Conditional probability exercise does not match my intuition
$begingroup$
I have been given the following exercise:
Consider the experiment of tossing a fair coin 7 times. Find the
probability of getting a prime number of heads given that heads occurs
on at least 6 of the tosses.
Intuitively, I would say that the only prime number possible, given that 6 heads occur is 7, hence the experiment returned heads either 6 or 7 times, thus the probability should be 1/2.
However, introducing the events:
$$
A := text {the experiment returned a prime number of heads} \
B:= text {the experiment returned at least 6 heads}
$$
then we are clearly asked for the conditional probability $mathbb{P}(A|B)$.
Let me compute:
$$
mathbb{P}(Acap B)=mathbb{P}(A)=frac{1}{2^7}.
$$
For $mathbb{P}(B)$, I would do:
$$
mathbb{P}(B) = frac{1}{2^7} + 7 frac{1}{2^7},
$$
where the first term in the sum considers the case that exactly 7 heads appear, while the second term considers the case with exactly 6 heads and exactly one tail (the tail can appear on any of the 7 tosses).
The result would give:
$$
mathbb{P}(A|B) = frac{mathbb{P}(Acap B)}{mathbb{P}(B)}=frac{1}{8},
$$
which however is in contrast with my first intuition. Is my computation correct?
probability
asked Jan 12 at 17:53
J. D.J. D.
1898
$endgroup$
add a comment |
$begingroup$
I have been given the following exercise:
Consider the experiment of tossing a fair coin 7 times. Find the
probability of getting a prime number of heads given that heads occurs
on at least 6 of the tosses.
Intuitively, I would say that the only prime number possible, given that 6 heads occur is 7, hence the experiment returned heads either 6 or 7 times, thus the probability should be 1/2.
However, introducing the events:
$$
A := text {the experiment returned a prime number of heads} \
B:= text {the experiment returned at least 6 heads}
$$
then we are clearly asked for the conditional probability $mathbb{P}(A|B)$.
Let me compute:
$$
mathbb{P}(Acap B)=mathbb{P}(A)=frac{1}{2^7}.
$$
For $mathbb{P}(B)$, I would do:
$$
mathbb{P}(B) = frac{1}{2^7} + 7 frac{1}{2^7},
$$
where the first term in the sum considers the case that exactly 7 heads appear, while the second term considers the case with exactly 6 heads and exactly one tail (the tail can appear on any of the 7 tosses).
The result would give:
$$
mathbb{P}(A|B) = frac{mathbb{P}(Acap B)}{mathbb{P}(B)}=frac{1}{8},
$$
which however is in contrast with my first intuition. Is my computation correct?
probability
asked Jan 12 at 17:53
J. D.J. D.
1898
$endgroup$
1
$begingroup$
This reminds me a bit of the argument that tomorrow the sun will either rise or not rise; therefore the probability that the sun will rise tomorrow is $1/2$.
$endgroup$
– littleO
Jan 12 at 18:04
add a comment |
$begingroup$
I have been given the following exercise:
Consider the experiment of tossing a fair coin 7 times. Find the
probability of getting a prime number of heads given that heads occurs
on at least 6 of the tosses.
Intuitively, I would say that the only prime number possible, given that 6 heads occur is 7, hence the experiment returned heads either 6 or 7 times, thus the probability should be 1/2.
However, introducing the events:
$$
A := text {the experiment returned a prime number of heads} \
B:= text {the experiment returned at least 6 heads}
$$
then we are clearly asked for the conditional probability $mathbb{P}(A|B)$.
Let me compute:
$$
mathbb{P}(Acap B)=mathbb{P}(A)=frac{1}{2^7}.
$$
For $mathbb{P}(B)$, I would do:
$$
mathbb{P}(B) = frac{1}{2^7} + 7 frac{1}{2^7},
$$
where the first term in the sum considers the case that exactly 7 heads appear, while the second term considers the case with exactly 6 heads and exactly one tail (the tail can appear on any of the 7 tosses).
The result would give:
$$
mathbb{P}(A|B) = frac{mathbb{P}(Acap B)}{mathbb{P}(B)}=frac{1}{8},
$$
which however is in contrast with my first intuition. Is my computation correct?
probability
asked Jan 12 at 17:53
J. D.J. D.
1898
$endgroup$
I have been given the following exercise:
Consider the experiment of tossing a fair coin 7 times. Find the
probability of getting a prime number of heads given that heads occurs
on at least 6 of the tosses.
Intuitively, I would say that the only prime number possible, given that 6 heads occur is 7, hence the experiment returned heads either 6 or 7 times, thus the probability should be 1/2.
However, introducing the events:
$$
A := text {the experiment returned a prime number of heads} \
B:= text {the experiment returned at least 6 heads}
$$
then we are clearly asked for the conditional probability $mathbb{P}(A|B)$.
Let me compute:
$$
mathbb{P}(Acap B)=mathbb{P}(A)=frac{1}{2^7}.
$$
For $mathbb{P}(B)$, I would do:
$$
mathbb{P}(B) = frac{1}{2^7} + 7 frac{1}{2^7},
$$
where the first term in the sum considers the case that exactly 7 heads appear, while the second term considers the case with exactly 6 heads and exactly one tail (the tail can appear on any of the 7 tosses).
The result would give:
$$
mathbb{P}(A|B) = frac{mathbb{P}(Acap B)}{mathbb{P}(B)}=frac{1}{8},
$$
which however is in contrast with my first intuition. Is my computation correct?
probability
probability
asked Jan 12 at 17:53
J. D.J. D.
1898
asked Jan 12 at 17:53
J. D.J. D.
1898
asked Jan 12 at 17:53
J. D.J. D.
1898
asked Jan 12 at 17:53
J. D.J. D.
1898
asked Jan 12 at 17:53
J. D.J. D.
1898
1898
1
$begingroup$
This reminds me a bit of the argument that tomorrow the sun will either rise or not rise; therefore the probability that the sun will rise tomorrow is $1/2$.
$endgroup$
– littleO
Jan 12 at 18:04
add a comment |
1
$begingroup$
This reminds me a bit of the argument that tomorrow the sun will either rise or not rise; therefore the probability that the sun will rise tomorrow is $1/2$.
$endgroup$
– littleO
Jan 12 at 18:04
1
1
$begingroup$
This reminds me a bit of the argument that tomorrow the sun will either rise or not rise; therefore the probability that the sun will rise tomorrow is $1/2$.
$endgroup$
– littleO
Jan 12 at 18:04
$begingroup$
This reminds me a bit of the argument that tomorrow the sun will either rise or not rise; therefore the probability that the sun will rise tomorrow is $1/2$.
$endgroup$
– littleO
Jan 12 at 18:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your intuition is correct this far:
Intuitively, I would say that the only prime number possible, given that 6 heads occur is 7, hence the experiment returned heads either 6 or 7 times,
Where you go wrong is in the part that comes after.
You assume since there are two possible outcomes,
they must be equally likely.
But you actually are much more likely to toss $6$ heads on $7$ coins than to toss all heads.
In fact, $6$ heads is exactly $7$ times as likely as all heads,
because only one sequence of tosses is all heads whereas there are $7$ different sequences that have exactly $6$ heads, and the sequences are equally likely.
answered Jan 12 at 18:00
David KDavid K
55.7k345121
$endgroup$
add a comment |
$begingroup$
Your intuition is wrong. I'm not sure how you are getting $1/2.$ You may be thinking that once we have $6$ heads, the probability is $1/2$ that the next toss will also be heads, but this is valid only if we are told that the first $6$ tosses (or some other specified $6$ tosses) are heads.
Otherwise, you may be thinking that since there are only two possibilities, ($6$ heads or $7$ heads) they are equally likely, but we are actually $7$ times as likely to get exactly $6$ heads as $7$ heads.
answered Jan 12 at 18:03
saulspatzsaulspatz
17.3k31435
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add a comment |
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2 Answers
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$begingroup$
Your intuition is correct this far:
Intuitively, I would say that the only prime number possible, given that 6 heads occur is 7, hence the experiment returned heads either 6 or 7 times,
Where you go wrong is in the part that comes after.
You assume since there are two possible outcomes,
they must be equally likely.
But you actually are much more likely to toss $6$ heads on $7$ coins than to toss all heads.
In fact, $6$ heads is exactly $7$ times as likely as all heads,
because only one sequence of tosses is all heads whereas there are $7$ different sequences that have exactly $6$ heads, and the sequences are equally likely.
answered Jan 12 at 18:00
David KDavid K
55.7k345121
$endgroup$
add a comment |
$begingroup$
Your intuition is correct this far:
Intuitively, I would say that the only prime number possible, given that 6 heads occur is 7, hence the experiment returned heads either 6 or 7 times,
Where you go wrong is in the part that comes after.
You assume since there are two possible outcomes,
they must be equally likely.
But you actually are much more likely to toss $6$ heads on $7$ coins than to toss all heads.
In fact, $6$ heads is exactly $7$ times as likely as all heads,
because only one sequence of tosses is all heads whereas there are $7$ different sequences that have exactly $6$ heads, and the sequences are equally likely.
answered Jan 12 at 18:00
David KDavid K
55.7k345121
$endgroup$
add a comment |
$begingroup$
Your intuition is correct this far:
Intuitively, I would say that the only prime number possible, given that 6 heads occur is 7, hence the experiment returned heads either 6 or 7 times,
Where you go wrong is in the part that comes after.
You assume since there are two possible outcomes,
they must be equally likely.
But you actually are much more likely to toss $6$ heads on $7$ coins than to toss all heads.
In fact, $6$ heads is exactly $7$ times as likely as all heads,
because only one sequence of tosses is all heads whereas there are $7$ different sequences that have exactly $6$ heads, and the sequences are equally likely.
answered Jan 12 at 18:00
David KDavid K
55.7k345121
$endgroup$
Your intuition is correct this far:
Intuitively, I would say that the only prime number possible, given that 6 heads occur is 7, hence the experiment returned heads either 6 or 7 times,
Where you go wrong is in the part that comes after.
You assume since there are two possible outcomes,
they must be equally likely.
But you actually are much more likely to toss $6$ heads on $7$ coins than to toss all heads.
In fact, $6$ heads is exactly $7$ times as likely as all heads,
because only one sequence of tosses is all heads whereas there are $7$ different sequences that have exactly $6$ heads, and the sequences are equally likely.
answered Jan 12 at 18:00
David KDavid K
55.7k345121
answered Jan 12 at 18:00
David KDavid K
55.7k345121
answered Jan 12 at 18:00
David KDavid K
55.7k345121
answered Jan 12 at 18:00
David KDavid K
55.7k345121
55.7k345121
add a comment |
add a comment |
$begingroup$
Your intuition is wrong. I'm not sure how you are getting $1/2.$ You may be thinking that once we have $6$ heads, the probability is $1/2$ that the next toss will also be heads, but this is valid only if we are told that the first $6$ tosses (or some other specified $6$ tosses) are heads.
Otherwise, you may be thinking that since there are only two possibilities, ($6$ heads or $7$ heads) they are equally likely, but we are actually $7$ times as likely to get exactly $6$ heads as $7$ heads.
answered Jan 12 at 18:03
saulspatzsaulspatz
17.3k31435
$endgroup$
add a comment |
$begingroup$
Your intuition is wrong. I'm not sure how you are getting $1/2.$ You may be thinking that once we have $6$ heads, the probability is $1/2$ that the next toss will also be heads, but this is valid only if we are told that the first $6$ tosses (or some other specified $6$ tosses) are heads.
Otherwise, you may be thinking that since there are only two possibilities, ($6$ heads or $7$ heads) they are equally likely, but we are actually $7$ times as likely to get exactly $6$ heads as $7$ heads.
answered Jan 12 at 18:03
saulspatzsaulspatz
17.3k31435
$endgroup$
add a comment |
$begingroup$
Your intuition is wrong. I'm not sure how you are getting $1/2.$ You may be thinking that once we have $6$ heads, the probability is $1/2$ that the next toss will also be heads, but this is valid only if we are told that the first $6$ tosses (or some other specified $6$ tosses) are heads.
Otherwise, you may be thinking that since there are only two possibilities, ($6$ heads or $7$ heads) they are equally likely, but we are actually $7$ times as likely to get exactly $6$ heads as $7$ heads.
answered Jan 12 at 18:03
saulspatzsaulspatz
17.3k31435
$endgroup$
Your intuition is wrong. I'm not sure how you are getting $1/2.$ You may be thinking that once we have $6$ heads, the probability is $1/2$ that the next toss will also be heads, but this is valid only if we are told that the first $6$ tosses (or some other specified $6$ tosses) are heads.
Otherwise, you may be thinking that since there are only two possibilities, ($6$ heads or $7$ heads) they are equally likely, but we are actually $7$ times as likely to get exactly $6$ heads as $7$ heads.
answered Jan 12 at 18:03
saulspatzsaulspatz
17.3k31435
answered Jan 12 at 18:03
saulspatzsaulspatz
17.3k31435
answered Jan 12 at 18:03
saulspatzsaulspatz
17.3k31435
answered Jan 12 at 18:03
saulspatzsaulspatz
17.3k31435
17.3k31435
add a comment |
add a comment |
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$begingroup$
This reminds me a bit of the argument that tomorrow the sun will either rise or not rise; therefore the probability that the sun will rise tomorrow is $1/2$.
$endgroup$
– littleO
Jan 12 at 18:04