Box contains 2 different coins, one is chosen randomly and tossed n times, head came all n times. Probability...
A box contains two coins: a regular coin and one fake two-headed coin (P(H)=1). One coin is choose at random and tossed $n$ times.
If the first n coin tosses result in heads, What is the probability that the $(n+1)^{th}$ coin toss will also result in heads?
My solution:
$$text{Required Probability} = frac{1}{2}times left(frac{1}{2}right)^n times frac{1}{2}+frac{1}{2} times 1^n times 1$$
probability
asked Aug 15 '16 at 17:59
amarVashishth
269218
add a comment |
A box contains two coins: a regular coin and one fake two-headed coin (P(H)=1). One coin is choose at random and tossed $n$ times.
If the first n coin tosses result in heads, What is the probability that the $(n+1)^{th}$ coin toss will also result in heads?
My solution:
$$text{Required Probability} = frac{1}{2}times left(frac{1}{2}right)^n times frac{1}{2}+frac{1}{2} times 1^n times 1$$
probability
asked Aug 15 '16 at 17:59
amarVashishth
269218
Your solution is correct.
– Mark Fischler
Aug 15 '16 at 18:08
4
This does not look correct to me. If $n$ is very large then you are sure that you must have the weighted coin, in which the desired probability is $1$. Your formula, however, tends to $frac 12$. Indeed, your value decreases as $n$ increases which makes no sense (to me).
– lulu
Aug 15 '16 at 18:11
1
@MarkFischler I expect you misread the question. The OP correctly calculates the probability of getting $n+1$ consecutive heads, but that is not at all what the question was asking.
– lulu
Aug 15 '16 at 18:16
add a comment |
A box contains two coins: a regular coin and one fake two-headed coin (P(H)=1). One coin is choose at random and tossed $n$ times.
If the first n coin tosses result in heads, What is the probability that the $(n+1)^{th}$ coin toss will also result in heads?
My solution:
$$text{Required Probability} = frac{1}{2}times left(frac{1}{2}right)^n times frac{1}{2}+frac{1}{2} times 1^n times 1$$
probability
asked Aug 15 '16 at 17:59
amarVashishth
269218
A box contains two coins: a regular coin and one fake two-headed coin (P(H)=1). One coin is choose at random and tossed $n$ times.
If the first n coin tosses result in heads, What is the probability that the $(n+1)^{th}$ coin toss will also result in heads?
My solution:
$$text{Required Probability} = frac{1}{2}times left(frac{1}{2}right)^n times frac{1}{2}+frac{1}{2} times 1^n times 1$$
probability
probability
asked Aug 15 '16 at 17:59
amarVashishth
269218
asked Aug 15 '16 at 17:59
amarVashishth
269218
asked Aug 15 '16 at 17:59
amarVashishth
269218
asked Aug 15 '16 at 17:59
amarVashishth
269218
asked Aug 15 '16 at 17:59
amarVashishth
269218
269218
Your solution is correct.
– Mark Fischler
Aug 15 '16 at 18:08
4
This does not look correct to me. If $n$ is very large then you are sure that you must have the weighted coin, in which the desired probability is $1$. Your formula, however, tends to $frac 12$. Indeed, your value decreases as $n$ increases which makes no sense (to me).
– lulu
Aug 15 '16 at 18:11
1
@MarkFischler I expect you misread the question. The OP correctly calculates the probability of getting $n+1$ consecutive heads, but that is not at all what the question was asking.
– lulu
Aug 15 '16 at 18:16
add a comment |
Your solution is correct.
– Mark Fischler
Aug 15 '16 at 18:08
4
This does not look correct to me. If $n$ is very large then you are sure that you must have the weighted coin, in which the desired probability is $1$. Your formula, however, tends to $frac 12$. Indeed, your value decreases as $n$ increases which makes no sense (to me).
– lulu
Aug 15 '16 at 18:11
1
@MarkFischler I expect you misread the question. The OP correctly calculates the probability of getting $n+1$ consecutive heads, but that is not at all what the question was asking.
– lulu
Aug 15 '16 at 18:16
Your solution is correct.
– Mark Fischler
Aug 15 '16 at 18:08
Your solution is correct.
– Mark Fischler
Aug 15 '16 at 18:08
4
4
This does not look correct to me. If $n$ is very large then you are sure that you must have the weighted coin, in which the desired probability is $1$. Your formula, however, tends to $frac 12$. Indeed, your value decreases as $n$ increases which makes no sense (to me).
– lulu
Aug 15 '16 at 18:11
This does not look correct to me. If $n$ is very large then you are sure that you must have the weighted coin, in which the desired probability is $1$. Your formula, however, tends to $frac 12$. Indeed, your value decreases as $n$ increases which makes no sense (to me).
– lulu
Aug 15 '16 at 18:11
1
1
@MarkFischler I expect you misread the question. The OP correctly calculates the probability of getting $n+1$ consecutive heads, but that is not at all what the question was asking.
– lulu
Aug 15 '16 at 18:16
@MarkFischler I expect you misread the question. The OP correctly calculates the probability of getting $n+1$ consecutive heads, but that is not at all what the question was asking.
– lulu
Aug 15 '16 at 18:16
add a comment |
2 Answers
2
active
oldest
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The probability of seeing $n$ heads in $n$ tosses of a fair coin is $2^{-n}$. Thus the total probability of seeing $n$ consecutive heads is $$frac 12times 2^{-n}+frac 12 times 1$$ Therefore, given that you have in fact observed $n$ consecutive heads the new estimate for the probability that you have the fair coin is $$frac {frac 12 times 2^{-n}}{frac 12times 2^{-n}+frac 12 times 1}=frac {1}{1+2^n}$$ and the probability that you have the fake coin is therefore $$frac {2^n}{1+2^n}$$ It follows that the probability that the next toss is also heads is $$boxed {frac 12times frac {1}{1+2^n}+frac {2^n}{1+2^n}=frac {1+2^{n+1}}{2+2^{n+1}}}$$
Note: as $n$ goes to $infty$ this tends to $1$, as it clearly should.
answered Aug 15 '16 at 18:15
lulu
39.2k24677
why is that the probability of a fair coin?
– Jorge Fernández
Aug 15 '16 at 18:21
@CarryonSmiling That's the portion of the winning results which is explained by the fair coin. Worth remarking that my final answer coincides with yours, so I expect we are saying the same thing in different words.
– lulu
Aug 15 '16 at 18:22
1
@CarryonSmiling On inspection, we are saying the same thing in the same words.
– lulu
Aug 15 '16 at 18:26
add a comment |
First we find the probability that the coin is fair.
To do this we use Bayes theorem, we want $P(A|B)$, where $A$ is the probability the coin is fair and $B$ is the probability the first $n$ flips are heads.
By Bayes theorem:
$P(A|B)=frac{P(B|A)P(A)}{P(B)}$, the numerator is $frac{1}{2^{n}}times frac{1}{2}$ and the denominator is $frac{1}{2}+frac{1}{2^n}times frac{1}{2}=frac{2^n+1}{2^{n+1}}$. hence $P(A|B)=frac{1}{2^{n}+1}$.
Therefore the final answer is:
$color {green}{frac{1}{2^n+1}times frac{1}{2}}+color{purple}{(1-frac{1}{2^n+1})}=1-frac{1}{2(2^n+1)}$
The green part is the probability that it is heads and that the coin is fair, and the purple part is the probability that it is heads and the coin is fake.
answered Aug 15 '16 at 18:16
Jorge Fernández
75.1k1190191
add a comment |
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2 Answers
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2 Answers
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active
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The probability of seeing $n$ heads in $n$ tosses of a fair coin is $2^{-n}$. Thus the total probability of seeing $n$ consecutive heads is $$frac 12times 2^{-n}+frac 12 times 1$$ Therefore, given that you have in fact observed $n$ consecutive heads the new estimate for the probability that you have the fair coin is $$frac {frac 12 times 2^{-n}}{frac 12times 2^{-n}+frac 12 times 1}=frac {1}{1+2^n}$$ and the probability that you have the fake coin is therefore $$frac {2^n}{1+2^n}$$ It follows that the probability that the next toss is also heads is $$boxed {frac 12times frac {1}{1+2^n}+frac {2^n}{1+2^n}=frac {1+2^{n+1}}{2+2^{n+1}}}$$
Note: as $n$ goes to $infty$ this tends to $1$, as it clearly should.
answered Aug 15 '16 at 18:15
lulu
39.2k24677
why is that the probability of a fair coin?
– Jorge Fernández
Aug 15 '16 at 18:21
@CarryonSmiling That's the portion of the winning results which is explained by the fair coin. Worth remarking that my final answer coincides with yours, so I expect we are saying the same thing in different words.
– lulu
Aug 15 '16 at 18:22
1
@CarryonSmiling On inspection, we are saying the same thing in the same words.
– lulu
Aug 15 '16 at 18:26
add a comment |
The probability of seeing $n$ heads in $n$ tosses of a fair coin is $2^{-n}$. Thus the total probability of seeing $n$ consecutive heads is $$frac 12times 2^{-n}+frac 12 times 1$$ Therefore, given that you have in fact observed $n$ consecutive heads the new estimate for the probability that you have the fair coin is $$frac {frac 12 times 2^{-n}}{frac 12times 2^{-n}+frac 12 times 1}=frac {1}{1+2^n}$$ and the probability that you have the fake coin is therefore $$frac {2^n}{1+2^n}$$ It follows that the probability that the next toss is also heads is $$boxed {frac 12times frac {1}{1+2^n}+frac {2^n}{1+2^n}=frac {1+2^{n+1}}{2+2^{n+1}}}$$
Note: as $n$ goes to $infty$ this tends to $1$, as it clearly should.
answered Aug 15 '16 at 18:15
lulu
39.2k24677
why is that the probability of a fair coin?
– Jorge Fernández
Aug 15 '16 at 18:21
@CarryonSmiling That's the portion of the winning results which is explained by the fair coin. Worth remarking that my final answer coincides with yours, so I expect we are saying the same thing in different words.
– lulu
Aug 15 '16 at 18:22
1
@CarryonSmiling On inspection, we are saying the same thing in the same words.
– lulu
Aug 15 '16 at 18:26
add a comment |
The probability of seeing $n$ heads in $n$ tosses of a fair coin is $2^{-n}$. Thus the total probability of seeing $n$ consecutive heads is $$frac 12times 2^{-n}+frac 12 times 1$$ Therefore, given that you have in fact observed $n$ consecutive heads the new estimate for the probability that you have the fair coin is $$frac {frac 12 times 2^{-n}}{frac 12times 2^{-n}+frac 12 times 1}=frac {1}{1+2^n}$$ and the probability that you have the fake coin is therefore $$frac {2^n}{1+2^n}$$ It follows that the probability that the next toss is also heads is $$boxed {frac 12times frac {1}{1+2^n}+frac {2^n}{1+2^n}=frac {1+2^{n+1}}{2+2^{n+1}}}$$
Note: as $n$ goes to $infty$ this tends to $1$, as it clearly should.
answered Aug 15 '16 at 18:15
lulu
39.2k24677
The probability of seeing $n$ heads in $n$ tosses of a fair coin is $2^{-n}$. Thus the total probability of seeing $n$ consecutive heads is $$frac 12times 2^{-n}+frac 12 times 1$$ Therefore, given that you have in fact observed $n$ consecutive heads the new estimate for the probability that you have the fair coin is $$frac {frac 12 times 2^{-n}}{frac 12times 2^{-n}+frac 12 times 1}=frac {1}{1+2^n}$$ and the probability that you have the fake coin is therefore $$frac {2^n}{1+2^n}$$ It follows that the probability that the next toss is also heads is $$boxed {frac 12times frac {1}{1+2^n}+frac {2^n}{1+2^n}=frac {1+2^{n+1}}{2+2^{n+1}}}$$
Note: as $n$ goes to $infty$ this tends to $1$, as it clearly should.
answered Aug 15 '16 at 18:15
lulu
39.2k24677
answered Aug 15 '16 at 18:15
lulu
39.2k24677
answered Aug 15 '16 at 18:15
lulu
39.2k24677
answered Aug 15 '16 at 18:15
lulu
39.2k24677
39.2k24677
why is that the probability of a fair coin?
– Jorge Fernández
Aug 15 '16 at 18:21
@CarryonSmiling That's the portion of the winning results which is explained by the fair coin. Worth remarking that my final answer coincides with yours, so I expect we are saying the same thing in different words.
– lulu
Aug 15 '16 at 18:22
1
@CarryonSmiling On inspection, we are saying the same thing in the same words.
– lulu
Aug 15 '16 at 18:26
add a comment |
why is that the probability of a fair coin?
– Jorge Fernández
Aug 15 '16 at 18:21
@CarryonSmiling That's the portion of the winning results which is explained by the fair coin. Worth remarking that my final answer coincides with yours, so I expect we are saying the same thing in different words.
– lulu
Aug 15 '16 at 18:22
1
@CarryonSmiling On inspection, we are saying the same thing in the same words.
– lulu
Aug 15 '16 at 18:26
why is that the probability of a fair coin?
– Jorge Fernández
Aug 15 '16 at 18:21
why is that the probability of a fair coin?
– Jorge Fernández
Aug 15 '16 at 18:21
@CarryonSmiling That's the portion of the winning results which is explained by the fair coin. Worth remarking that my final answer coincides with yours, so I expect we are saying the same thing in different words.
– lulu
Aug 15 '16 at 18:22
@CarryonSmiling That's the portion of the winning results which is explained by the fair coin. Worth remarking that my final answer coincides with yours, so I expect we are saying the same thing in different words.
– lulu
Aug 15 '16 at 18:22
1
1
@CarryonSmiling On inspection, we are saying the same thing in the same words.
– lulu
Aug 15 '16 at 18:26
@CarryonSmiling On inspection, we are saying the same thing in the same words.
– lulu
Aug 15 '16 at 18:26
add a comment |
First we find the probability that the coin is fair.
To do this we use Bayes theorem, we want $P(A|B)$, where $A$ is the probability the coin is fair and $B$ is the probability the first $n$ flips are heads.
By Bayes theorem:
$P(A|B)=frac{P(B|A)P(A)}{P(B)}$, the numerator is $frac{1}{2^{n}}times frac{1}{2}$ and the denominator is $frac{1}{2}+frac{1}{2^n}times frac{1}{2}=frac{2^n+1}{2^{n+1}}$. hence $P(A|B)=frac{1}{2^{n}+1}$.
Therefore the final answer is:
$color {green}{frac{1}{2^n+1}times frac{1}{2}}+color{purple}{(1-frac{1}{2^n+1})}=1-frac{1}{2(2^n+1)}$
The green part is the probability that it is heads and that the coin is fair, and the purple part is the probability that it is heads and the coin is fake.
answered Aug 15 '16 at 18:16
Jorge Fernández
75.1k1190191
add a comment |
First we find the probability that the coin is fair.
To do this we use Bayes theorem, we want $P(A|B)$, where $A$ is the probability the coin is fair and $B$ is the probability the first $n$ flips are heads.
By Bayes theorem:
$P(A|B)=frac{P(B|A)P(A)}{P(B)}$, the numerator is $frac{1}{2^{n}}times frac{1}{2}$ and the denominator is $frac{1}{2}+frac{1}{2^n}times frac{1}{2}=frac{2^n+1}{2^{n+1}}$. hence $P(A|B)=frac{1}{2^{n}+1}$.
Therefore the final answer is:
$color {green}{frac{1}{2^n+1}times frac{1}{2}}+color{purple}{(1-frac{1}{2^n+1})}=1-frac{1}{2(2^n+1)}$
The green part is the probability that it is heads and that the coin is fair, and the purple part is the probability that it is heads and the coin is fake.
answered Aug 15 '16 at 18:16
Jorge Fernández
75.1k1190191
add a comment |
First we find the probability that the coin is fair.
To do this we use Bayes theorem, we want $P(A|B)$, where $A$ is the probability the coin is fair and $B$ is the probability the first $n$ flips are heads.
By Bayes theorem:
$P(A|B)=frac{P(B|A)P(A)}{P(B)}$, the numerator is $frac{1}{2^{n}}times frac{1}{2}$ and the denominator is $frac{1}{2}+frac{1}{2^n}times frac{1}{2}=frac{2^n+1}{2^{n+1}}$. hence $P(A|B)=frac{1}{2^{n}+1}$.
Therefore the final answer is:
$color {green}{frac{1}{2^n+1}times frac{1}{2}}+color{purple}{(1-frac{1}{2^n+1})}=1-frac{1}{2(2^n+1)}$
The green part is the probability that it is heads and that the coin is fair, and the purple part is the probability that it is heads and the coin is fake.
answered Aug 15 '16 at 18:16
Jorge Fernández
75.1k1190191
First we find the probability that the coin is fair.
To do this we use Bayes theorem, we want $P(A|B)$, where $A$ is the probability the coin is fair and $B$ is the probability the first $n$ flips are heads.
By Bayes theorem:
$P(A|B)=frac{P(B|A)P(A)}{P(B)}$, the numerator is $frac{1}{2^{n}}times frac{1}{2}$ and the denominator is $frac{1}{2}+frac{1}{2^n}times frac{1}{2}=frac{2^n+1}{2^{n+1}}$. hence $P(A|B)=frac{1}{2^{n}+1}$.
Therefore the final answer is:
$color {green}{frac{1}{2^n+1}times frac{1}{2}}+color{purple}{(1-frac{1}{2^n+1})}=1-frac{1}{2(2^n+1)}$
The green part is the probability that it is heads and that the coin is fair, and the purple part is the probability that it is heads and the coin is fake.
answered Aug 15 '16 at 18:16
Jorge Fernández
75.1k1190191
answered Aug 15 '16 at 18:16
Jorge Fernández
75.1k1190191
answered Aug 15 '16 at 18:16
Jorge Fernández
75.1k1190191
answered Aug 15 '16 at 18:16
Jorge Fernández
75.1k1190191
75.1k1190191
add a comment |
add a comment |
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Your solution is correct.
– Mark Fischler
Aug 15 '16 at 18:08
4
This does not look correct to me. If $n$ is very large then you are sure that you must have the weighted coin, in which the desired probability is $1$. Your formula, however, tends to $frac 12$. Indeed, your value decreases as $n$ increases which makes no sense (to me).
– lulu
Aug 15 '16 at 18:11
1
@MarkFischler I expect you misread the question. The OP correctly calculates the probability of getting $n+1$ consecutive heads, but that is not at all what the question was asking.
– lulu
Aug 15 '16 at 18:16