Independence of coin flipscoin flips and markov chainProbability of exact match of peak of expected distribution for 1000 fair coin flips.Finding probability in the case of a biased coinUnfair coin tossed twiceA balanced coin is tossed four times.Markov inequality with coin flipsThe probability that a tossed coin lands heads is p. What are the possible values of p, and which of these values is plausible for a physical coin?Probability of a biased coin obtaining its second heads or seconds tails on the $6^textth$ tossWhat is the Probability that coin is tossed three timesDetermine a probability of coin is fair in repeated trial
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Independence of coin flips
coin flips and markov chainProbability of exact match of peak of expected distribution for 1000 fair coin flips.Finding probability in the case of a biased coinUnfair coin tossed twiceA balanced coin is tossed four times.Markov inequality with coin flipsThe probability that a tossed coin lands heads is p. What are the possible values of p, and which of these values is plausible for a physical coin?Probability of a biased coin obtaining its second heads or seconds tails on the $6^textth$ tossWhat is the Probability that coin is tossed three timesDetermine a probability of coin is fair in repeated trial
$begingroup$
A fair coin is tossed three times in succession. If at least one of
the tosses has resulted in Heads, what is the probability that at
least one of the tosses resulted in Tails?
My argument and answer: The coin was flipped thrice, and one of them was heads. So we have two unknown trials. The coin flips are all independent of each other, and so there is no useful information to be derived from the fact that one of them was heads. The probability of getting at least one Tails in these two trials is $ frac 12 + frac 12 - frac 14 = frac 34 $.
The given answer: $ frac 67 $. The answer proceeds as follows: Initially the sample space consists of 8 events. We now know that one of those events can't happen (TTT can't happen because one of them were heads).6 of the remaining 7 events have at least one tail, and so the probability is $ frac 67 $.
Why is my answer wrong? What am I missing?
probability
$endgroup$
add a comment |
$begingroup$
A fair coin is tossed three times in succession. If at least one of
the tosses has resulted in Heads, what is the probability that at
least one of the tosses resulted in Tails?
My argument and answer: The coin was flipped thrice, and one of them was heads. So we have two unknown trials. The coin flips are all independent of each other, and so there is no useful information to be derived from the fact that one of them was heads. The probability of getting at least one Tails in these two trials is $ frac 12 + frac 12 - frac 14 = frac 34 $.
The given answer: $ frac 67 $. The answer proceeds as follows: Initially the sample space consists of 8 events. We now know that one of those events can't happen (TTT can't happen because one of them were heads).6 of the remaining 7 events have at least one tail, and so the probability is $ frac 67 $.
Why is my answer wrong? What am I missing?
probability
$endgroup$
$begingroup$
Your answer is correct.
$endgroup$
– Jorge Fernández Hidalgo
Mar 28 at 21:08
2
$begingroup$
Your answer of 3/4 would be right, for example, if you tossed a penny, a nickel and a dime and were told "the penny came up heads." Same if you were told "nickel was H" or if you were told "dime was H". However these three conditions are not disjoint, so the relationship of the event "at least one T" to their union --i.e. the event "At least one H" -- is not the same as it is to the individual conditions, even though that relationship (i.e the conditional probability) is the same for all three of the individual conditions. Given "At least 1 H", 6 out of 7 equally likely outcomes have a T.
$endgroup$
– Ned
Mar 28 at 21:30
add a comment |
$begingroup$
A fair coin is tossed three times in succession. If at least one of
the tosses has resulted in Heads, what is the probability that at
least one of the tosses resulted in Tails?
My argument and answer: The coin was flipped thrice, and one of them was heads. So we have two unknown trials. The coin flips are all independent of each other, and so there is no useful information to be derived from the fact that one of them was heads. The probability of getting at least one Tails in these two trials is $ frac 12 + frac 12 - frac 14 = frac 34 $.
The given answer: $ frac 67 $. The answer proceeds as follows: Initially the sample space consists of 8 events. We now know that one of those events can't happen (TTT can't happen because one of them were heads).6 of the remaining 7 events have at least one tail, and so the probability is $ frac 67 $.
Why is my answer wrong? What am I missing?
probability
$endgroup$
A fair coin is tossed three times in succession. If at least one of
the tosses has resulted in Heads, what is the probability that at
least one of the tosses resulted in Tails?
My argument and answer: The coin was flipped thrice, and one of them was heads. So we have two unknown trials. The coin flips are all independent of each other, and so there is no useful information to be derived from the fact that one of them was heads. The probability of getting at least one Tails in these two trials is $ frac 12 + frac 12 - frac 14 = frac 34 $.
The given answer: $ frac 67 $. The answer proceeds as follows: Initially the sample space consists of 8 events. We now know that one of those events can't happen (TTT can't happen because one of them were heads).6 of the remaining 7 events have at least one tail, and so the probability is $ frac 67 $.
Why is my answer wrong? What am I missing?
probability
probability
asked Mar 28 at 21:03
WorldGovWorldGov
345211
345211
$begingroup$
Your answer is correct.
$endgroup$
– Jorge Fernández Hidalgo
Mar 28 at 21:08
2
$begingroup$
Your answer of 3/4 would be right, for example, if you tossed a penny, a nickel and a dime and were told "the penny came up heads." Same if you were told "nickel was H" or if you were told "dime was H". However these three conditions are not disjoint, so the relationship of the event "at least one T" to their union --i.e. the event "At least one H" -- is not the same as it is to the individual conditions, even though that relationship (i.e the conditional probability) is the same for all three of the individual conditions. Given "At least 1 H", 6 out of 7 equally likely outcomes have a T.
$endgroup$
– Ned
Mar 28 at 21:30
add a comment |
$begingroup$
Your answer is correct.
$endgroup$
– Jorge Fernández Hidalgo
Mar 28 at 21:08
2
$begingroup$
Your answer of 3/4 would be right, for example, if you tossed a penny, a nickel and a dime and were told "the penny came up heads." Same if you were told "nickel was H" or if you were told "dime was H". However these three conditions are not disjoint, so the relationship of the event "at least one T" to their union --i.e. the event "At least one H" -- is not the same as it is to the individual conditions, even though that relationship (i.e the conditional probability) is the same for all three of the individual conditions. Given "At least 1 H", 6 out of 7 equally likely outcomes have a T.
$endgroup$
– Ned
Mar 28 at 21:30
$begingroup$
Your answer is correct.
$endgroup$
– Jorge Fernández Hidalgo
Mar 28 at 21:08
$begingroup$
Your answer is correct.
$endgroup$
– Jorge Fernández Hidalgo
Mar 28 at 21:08
2
2
$begingroup$
Your answer of 3/4 would be right, for example, if you tossed a penny, a nickel and a dime and were told "the penny came up heads." Same if you were told "nickel was H" or if you were told "dime was H". However these three conditions are not disjoint, so the relationship of the event "at least one T" to their union --i.e. the event "At least one H" -- is not the same as it is to the individual conditions, even though that relationship (i.e the conditional probability) is the same for all three of the individual conditions. Given "At least 1 H", 6 out of 7 equally likely outcomes have a T.
$endgroup$
– Ned
Mar 28 at 21:30
$begingroup$
Your answer of 3/4 would be right, for example, if you tossed a penny, a nickel and a dime and were told "the penny came up heads." Same if you were told "nickel was H" or if you were told "dime was H". However these three conditions are not disjoint, so the relationship of the event "at least one T" to their union --i.e. the event "At least one H" -- is not the same as it is to the individual conditions, even though that relationship (i.e the conditional probability) is the same for all three of the individual conditions. Given "At least 1 H", 6 out of 7 equally likely outcomes have a T.
$endgroup$
– Ned
Mar 28 at 21:30
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The answer $frac67$ is correct. Whatever the issue in your reasoning, I feel it must lie in the statement "there is no useful information" in saying one of the 3 flips was heads. To me it seems there is a difference between reading off the results of coins already flipped and predicting (as you do in your argument) the results of future flips based on previous ones. Clearly, as you mention, each flip is independent, but perhaps information about some of the flips is useful if all of the flips have already been made.
To illustrate this point, I think it helps to take this problem to the extreme. Suppose we perform a million coin flips. I don't tell you the results of each coin flip, but I do tell you that 999,999 of the flips yielded heads. I now ask you to guess the result of the remaining flip. What should you answer? (I recommend thinking about this yourself, then look at my answer below).
|
|
|
|
|
|
|
|
Answer: You may suspect that you have a 50-50 shot, but that is not the case. The crucial point is I did not tell you which of the 999,999 coins were heads. Had I made this specification it would be equally likely. However, there are, in fact, one million possible sequences of flips in which only one is tails. Here they are:
beginequation
T, H, H, H, ..., H \
H, T, H, H, ..., H \
H, H, T, H, ..., H \
. \
. \
. \
H, H, H, H, ..., T \
endequation
In contrast, only one sequence has all heads. Assuming sequences of flips are equally likely, we must conclude that guessing tails is the appropriate choice. Notice that we cannot solve this problem correctly by reducing the size of our string of flips to one, similar to your method. We had to consider the flips as part of a larger sequence. I hope this helps!
$endgroup$
$begingroup$
Thank you. Could you please help me strengthen my understanding of this? Suppose someone says the first coin flip was Heads, and asked me what the probability of getting at least one tails in the next two flips was, would my answers (0.75) be corrected then?
$endgroup$
– WorldGov
Mar 29 at 7:39
1
$begingroup$
Yup, that's correct!
$endgroup$
– Jacob
2 days ago
add a comment |
$begingroup$
The coin flips are all independent of each other,
Here's the error in your reasoning.
The coin flips are not conditionally independent under constraint that at least one is a head.
Let $H_k$ be the event that trial $k$ shows a head, $T_k$ be the event that it is a tail, and the evidence that at least one among them shows a head is: $E = H_1cup H_2cup H_3$
Obviously under that condition, each particular trial will still have a non-zero probability that it might be a tail.
$$mathsf P(T_1mid E)~mathsf P(T_2mid E)~mathsf P(T_3mid E) >0$$
However, there the conditional probability that all trials are tails is zero under the condition that at least one of them is a head.
$$mathsf P(T_1cap T_2cap T_3mid E)=0$$
$endgroup$
add a comment |
$begingroup$
Well, let's look at the problem. It's asking the odds of flipping any tail given that you flipped at least one head, or $P(T>0|X>0)$ Using the definition of conditional probability, we can say that $P(T>0|X>0)=P(T>0,X>0)/P(X>0)$. Then, using some identities, we have that $P(T>0,X>0)=1-P(T=0)-P(X=0)$ and $P(X>0)=1-P(X=0)$.
To put it in words, the probability that our coin toss triple contains one head and one tail is 1 minus the odds of it containing no heads or no tails, and the probability that it held at least one head is 1 minus the odds that it contained no heads.
Then, doing the remaining math:
$$frac1-P(T=0)-P(X=0)1-P(X=0)=frac1-frac18-frac181-frac18=fracfrac68frac78=frac67$$
Your problem is that you specifically looked at the case when one head was flipped- you did not account for other cases.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer $frac67$ is correct. Whatever the issue in your reasoning, I feel it must lie in the statement "there is no useful information" in saying one of the 3 flips was heads. To me it seems there is a difference between reading off the results of coins already flipped and predicting (as you do in your argument) the results of future flips based on previous ones. Clearly, as you mention, each flip is independent, but perhaps information about some of the flips is useful if all of the flips have already been made.
To illustrate this point, I think it helps to take this problem to the extreme. Suppose we perform a million coin flips. I don't tell you the results of each coin flip, but I do tell you that 999,999 of the flips yielded heads. I now ask you to guess the result of the remaining flip. What should you answer? (I recommend thinking about this yourself, then look at my answer below).
|
|
|
|
|
|
|
|
Answer: You may suspect that you have a 50-50 shot, but that is not the case. The crucial point is I did not tell you which of the 999,999 coins were heads. Had I made this specification it would be equally likely. However, there are, in fact, one million possible sequences of flips in which only one is tails. Here they are:
beginequation
T, H, H, H, ..., H \
H, T, H, H, ..., H \
H, H, T, H, ..., H \
. \
. \
. \
H, H, H, H, ..., T \
endequation
In contrast, only one sequence has all heads. Assuming sequences of flips are equally likely, we must conclude that guessing tails is the appropriate choice. Notice that we cannot solve this problem correctly by reducing the size of our string of flips to one, similar to your method. We had to consider the flips as part of a larger sequence. I hope this helps!
$endgroup$
$begingroup$
Thank you. Could you please help me strengthen my understanding of this? Suppose someone says the first coin flip was Heads, and asked me what the probability of getting at least one tails in the next two flips was, would my answers (0.75) be corrected then?
$endgroup$
– WorldGov
Mar 29 at 7:39
1
$begingroup$
Yup, that's correct!
$endgroup$
– Jacob
2 days ago
add a comment |
$begingroup$
The answer $frac67$ is correct. Whatever the issue in your reasoning, I feel it must lie in the statement "there is no useful information" in saying one of the 3 flips was heads. To me it seems there is a difference between reading off the results of coins already flipped and predicting (as you do in your argument) the results of future flips based on previous ones. Clearly, as you mention, each flip is independent, but perhaps information about some of the flips is useful if all of the flips have already been made.
To illustrate this point, I think it helps to take this problem to the extreme. Suppose we perform a million coin flips. I don't tell you the results of each coin flip, but I do tell you that 999,999 of the flips yielded heads. I now ask you to guess the result of the remaining flip. What should you answer? (I recommend thinking about this yourself, then look at my answer below).
|
|
|
|
|
|
|
|
Answer: You may suspect that you have a 50-50 shot, but that is not the case. The crucial point is I did not tell you which of the 999,999 coins were heads. Had I made this specification it would be equally likely. However, there are, in fact, one million possible sequences of flips in which only one is tails. Here they are:
beginequation
T, H, H, H, ..., H \
H, T, H, H, ..., H \
H, H, T, H, ..., H \
. \
. \
. \
H, H, H, H, ..., T \
endequation
In contrast, only one sequence has all heads. Assuming sequences of flips are equally likely, we must conclude that guessing tails is the appropriate choice. Notice that we cannot solve this problem correctly by reducing the size of our string of flips to one, similar to your method. We had to consider the flips as part of a larger sequence. I hope this helps!
$endgroup$
$begingroup$
Thank you. Could you please help me strengthen my understanding of this? Suppose someone says the first coin flip was Heads, and asked me what the probability of getting at least one tails in the next two flips was, would my answers (0.75) be corrected then?
$endgroup$
– WorldGov
Mar 29 at 7:39
1
$begingroup$
Yup, that's correct!
$endgroup$
– Jacob
2 days ago
add a comment |
$begingroup$
The answer $frac67$ is correct. Whatever the issue in your reasoning, I feel it must lie in the statement "there is no useful information" in saying one of the 3 flips was heads. To me it seems there is a difference between reading off the results of coins already flipped and predicting (as you do in your argument) the results of future flips based on previous ones. Clearly, as you mention, each flip is independent, but perhaps information about some of the flips is useful if all of the flips have already been made.
To illustrate this point, I think it helps to take this problem to the extreme. Suppose we perform a million coin flips. I don't tell you the results of each coin flip, but I do tell you that 999,999 of the flips yielded heads. I now ask you to guess the result of the remaining flip. What should you answer? (I recommend thinking about this yourself, then look at my answer below).
|
|
|
|
|
|
|
|
Answer: You may suspect that you have a 50-50 shot, but that is not the case. The crucial point is I did not tell you which of the 999,999 coins were heads. Had I made this specification it would be equally likely. However, there are, in fact, one million possible sequences of flips in which only one is tails. Here they are:
beginequation
T, H, H, H, ..., H \
H, T, H, H, ..., H \
H, H, T, H, ..., H \
. \
. \
. \
H, H, H, H, ..., T \
endequation
In contrast, only one sequence has all heads. Assuming sequences of flips are equally likely, we must conclude that guessing tails is the appropriate choice. Notice that we cannot solve this problem correctly by reducing the size of our string of flips to one, similar to your method. We had to consider the flips as part of a larger sequence. I hope this helps!
$endgroup$
The answer $frac67$ is correct. Whatever the issue in your reasoning, I feel it must lie in the statement "there is no useful information" in saying one of the 3 flips was heads. To me it seems there is a difference between reading off the results of coins already flipped and predicting (as you do in your argument) the results of future flips based on previous ones. Clearly, as you mention, each flip is independent, but perhaps information about some of the flips is useful if all of the flips have already been made.
To illustrate this point, I think it helps to take this problem to the extreme. Suppose we perform a million coin flips. I don't tell you the results of each coin flip, but I do tell you that 999,999 of the flips yielded heads. I now ask you to guess the result of the remaining flip. What should you answer? (I recommend thinking about this yourself, then look at my answer below).
|
|
|
|
|
|
|
|
Answer: You may suspect that you have a 50-50 shot, but that is not the case. The crucial point is I did not tell you which of the 999,999 coins were heads. Had I made this specification it would be equally likely. However, there are, in fact, one million possible sequences of flips in which only one is tails. Here they are:
beginequation
T, H, H, H, ..., H \
H, T, H, H, ..., H \
H, H, T, H, ..., H \
. \
. \
. \
H, H, H, H, ..., T \
endequation
In contrast, only one sequence has all heads. Assuming sequences of flips are equally likely, we must conclude that guessing tails is the appropriate choice. Notice that we cannot solve this problem correctly by reducing the size of our string of flips to one, similar to your method. We had to consider the flips as part of a larger sequence. I hope this helps!
answered Mar 28 at 22:04
JacobJacob
363
363
$begingroup$
Thank you. Could you please help me strengthen my understanding of this? Suppose someone says the first coin flip was Heads, and asked me what the probability of getting at least one tails in the next two flips was, would my answers (0.75) be corrected then?
$endgroup$
– WorldGov
Mar 29 at 7:39
1
$begingroup$
Yup, that's correct!
$endgroup$
– Jacob
2 days ago
add a comment |
$begingroup$
Thank you. Could you please help me strengthen my understanding of this? Suppose someone says the first coin flip was Heads, and asked me what the probability of getting at least one tails in the next two flips was, would my answers (0.75) be corrected then?
$endgroup$
– WorldGov
Mar 29 at 7:39
1
$begingroup$
Yup, that's correct!
$endgroup$
– Jacob
2 days ago
$begingroup$
Thank you. Could you please help me strengthen my understanding of this? Suppose someone says the first coin flip was Heads, and asked me what the probability of getting at least one tails in the next two flips was, would my answers (0.75) be corrected then?
$endgroup$
– WorldGov
Mar 29 at 7:39
$begingroup$
Thank you. Could you please help me strengthen my understanding of this? Suppose someone says the first coin flip was Heads, and asked me what the probability of getting at least one tails in the next two flips was, would my answers (0.75) be corrected then?
$endgroup$
– WorldGov
Mar 29 at 7:39
1
1
$begingroup$
Yup, that's correct!
$endgroup$
– Jacob
2 days ago
$begingroup$
Yup, that's correct!
$endgroup$
– Jacob
2 days ago
add a comment |
$begingroup$
The coin flips are all independent of each other,
Here's the error in your reasoning.
The coin flips are not conditionally independent under constraint that at least one is a head.
Let $H_k$ be the event that trial $k$ shows a head, $T_k$ be the event that it is a tail, and the evidence that at least one among them shows a head is: $E = H_1cup H_2cup H_3$
Obviously under that condition, each particular trial will still have a non-zero probability that it might be a tail.
$$mathsf P(T_1mid E)~mathsf P(T_2mid E)~mathsf P(T_3mid E) >0$$
However, there the conditional probability that all trials are tails is zero under the condition that at least one of them is a head.
$$mathsf P(T_1cap T_2cap T_3mid E)=0$$
$endgroup$
add a comment |
$begingroup$
The coin flips are all independent of each other,
Here's the error in your reasoning.
The coin flips are not conditionally independent under constraint that at least one is a head.
Let $H_k$ be the event that trial $k$ shows a head, $T_k$ be the event that it is a tail, and the evidence that at least one among them shows a head is: $E = H_1cup H_2cup H_3$
Obviously under that condition, each particular trial will still have a non-zero probability that it might be a tail.
$$mathsf P(T_1mid E)~mathsf P(T_2mid E)~mathsf P(T_3mid E) >0$$
However, there the conditional probability that all trials are tails is zero under the condition that at least one of them is a head.
$$mathsf P(T_1cap T_2cap T_3mid E)=0$$
$endgroup$
add a comment |
$begingroup$
The coin flips are all independent of each other,
Here's the error in your reasoning.
The coin flips are not conditionally independent under constraint that at least one is a head.
Let $H_k$ be the event that trial $k$ shows a head, $T_k$ be the event that it is a tail, and the evidence that at least one among them shows a head is: $E = H_1cup H_2cup H_3$
Obviously under that condition, each particular trial will still have a non-zero probability that it might be a tail.
$$mathsf P(T_1mid E)~mathsf P(T_2mid E)~mathsf P(T_3mid E) >0$$
However, there the conditional probability that all trials are tails is zero under the condition that at least one of them is a head.
$$mathsf P(T_1cap T_2cap T_3mid E)=0$$
$endgroup$
The coin flips are all independent of each other,
Here's the error in your reasoning.
The coin flips are not conditionally independent under constraint that at least one is a head.
Let $H_k$ be the event that trial $k$ shows a head, $T_k$ be the event that it is a tail, and the evidence that at least one among them shows a head is: $E = H_1cup H_2cup H_3$
Obviously under that condition, each particular trial will still have a non-zero probability that it might be a tail.
$$mathsf P(T_1mid E)~mathsf P(T_2mid E)~mathsf P(T_3mid E) >0$$
However, there the conditional probability that all trials are tails is zero under the condition that at least one of them is a head.
$$mathsf P(T_1cap T_2cap T_3mid E)=0$$
answered Mar 28 at 23:24
Graham KempGraham Kemp
87.6k43578
87.6k43578
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Well, let's look at the problem. It's asking the odds of flipping any tail given that you flipped at least one head, or $P(T>0|X>0)$ Using the definition of conditional probability, we can say that $P(T>0|X>0)=P(T>0,X>0)/P(X>0)$. Then, using some identities, we have that $P(T>0,X>0)=1-P(T=0)-P(X=0)$ and $P(X>0)=1-P(X=0)$.
To put it in words, the probability that our coin toss triple contains one head and one tail is 1 minus the odds of it containing no heads or no tails, and the probability that it held at least one head is 1 minus the odds that it contained no heads.
Then, doing the remaining math:
$$frac1-P(T=0)-P(X=0)1-P(X=0)=frac1-frac18-frac181-frac18=fracfrac68frac78=frac67$$
Your problem is that you specifically looked at the case when one head was flipped- you did not account for other cases.
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add a comment |
$begingroup$
Well, let's look at the problem. It's asking the odds of flipping any tail given that you flipped at least one head, or $P(T>0|X>0)$ Using the definition of conditional probability, we can say that $P(T>0|X>0)=P(T>0,X>0)/P(X>0)$. Then, using some identities, we have that $P(T>0,X>0)=1-P(T=0)-P(X=0)$ and $P(X>0)=1-P(X=0)$.
To put it in words, the probability that our coin toss triple contains one head and one tail is 1 minus the odds of it containing no heads or no tails, and the probability that it held at least one head is 1 minus the odds that it contained no heads.
Then, doing the remaining math:
$$frac1-P(T=0)-P(X=0)1-P(X=0)=frac1-frac18-frac181-frac18=fracfrac68frac78=frac67$$
Your problem is that you specifically looked at the case when one head was flipped- you did not account for other cases.
$endgroup$
add a comment |
$begingroup$
Well, let's look at the problem. It's asking the odds of flipping any tail given that you flipped at least one head, or $P(T>0|X>0)$ Using the definition of conditional probability, we can say that $P(T>0|X>0)=P(T>0,X>0)/P(X>0)$. Then, using some identities, we have that $P(T>0,X>0)=1-P(T=0)-P(X=0)$ and $P(X>0)=1-P(X=0)$.
To put it in words, the probability that our coin toss triple contains one head and one tail is 1 minus the odds of it containing no heads or no tails, and the probability that it held at least one head is 1 minus the odds that it contained no heads.
Then, doing the remaining math:
$$frac1-P(T=0)-P(X=0)1-P(X=0)=frac1-frac18-frac181-frac18=fracfrac68frac78=frac67$$
Your problem is that you specifically looked at the case when one head was flipped- you did not account for other cases.
$endgroup$
Well, let's look at the problem. It's asking the odds of flipping any tail given that you flipped at least one head, or $P(T>0|X>0)$ Using the definition of conditional probability, we can say that $P(T>0|X>0)=P(T>0,X>0)/P(X>0)$. Then, using some identities, we have that $P(T>0,X>0)=1-P(T=0)-P(X=0)$ and $P(X>0)=1-P(X=0)$.
To put it in words, the probability that our coin toss triple contains one head and one tail is 1 minus the odds of it containing no heads or no tails, and the probability that it held at least one head is 1 minus the odds that it contained no heads.
Then, doing the remaining math:
$$frac1-P(T=0)-P(X=0)1-P(X=0)=frac1-frac18-frac181-frac18=fracfrac68frac78=frac67$$
Your problem is that you specifically looked at the case when one head was flipped- you did not account for other cases.
edited yesterday
answered Mar 28 at 21:29
IspilIspil
6071512
6071512
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Your answer is correct.
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– Jorge Fernández Hidalgo
Mar 28 at 21:08
2
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Your answer of 3/4 would be right, for example, if you tossed a penny, a nickel and a dime and were told "the penny came up heads." Same if you were told "nickel was H" or if you were told "dime was H". However these three conditions are not disjoint, so the relationship of the event "at least one T" to their union --i.e. the event "At least one H" -- is not the same as it is to the individual conditions, even though that relationship (i.e the conditional probability) is the same for all three of the individual conditions. Given "At least 1 H", 6 out of 7 equally likely outcomes have a T.
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– Ned
Mar 28 at 21:30