have two aces.
$endgroup$
– q126y
Jan 9 at 9:24
$begingroup$
@Peter I know the answer comes out to be 1/33 using bayes' theorem. I want to find gaps in my understanding by approaching it differently
$endgroup$
– q126y
Jan 9 at 9:26
$begingroup$
@Peter I know that. I have mentioned that answers to the question are available here math.stackexchange.com/questions/1390830/… I want to get more grip on my understanding by approaching it in this manner. What am I doing wrong with this approach? How should I correct it? Is formulation of the problem wrong, or something else. Thanks Edit: or why this approach doesn't work.
$endgroup$
– q126y
Jan 9 at 9:30
$begingroup$
OK, so you only want to analyze your concrete approach - You used the conditional probability within the formula for the probability of $P(Acup Bcup Ccup D)$. This might be the issue.
$endgroup$
– Peter
Jan 9 at 9:37
|
show 1 more comment
$begingroup$
Where does $3/51$ come from. Surely the probability of drawing the ace of spades is $1/26$?
$endgroup$
– Lord Shark the Unknown
Jan 9 at 9:23
$begingroup$
@LordSharktheUnknown Given we have ace of spades, we have 3/51 probability that we will have two aces.
$endgroup$
– q126y
Jan 9 at 9:24
$begingroup$
@Peter I know the answer comes out to be 1/33 using bayes' theorem. I want to find gaps in my understanding by approaching it differently
$endgroup$
– q126y
Jan 9 at 9:26
$begingroup$
@Peter I know that. I have mentioned that answers to the question are available here math.stackexchange.com/questions/1390830/… I want to get more grip on my understanding by approaching it in this manner. What am I doing wrong with this approach? How should I correct it? Is formulation of the problem wrong, or something else. Thanks Edit: or why this approach doesn't work.
$endgroup$
– q126y
Jan 9 at 9:30
$begingroup$
OK, so you only want to analyze your concrete approach - You used the conditional probability within the formula for the probability of $P(Acup Bcup Ccup D)$. This might be the issue.
$endgroup$
– Peter
Jan 9 at 9:37
$begingroup$
Where does $3/51$ come from. Surely the probability of drawing the ace of spades is $1/26$?
$endgroup$
– Lord Shark the Unknown
Jan 9 at 9:23
$begingroup$
Where does $3/51$ come from. Surely the probability of drawing the ace of spades is $1/26$?
$endgroup$
– Lord Shark the Unknown
Jan 9 at 9:23
$begingroup$
@LordSharktheUnknown Given we have ace of spades, we have 3/51 probability that we will have two aces.
$endgroup$
– q126y
Jan 9 at 9:24
$begingroup$
@LordSharktheUnknown Given we have ace of spades, we have 3/51 probability that we will have two aces.
$endgroup$
– q126y
Jan 9 at 9:24
$begingroup$
@Peter I know the answer comes out to be 1/33 using bayes' theorem. I want to find gaps in my understanding by approaching it differently
$endgroup$
– q126y
Jan 9 at 9:26
$begingroup$
@Peter I know the answer comes out to be 1/33 using bayes' theorem. I want to find gaps in my understanding by approaching it differently
$endgroup$
– q126y
Jan 9 at 9:26
$begingroup$
@Peter I know that. I have mentioned that answers to the question are available here math.stackexchange.com/questions/1390830/… I want to get more grip on my understanding by approaching it in this manner. What am I doing wrong with this approach? How should I correct it? Is formulation of the problem wrong, or something else. Thanks Edit: or why this approach doesn't work.
$endgroup$
– q126y
Jan 9 at 9:30
$begingroup$
@Peter I know that. I have mentioned that answers to the question are available here math.stackexchange.com/questions/1390830/… I want to get more grip on my understanding by approaching it in this manner. What am I doing wrong with this approach? How should I correct it? Is formulation of the problem wrong, or something else. Thanks Edit: or why this approach doesn't work.
$endgroup$
– q126y
Jan 9 at 9:30
$begingroup$
OK, so you only want to analyze your concrete approach - You used the conditional probability within the formula for the probability of $P(Acup Bcup Ccup D)$. This might be the issue.
$endgroup$
– Peter
Jan 9 at 9:37
$begingroup$
OK, so you only want to analyze your concrete approach - You used the conditional probability within the formula for the probability of $P(Acup Bcup Ccup D)$. This might be the issue.
$endgroup$
– Peter
Jan 9 at 9:37
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Your "$P(A)=frac{3}{51}$" is a conditional probability; the probability that we have a second ace, given that we have the ace of spades. Now you want to add that to $P(B)$, the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful. Those events don't live in the same space, because they don't have the same condition. Then, of course, $P(Acap B)$ is nonsense, because $A$ and $B$ aren't in the same space. Conditional probabilities create a new probability space by taking a slice of the old one and scaling up the measure - but of course, that means it's a different space for every condition.
Then again, when you wrote
Let us denote A = event that we get ace from first suite in case of success, i.e. when two aces are drawn.
That suggests that $A$ is the event that one of or two cards is the ace of spades given that both are aces. Now we're basically in a deck of four cards - we pick two, so the probability of getting a particular card is $frac12$. That doesn't seem like a particularly fruitful approach.
I've answered based on my best understanding of the post, but there's enough self-contradiction in there to muddy everything severely. You definitely need to work on writing clearly and saying what you mean.
answered Jan 9 at 9:46
jmerryjmerry
6,437718
$endgroup$
$begingroup$
Thanks sir. English is my second language. What i meant was the problem says "given atleast one ace", then event A= the given ace = ace of first suite. "Now you want to add that to P(B), the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful." I wanted to take 4 cases, for the "given ace", as the "given ace" can be of any one of 4 suits, and wanted to union all 4 cases. How should we approach the problem one suite at a time for the "given ace"(which I tried)?
$endgroup$
– q126y
Jan 9 at 9:57
$begingroup$
Honestly, all I can see is the conventional method. We find the probability that two random cards have at least one ace (possibly with an inclusion-exclusion sum based on which aces appear), the probability that the two cards are both aces, and then we divide to get the conditional probability asked for. It's hard to get anywhere by conditioning on something we haven't measured yet, and conditioning on other things is just asking for confusion.
$endgroup$
– jmerry
Jan 9 at 10:23
$begingroup$
Yes straightforward is the right way to go. I as a beginner try to see why other approaches fail, that provides valuable insights that are not presented in ready made form in the texts. Thank you.
$endgroup$
– q126y
Jan 9 at 10:28
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your "$P(A)=frac{3}{51}$" is a conditional probability; the probability that we have a second ace, given that we have the ace of spades. Now you want to add that to $P(B)$, the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful. Those events don't live in the same space, because they don't have the same condition. Then, of course, $P(Acap B)$ is nonsense, because $A$ and $B$ aren't in the same space. Conditional probabilities create a new probability space by taking a slice of the old one and scaling up the measure - but of course, that means it's a different space for every condition.
Then again, when you wrote
Let us denote A = event that we get ace from first suite in case of success, i.e. when two aces are drawn.
That suggests that $A$ is the event that one of or two cards is the ace of spades given that both are aces. Now we're basically in a deck of four cards - we pick two, so the probability of getting a particular card is $frac12$. That doesn't seem like a particularly fruitful approach.
I've answered based on my best understanding of the post, but there's enough self-contradiction in there to muddy everything severely. You definitely need to work on writing clearly and saying what you mean.
answered Jan 9 at 9:46
jmerryjmerry
6,437718
$endgroup$
$begingroup$
Thanks sir. English is my second language. What i meant was the problem says "given atleast one ace", then event A= the given ace = ace of first suite. "Now you want to add that to P(B), the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful." I wanted to take 4 cases, for the "given ace", as the "given ace" can be of any one of 4 suits, and wanted to union all 4 cases. How should we approach the problem one suite at a time for the "given ace"(which I tried)?
$endgroup$
– q126y
Jan 9 at 9:57
$begingroup$
Honestly, all I can see is the conventional method. We find the probability that two random cards have at least one ace (possibly with an inclusion-exclusion sum based on which aces appear), the probability that the two cards are both aces, and then we divide to get the conditional probability asked for. It's hard to get anywhere by conditioning on something we haven't measured yet, and conditioning on other things is just asking for confusion.
$endgroup$
– jmerry
Jan 9 at 10:23
$begingroup$
Yes straightforward is the right way to go. I as a beginner try to see why other approaches fail, that provides valuable insights that are not presented in ready made form in the texts. Thank you.
$endgroup$
– q126y
Jan 9 at 10:28
add a comment |
$begingroup$
Your "$P(A)=frac{3}{51}$" is a conditional probability; the probability that we have a second ace, given that we have the ace of spades. Now you want to add that to $P(B)$, the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful. Those events don't live in the same space, because they don't have the same condition. Then, of course, $P(Acap B)$ is nonsense, because $A$ and $B$ aren't in the same space. Conditional probabilities create a new probability space by taking a slice of the old one and scaling up the measure - but of course, that means it's a different space for every condition.
Then again, when you wrote
Let us denote A = event that we get ace from first suite in case of success, i.e. when two aces are drawn.
That suggests that $A$ is the event that one of or two cards is the ace of spades given that both are aces. Now we're basically in a deck of four cards - we pick two, so the probability of getting a particular card is $frac12$. That doesn't seem like a particularly fruitful approach.
I've answered based on my best understanding of the post, but there's enough self-contradiction in there to muddy everything severely. You definitely need to work on writing clearly and saying what you mean.
answered Jan 9 at 9:46
jmerryjmerry
6,437718
$endgroup$
$begingroup$
Thanks sir. English is my second language. What i meant was the problem says "given atleast one ace", then event A= the given ace = ace of first suite. "Now you want to add that to P(B), the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful." I wanted to take 4 cases, for the "given ace", as the "given ace" can be of any one of 4 suits, and wanted to union all 4 cases. How should we approach the problem one suite at a time for the "given ace"(which I tried)?
$endgroup$
– q126y
Jan 9 at 9:57
$begingroup$
Honestly, all I can see is the conventional method. We find the probability that two random cards have at least one ace (possibly with an inclusion-exclusion sum based on which aces appear), the probability that the two cards are both aces, and then we divide to get the conditional probability asked for. It's hard to get anywhere by conditioning on something we haven't measured yet, and conditioning on other things is just asking for confusion.
$endgroup$
– jmerry
Jan 9 at 10:23
$begingroup$
Yes straightforward is the right way to go. I as a beginner try to see why other approaches fail, that provides valuable insights that are not presented in ready made form in the texts. Thank you.
$endgroup$
– q126y
Jan 9 at 10:28
add a comment |
$begingroup$
Your "$P(A)=frac{3}{51}$" is a conditional probability; the probability that we have a second ace, given that we have the ace of spades. Now you want to add that to $P(B)$, the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful. Those events don't live in the same space, because they don't have the same condition. Then, of course, $P(Acap B)$ is nonsense, because $A$ and $B$ aren't in the same space. Conditional probabilities create a new probability space by taking a slice of the old one and scaling up the measure - but of course, that means it's a different space for every condition.
Then again, when you wrote
Let us denote A = event that we get ace from first suite in case of success, i.e. when two aces are drawn.
That suggests that $A$ is the event that one of or two cards is the ace of spades given that both are aces. Now we're basically in a deck of four cards - we pick two, so the probability of getting a particular card is $frac12$. That doesn't seem like a particularly fruitful approach.
I've answered based on my best understanding of the post, but there's enough self-contradiction in there to muddy everything severely. You definitely need to work on writing clearly and saying what you mean.
answered Jan 9 at 9:46
jmerryjmerry
6,437718
$endgroup$
Your "$P(A)=frac{3}{51}$" is a conditional probability; the probability that we have a second ace, given that we have the ace of spades. Now you want to add that to $P(B)$, the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful. Those events don't live in the same space, because they don't have the same condition. Then, of course, $P(Acap B)$ is nonsense, because $A$ and $B$ aren't in the same space. Conditional probabilities create a new probability space by taking a slice of the old one and scaling up the measure - but of course, that means it's a different space for every condition.
Then again, when you wrote
Let us denote A = event that we get ace from first suite in case of success, i.e. when two aces are drawn.
That suggests that $A$ is the event that one of or two cards is the ace of spades given that both are aces. Now we're basically in a deck of four cards - we pick two, so the probability of getting a particular card is $frac12$. That doesn't seem like a particularly fruitful approach.
I've answered based on my best understanding of the post, but there's enough self-contradiction in there to muddy everything severely. You definitely need to work on writing clearly and saying what you mean.
answered Jan 9 at 9:46
jmerryjmerry
6,437718
answered Jan 9 at 9:46
jmerryjmerry
6,437718
answered Jan 9 at 9:46
jmerryjmerry
6,437718
answered Jan 9 at 9:46
jmerryjmerry
6,437718
6,437718
$begingroup$
Thanks sir. English is my second language. What i meant was the problem says "given atleast one ace", then event A= the given ace = ace of first suite. "Now you want to add that to P(B), the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful." I wanted to take 4 cases, for the "given ace", as the "given ace" can be of any one of 4 suits, and wanted to union all 4 cases. How should we approach the problem one suite at a time for the "given ace"(which I tried)?
$endgroup$
– q126y
Jan 9 at 9:57
$begingroup$
Honestly, all I can see is the conventional method. We find the probability that two random cards have at least one ace (possibly with an inclusion-exclusion sum based on which aces appear), the probability that the two cards are both aces, and then we divide to get the conditional probability asked for. It's hard to get anywhere by conditioning on something we haven't measured yet, and conditioning on other things is just asking for confusion.
$endgroup$
– jmerry
Jan 9 at 10:23
$begingroup$
Yes straightforward is the right way to go. I as a beginner try to see why other approaches fail, that provides valuable insights that are not presented in ready made form in the texts. Thank you.
$endgroup$
– q126y
Jan 9 at 10:28
add a comment |
$begingroup$
Thanks sir. English is my second language. What i meant was the problem says "given atleast one ace", then event A= the given ace = ace of first suite. "Now you want to add that to P(B), the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful." I wanted to take 4 cases, for the "given ace", as the "given ace" can be of any one of 4 suits, and wanted to union all 4 cases. How should we approach the problem one suite at a time for the "given ace"(which I tried)?
$endgroup$
– q126y
Jan 9 at 9:57
$begingroup$
Honestly, all I can see is the conventional method. We find the probability that two random cards have at least one ace (possibly with an inclusion-exclusion sum based on which aces appear), the probability that the two cards are both aces, and then we divide to get the conditional probability asked for. It's hard to get anywhere by conditioning on something we haven't measured yet, and conditioning on other things is just asking for confusion.
$endgroup$
– jmerry
Jan 9 at 10:23
$begingroup$
Yes straightforward is the right way to go. I as a beginner try to see why other approaches fail, that provides valuable insights that are not presented in ready made form in the texts. Thank you.
$endgroup$
– q126y
Jan 9 at 10:28
$begingroup$
Thanks sir. English is my second language. What i meant was the problem says "given atleast one ace", then event A= the given ace = ace of first suite. "Now you want to add that to P(B), the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful." I wanted to take 4 cases, for the "given ace", as the "given ace" can be of any one of 4 suits, and wanted to union all 4 cases. How should we approach the problem one suite at a time for the "given ace"(which I tried)?
$endgroup$
– q126y
Jan 9 at 9:57
$begingroup$
Thanks sir. English is my second language. What i meant was the problem says "given atleast one ace", then event A= the given ace = ace of first suite. "Now you want to add that to P(B), the probability that we have a second ace given that we have the ace of clubs. This sum ... doesn't represent anything meaningful." I wanted to take 4 cases, for the "given ace", as the "given ace" can be of any one of 4 suits, and wanted to union all 4 cases. How should we approach the problem one suite at a time for the "given ace"(which I tried)?
$endgroup$
– q126y
Jan 9 at 9:57
$begingroup$
Honestly, all I can see is the conventional method. We find the probability that two random cards have at least one ace (possibly with an inclusion-exclusion sum based on which aces appear), the probability that the two cards are both aces, and then we divide to get the conditional probability asked for. It's hard to get anywhere by conditioning on something we haven't measured yet, and conditioning on other things is just asking for confusion.
$endgroup$
– jmerry
Jan 9 at 10:23
$begingroup$
Honestly, all I can see is the conventional method. We find the probability that two random cards have at least one ace (possibly with an inclusion-exclusion sum based on which aces appear), the probability that the two cards are both aces, and then we divide to get the conditional probability asked for. It's hard to get anywhere by conditioning on something we haven't measured yet, and conditioning on other things is just asking for confusion.
$endgroup$
– jmerry
Jan 9 at 10:23
$begingroup$
Yes straightforward is the right way to go. I as a beginner try to see why other approaches fail, that provides valuable insights that are not presented in ready made form in the texts. Thank you.
$endgroup$
– q126y
Jan 9 at 10:28
$begingroup$
Yes straightforward is the right way to go. I as a beginner try to see why other approaches fail, that provides valuable insights that are not presented in ready made form in the texts. Thank you.
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– q126y
Jan 9 at 10:28
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$begingroup$
Where does $3/51$ come from. Surely the probability of drawing the ace of spades is $1/26$?
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– Lord Shark the Unknown
Jan 9 at 9:23
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@LordSharktheUnknown Given we have ace of spades, we have 3/51 probability that we will have two aces.
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– q126y
Jan 9 at 9:24
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@Peter I know the answer comes out to be 1/33 using bayes' theorem. I want to find gaps in my understanding by approaching it differently
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– q126y
Jan 9 at 9:26
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@Peter I know that. I have mentioned that answers to the question are available here math.stackexchange.com/questions/1390830/… I want to get more grip on my understanding by approaching it in this manner. What am I doing wrong with this approach? How should I correct it? Is formulation of the problem wrong, or something else. Thanks Edit: or why this approach doesn't work.
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– q126y
Jan 9 at 9:30
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OK, so you only want to analyze your concrete approach - You used the conditional probability within the formula for the probability of $P(Acup Bcup Ccup D)$. This might be the issue.
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– Peter
Jan 9 at 9:37