Verifying solution: Twelve fair coins are flipped
I need to know if I did this problem correctly or incorrectly.
Twelve fair coins are flipped.
(a) What is the expected number of heads that will be obtained?
if a coin is tossed 12 times, the maximum probability of getting heads is 12.
(b) What is the variance in the expected number of heads?
E[X]= (1)(1/2)+(2)(1/2)+(3)(1/2)+(4)(1/2)+(5)(1/2)+(6)(1/2)+(7)(1/2)+(8)(1/2)+(9)(1/2)+(10)(1/2)+(11)(1/2)+(12)(1/2) = 39
E[X^2]= (1)^2*(1/2)+(2)^2*(1/2)+(3)^2*(1/2)+(4)^2*(1/2)+(5)^2*(1/2)+(6)^2*(1/2)+(7)^2*(1/2)+(8)^2*(1/2)+(9)^2*(1/2)+(10)^2*(1/2)+(11)^2*(1/2)+(12)^2*(1/2) = 325
Var(X)=(E[X^2]-µ^2)= 325-(39)^2= 325-1521 = -1196
probability combinatorics probability-theory probability-distributions permutations
asked Dec 12 '18 at 21:49
Daniel WDaniel W
1
|
show 6 more comments
I need to know if I did this problem correctly or incorrectly.
Twelve fair coins are flipped.
(a) What is the expected number of heads that will be obtained?
if a coin is tossed 12 times, the maximum probability of getting heads is 12.
(b) What is the variance in the expected number of heads?
E[X]= (1)(1/2)+(2)(1/2)+(3)(1/2)+(4)(1/2)+(5)(1/2)+(6)(1/2)+(7)(1/2)+(8)(1/2)+(9)(1/2)+(10)(1/2)+(11)(1/2)+(12)(1/2) = 39
E[X^2]= (1)^2*(1/2)+(2)^2*(1/2)+(3)^2*(1/2)+(4)^2*(1/2)+(5)^2*(1/2)+(6)^2*(1/2)+(7)^2*(1/2)+(8)^2*(1/2)+(9)^2*(1/2)+(10)^2*(1/2)+(11)^2*(1/2)+(12)^2*(1/2) = 325
Var(X)=(E[X^2]-µ^2)= 325-(39)^2= 325-1521 = -1196
probability combinatorics probability-theory probability-distributions permutations
asked Dec 12 '18 at 21:49
Daniel WDaniel W
1
3
Sorry...out of $12$ tosses you expect to get $39$ Heads?
– lulu
Dec 12 '18 at 21:50
2
negative variance ... ugh, can't unsee
– Makina
Dec 12 '18 at 21:51
2
And you think the variance is negative? I think you should review the basic properties of average and variance.
– lulu
Dec 12 '18 at 21:52
3
The maximum probability of getting head is 12? Probabilities must be between 0 and 1.
– gd1035
Dec 12 '18 at 21:53
2
"What is the expected number of heads that will be obtained? ... if a coin is tossed 12 times, the maximum probability of getting heads is 12" You are using words in very incorrect ways. It makes me think you do not actually know what those words mean. You mean to say if a coin is tossed 12 times, the maximum number of heads possible is 12. This, however, has very little to do with the problem of finding the expected number of heads that you will see.
– JMoravitz
Dec 12 '18 at 21:54
|
show 6 more comments
I need to know if I did this problem correctly or incorrectly.
Twelve fair coins are flipped.
(a) What is the expected number of heads that will be obtained?
if a coin is tossed 12 times, the maximum probability of getting heads is 12.
(b) What is the variance in the expected number of heads?
E[X]= (1)(1/2)+(2)(1/2)+(3)(1/2)+(4)(1/2)+(5)(1/2)+(6)(1/2)+(7)(1/2)+(8)(1/2)+(9)(1/2)+(10)(1/2)+(11)(1/2)+(12)(1/2) = 39
E[X^2]= (1)^2*(1/2)+(2)^2*(1/2)+(3)^2*(1/2)+(4)^2*(1/2)+(5)^2*(1/2)+(6)^2*(1/2)+(7)^2*(1/2)+(8)^2*(1/2)+(9)^2*(1/2)+(10)^2*(1/2)+(11)^2*(1/2)+(12)^2*(1/2) = 325
Var(X)=(E[X^2]-µ^2)= 325-(39)^2= 325-1521 = -1196
probability combinatorics probability-theory probability-distributions permutations
asked Dec 12 '18 at 21:49
Daniel WDaniel W
1
I need to know if I did this problem correctly or incorrectly.
Twelve fair coins are flipped.
(a) What is the expected number of heads that will be obtained?
if a coin is tossed 12 times, the maximum probability of getting heads is 12.
(b) What is the variance in the expected number of heads?
E[X]= (1)(1/2)+(2)(1/2)+(3)(1/2)+(4)(1/2)+(5)(1/2)+(6)(1/2)+(7)(1/2)+(8)(1/2)+(9)(1/2)+(10)(1/2)+(11)(1/2)+(12)(1/2) = 39
E[X^2]= (1)^2*(1/2)+(2)^2*(1/2)+(3)^2*(1/2)+(4)^2*(1/2)+(5)^2*(1/2)+(6)^2*(1/2)+(7)^2*(1/2)+(8)^2*(1/2)+(9)^2*(1/2)+(10)^2*(1/2)+(11)^2*(1/2)+(12)^2*(1/2) = 325
Var(X)=(E[X^2]-µ^2)= 325-(39)^2= 325-1521 = -1196
probability combinatorics probability-theory probability-distributions permutations
probability combinatorics probability-theory probability-distributions permutations
asked Dec 12 '18 at 21:49
Daniel WDaniel W
1
asked Dec 12 '18 at 21:49
Daniel WDaniel W
1
asked Dec 12 '18 at 21:49
Daniel WDaniel W
1
asked Dec 12 '18 at 21:49
Daniel WDaniel W
1
asked Dec 12 '18 at 21:49
Daniel WDaniel W
1
1
3
Sorry...out of $12$ tosses you expect to get $39$ Heads?
– lulu
Dec 12 '18 at 21:50
2
negative variance ... ugh, can't unsee
– Makina
Dec 12 '18 at 21:51
2
And you think the variance is negative? I think you should review the basic properties of average and variance.
– lulu
Dec 12 '18 at 21:52
3
The maximum probability of getting head is 12? Probabilities must be between 0 and 1.
– gd1035
Dec 12 '18 at 21:53
2
"What is the expected number of heads that will be obtained? ... if a coin is tossed 12 times, the maximum probability of getting heads is 12" You are using words in very incorrect ways. It makes me think you do not actually know what those words mean. You mean to say if a coin is tossed 12 times, the maximum number of heads possible is 12. This, however, has very little to do with the problem of finding the expected number of heads that you will see.
– JMoravitz
Dec 12 '18 at 21:54
|
show 6 more comments
3
Sorry...out of $12$ tosses you expect to get $39$ Heads?
– lulu
Dec 12 '18 at 21:50
2
negative variance ... ugh, can't unsee
– Makina
Dec 12 '18 at 21:51
2
And you think the variance is negative? I think you should review the basic properties of average and variance.
– lulu
Dec 12 '18 at 21:52
3
The maximum probability of getting head is 12? Probabilities must be between 0 and 1.
– gd1035
Dec 12 '18 at 21:53
2
"What is the expected number of heads that will be obtained? ... if a coin is tossed 12 times, the maximum probability of getting heads is 12" You are using words in very incorrect ways. It makes me think you do not actually know what those words mean. You mean to say if a coin is tossed 12 times, the maximum number of heads possible is 12. This, however, has very little to do with the problem of finding the expected number of heads that you will see.
– JMoravitz
Dec 12 '18 at 21:54
3
3
Sorry...out of $12$ tosses you expect to get $39$ Heads?
– lulu
Dec 12 '18 at 21:50
Sorry...out of $12$ tosses you expect to get $39$ Heads?
– lulu
Dec 12 '18 at 21:50
2
2
negative variance ... ugh, can't unsee
– Makina
Dec 12 '18 at 21:51
negative variance ... ugh, can't unsee
– Makina
Dec 12 '18 at 21:51
2
2
And you think the variance is negative? I think you should review the basic properties of average and variance.
– lulu
Dec 12 '18 at 21:52
And you think the variance is negative? I think you should review the basic properties of average and variance.
– lulu
Dec 12 '18 at 21:52
3
3
The maximum probability of getting head is 12? Probabilities must be between 0 and 1.
– gd1035
Dec 12 '18 at 21:53
The maximum probability of getting head is 12? Probabilities must be between 0 and 1.
– gd1035
Dec 12 '18 at 21:53
2
2
"What is the expected number of heads that will be obtained? ... if a coin is tossed 12 times, the maximum probability of getting heads is 12" You are using words in very incorrect ways. It makes me think you do not actually know what those words mean. You mean to say if a coin is tossed 12 times, the maximum number of heads possible is 12. This, however, has very little to do with the problem of finding the expected number of heads that you will see.
– JMoravitz
Dec 12 '18 at 21:54
"What is the expected number of heads that will be obtained? ... if a coin is tossed 12 times, the maximum probability of getting heads is 12" You are using words in very incorrect ways. It makes me think you do not actually know what those words mean. You mean to say if a coin is tossed 12 times, the maximum number of heads possible is 12. This, however, has very little to do with the problem of finding the expected number of heads that you will see.
– JMoravitz
Dec 12 '18 at 21:54
|
show 6 more comments
1 Answer
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Your attempted answers are incorrect and should be obvious that they are incorrect for a number of reasons. Think about what it is saying. "If I flip 12 coins, I expect around 39 of those coins to be heads" can't be true. Further, variances are never negative numbers.
Hint:
This follows a binomial distribution.
The probability of getting exactly $k$ heads in your experiment is instead $$Pr(X=k)=binom{12}{k}left(frac{1}{2}right)^kleft(frac{1}{2}right)^{12-k}$$
Use this to properly calculate the correct probabilities for $X$ to be equal to $0,1,2,dots,12$ and fix your calculations for $E[X]$ and later for $E[X^2]$.
$E[X] = 0times Pr(X=0)+1times Pr(X=1)+2times Pr(X=2)+dots+12times Pr(X=12)$, you incorrectly filled in $frac{1}{2}$ for each of these probabilities.
Alternate solution hint:
You could approach via indicator variables quite easily by noting that the random variable $X$ which counts the total number of heads is the sum of indicator variables $X_1,X_2,X_3,dots,X_{12}$ where $X_i$ is a random variable which takes the value of $0$ if the $i$'th coin is a tail and $1$ if the $i$'th coin is a head. From there you can use linearity of expectation for an almost immediate solution.
$E[X] = E[X_1+X_2+X_3+dots+X_{12}]=E[X_1]+E[X_2]+dots+E[X_{12}]$ and note that $E[X_1]=Pr(X_1=1)$ is simply the probability that the first coin is a head. Similarly for the others.
answered Dec 12 '18 at 22:10
JMoravitzJMoravitz
46.5k33785
add a comment |
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1 Answer
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Your attempted answers are incorrect and should be obvious that they are incorrect for a number of reasons. Think about what it is saying. "If I flip 12 coins, I expect around 39 of those coins to be heads" can't be true. Further, variances are never negative numbers.
Hint:
This follows a binomial distribution.
The probability of getting exactly $k$ heads in your experiment is instead $$Pr(X=k)=binom{12}{k}left(frac{1}{2}right)^kleft(frac{1}{2}right)^{12-k}$$
Use this to properly calculate the correct probabilities for $X$ to be equal to $0,1,2,dots,12$ and fix your calculations for $E[X]$ and later for $E[X^2]$.
$E[X] = 0times Pr(X=0)+1times Pr(X=1)+2times Pr(X=2)+dots+12times Pr(X=12)$, you incorrectly filled in $frac{1}{2}$ for each of these probabilities.
Alternate solution hint:
You could approach via indicator variables quite easily by noting that the random variable $X$ which counts the total number of heads is the sum of indicator variables $X_1,X_2,X_3,dots,X_{12}$ where $X_i$ is a random variable which takes the value of $0$ if the $i$'th coin is a tail and $1$ if the $i$'th coin is a head. From there you can use linearity of expectation for an almost immediate solution.
$E[X] = E[X_1+X_2+X_3+dots+X_{12}]=E[X_1]+E[X_2]+dots+E[X_{12}]$ and note that $E[X_1]=Pr(X_1=1)$ is simply the probability that the first coin is a head. Similarly for the others.
answered Dec 12 '18 at 22:10
JMoravitzJMoravitz
46.5k33785
add a comment |
Your attempted answers are incorrect and should be obvious that they are incorrect for a number of reasons. Think about what it is saying. "If I flip 12 coins, I expect around 39 of those coins to be heads" can't be true. Further, variances are never negative numbers.
Hint:
This follows a binomial distribution.
The probability of getting exactly $k$ heads in your experiment is instead $$Pr(X=k)=binom{12}{k}left(frac{1}{2}right)^kleft(frac{1}{2}right)^{12-k}$$
Use this to properly calculate the correct probabilities for $X$ to be equal to $0,1,2,dots,12$ and fix your calculations for $E[X]$ and later for $E[X^2]$.
$E[X] = 0times Pr(X=0)+1times Pr(X=1)+2times Pr(X=2)+dots+12times Pr(X=12)$, you incorrectly filled in $frac{1}{2}$ for each of these probabilities.
Alternate solution hint:
You could approach via indicator variables quite easily by noting that the random variable $X$ which counts the total number of heads is the sum of indicator variables $X_1,X_2,X_3,dots,X_{12}$ where $X_i$ is a random variable which takes the value of $0$ if the $i$'th coin is a tail and $1$ if the $i$'th coin is a head. From there you can use linearity of expectation for an almost immediate solution.
$E[X] = E[X_1+X_2+X_3+dots+X_{12}]=E[X_1]+E[X_2]+dots+E[X_{12}]$ and note that $E[X_1]=Pr(X_1=1)$ is simply the probability that the first coin is a head. Similarly for the others.
answered Dec 12 '18 at 22:10
JMoravitzJMoravitz
46.5k33785
add a comment |
Your attempted answers are incorrect and should be obvious that they are incorrect for a number of reasons. Think about what it is saying. "If I flip 12 coins, I expect around 39 of those coins to be heads" can't be true. Further, variances are never negative numbers.
Hint:
This follows a binomial distribution.
The probability of getting exactly $k$ heads in your experiment is instead $$Pr(X=k)=binom{12}{k}left(frac{1}{2}right)^kleft(frac{1}{2}right)^{12-k}$$
Use this to properly calculate the correct probabilities for $X$ to be equal to $0,1,2,dots,12$ and fix your calculations for $E[X]$ and later for $E[X^2]$.
$E[X] = 0times Pr(X=0)+1times Pr(X=1)+2times Pr(X=2)+dots+12times Pr(X=12)$, you incorrectly filled in $frac{1}{2}$ for each of these probabilities.
Alternate solution hint:
You could approach via indicator variables quite easily by noting that the random variable $X$ which counts the total number of heads is the sum of indicator variables $X_1,X_2,X_3,dots,X_{12}$ where $X_i$ is a random variable which takes the value of $0$ if the $i$'th coin is a tail and $1$ if the $i$'th coin is a head. From there you can use linearity of expectation for an almost immediate solution.
$E[X] = E[X_1+X_2+X_3+dots+X_{12}]=E[X_1]+E[X_2]+dots+E[X_{12}]$ and note that $E[X_1]=Pr(X_1=1)$ is simply the probability that the first coin is a head. Similarly for the others.
answered Dec 12 '18 at 22:10
JMoravitzJMoravitz
46.5k33785
Your attempted answers are incorrect and should be obvious that they are incorrect for a number of reasons. Think about what it is saying. "If I flip 12 coins, I expect around 39 of those coins to be heads" can't be true. Further, variances are never negative numbers.
Hint:
This follows a binomial distribution.
The probability of getting exactly $k$ heads in your experiment is instead $$Pr(X=k)=binom{12}{k}left(frac{1}{2}right)^kleft(frac{1}{2}right)^{12-k}$$
Use this to properly calculate the correct probabilities for $X$ to be equal to $0,1,2,dots,12$ and fix your calculations for $E[X]$ and later for $E[X^2]$.
$E[X] = 0times Pr(X=0)+1times Pr(X=1)+2times Pr(X=2)+dots+12times Pr(X=12)$, you incorrectly filled in $frac{1}{2}$ for each of these probabilities.
Alternate solution hint:
You could approach via indicator variables quite easily by noting that the random variable $X$ which counts the total number of heads is the sum of indicator variables $X_1,X_2,X_3,dots,X_{12}$ where $X_i$ is a random variable which takes the value of $0$ if the $i$'th coin is a tail and $1$ if the $i$'th coin is a head. From there you can use linearity of expectation for an almost immediate solution.
$E[X] = E[X_1+X_2+X_3+dots+X_{12}]=E[X_1]+E[X_2]+dots+E[X_{12}]$ and note that $E[X_1]=Pr(X_1=1)$ is simply the probability that the first coin is a head. Similarly for the others.
answered Dec 12 '18 at 22:10
JMoravitzJMoravitz
46.5k33785
answered Dec 12 '18 at 22:10
JMoravitzJMoravitz
46.5k33785
answered Dec 12 '18 at 22:10
JMoravitzJMoravitz
46.5k33785
answered Dec 12 '18 at 22:10
JMoravitzJMoravitz
46.5k33785
46.5k33785
add a comment |
add a comment |
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3
Sorry...out of $12$ tosses you expect to get $39$ Heads?
– lulu
Dec 12 '18 at 21:50
2
negative variance ... ugh, can't unsee
– Makina
Dec 12 '18 at 21:51
2
And you think the variance is negative? I think you should review the basic properties of average and variance.
– lulu
Dec 12 '18 at 21:52
3
The maximum probability of getting head is 12? Probabilities must be between 0 and 1.
– gd1035
Dec 12 '18 at 21:53
2
"What is the expected number of heads that will be obtained? ... if a coin is tossed 12 times, the maximum probability of getting heads is 12" You are using words in very incorrect ways. It makes me think you do not actually know what those words mean. You mean to say if a coin is tossed 12 times, the maximum number of heads possible is 12. This, however, has very little to do with the problem of finding the expected number of heads that you will see.
– JMoravitz
Dec 12 '18 at 21:54