Mixing
Tossing a coin (5,10) and throwing a die (1-6) at the same time twice
$begingroup$
You flip a coin having a 5 on one side and a 10 on the other. At the same time, you throw a normal 6-sided die. So there are 12 possible combinations. You add the two numbers. This gives 10 possible unique results and two possibilities to get 11 (5+6 and 10+1).
If you perform the experiment twice, what is the probability to get 11 at least once?
I thought that this has to be calculated by using the counter-probability. The probability to get 11 is $frac{2}{12}$ or $frac{1}{6}$.
So the counter-probability when doing the experiment twice is $frac{5}{6}^2 = frac{25}{36}$.
Therefore the probability to get 11 at least once should be $frac{11}{36}$ (30.55 %).
But this seems to be incorrect. What am I doing wrong?
probability dice
edited Jan 21 at 15:43
Especially Lime
22.3k22858
asked Jan 21 at 14:40
KranzeKranze
84
$endgroup$
|
show 3 more comments
$begingroup$
You flip a coin having a 5 on one side and a 10 on the other. At the same time, you throw a normal 6-sided die. So there are 12 possible combinations. You add the two numbers. This gives 10 possible unique results and two possibilities to get 11 (5+6 and 10+1).
If you perform the experiment twice, what is the probability to get 11 at least once?
I thought that this has to be calculated by using the counter-probability. The probability to get 11 is $frac{2}{12}$ or $frac{1}{6}$.
So the counter-probability when doing the experiment twice is $frac{5}{6}^2 = frac{25}{36}$.
Therefore the probability to get 11 at least once should be $frac{11}{36}$ (30.55 %).
But this seems to be incorrect. What am I doing wrong?
probability dice
edited Jan 21 at 15:43
Especially Lime
22.3k22858
asked Jan 21 at 14:40
KranzeKranze
84
$endgroup$
1
$begingroup$
What is the correct answer? Unless you have mis-stated the question, this seems correct to me.
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 14:43
$begingroup$
I wrote a little simulation in Python and got approx. 44% chance. Despite this, the professor said that the solution is wrong.
$endgroup$
– Kranze
Jan 21 at 15:05
$begingroup$
Could you recheck the question. I don't see any way to claim that you are wrong.
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 15:11
$begingroup$
Here is my script: <!-- language: lang-py --> import random def throw(): '''coin toss''' coin = random.choice([5, 10]) die = random.choice([1,2,3,4,5,6]) return bool(coin+die==11) def main(): '''perform experiment''' t, f = 0, 0 # true , false counter for _ in range(1000000): throw1 = throw() throw2 = throw() if throw1 or throw2: t+=1 else: f+=1 print(t/f)
$endgroup$
– Kranze
Jan 21 at 15:20
$begingroup$
I hate the fact that I do not know any codes. Can you tell me how I can implement this myself?
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 15:25
|
show 3 more comments
$begingroup$
You flip a coin having a 5 on one side and a 10 on the other. At the same time, you throw a normal 6-sided die. So there are 12 possible combinations. You add the two numbers. This gives 10 possible unique results and two possibilities to get 11 (5+6 and 10+1).
If you perform the experiment twice, what is the probability to get 11 at least once?
I thought that this has to be calculated by using the counter-probability. The probability to get 11 is $frac{2}{12}$ or $frac{1}{6}$.
So the counter-probability when doing the experiment twice is $frac{5}{6}^2 = frac{25}{36}$.
Therefore the probability to get 11 at least once should be $frac{11}{36}$ (30.55 %).
But this seems to be incorrect. What am I doing wrong?
probability dice
edited Jan 21 at 15:43
Especially Lime
22.3k22858
asked Jan 21 at 14:40
KranzeKranze
84
$endgroup$
You flip a coin having a 5 on one side and a 10 on the other. At the same time, you throw a normal 6-sided die. So there are 12 possible combinations. You add the two numbers. This gives 10 possible unique results and two possibilities to get 11 (5+6 and 10+1).
If you perform the experiment twice, what is the probability to get 11 at least once?
I thought that this has to be calculated by using the counter-probability. The probability to get 11 is $frac{2}{12}$ or $frac{1}{6}$.
So the counter-probability when doing the experiment twice is $frac{5}{6}^2 = frac{25}{36}$.
Therefore the probability to get 11 at least once should be $frac{11}{36}$ (30.55 %).
But this seems to be incorrect. What am I doing wrong?
probability dice
probability dice
edited Jan 21 at 15:43
Especially Lime
22.3k22858
asked Jan 21 at 14:40
KranzeKranze
84
edited Jan 21 at 15:43
Especially Lime
22.3k22858
asked Jan 21 at 14:40
KranzeKranze
84
edited Jan 21 at 15:43
Especially Lime
22.3k22858
edited Jan 21 at 15:43
Especially Lime
22.3k22858
edited Jan 21 at 15:43
Especially Lime
22.3k22858
22.3k22858
asked Jan 21 at 14:40
KranzeKranze
84
asked Jan 21 at 14:40
KranzeKranze
84
asked Jan 21 at 14:40
KranzeKranze
84
84
1
$begingroup$
What is the correct answer? Unless you have mis-stated the question, this seems correct to me.
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 14:43
$begingroup$
I wrote a little simulation in Python and got approx. 44% chance. Despite this, the professor said that the solution is wrong.
$endgroup$
– Kranze
Jan 21 at 15:05
$begingroup$
Could you recheck the question. I don't see any way to claim that you are wrong.
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 15:11
$begingroup$
Here is my script: <!-- language: lang-py --> import random def throw(): '''coin toss''' coin = random.choice([5, 10]) die = random.choice([1,2,3,4,5,6]) return bool(coin+die==11) def main(): '''perform experiment''' t, f = 0, 0 # true , false counter for _ in range(1000000): throw1 = throw() throw2 = throw() if throw1 or throw2: t+=1 else: f+=1 print(t/f)
$endgroup$
– Kranze
Jan 21 at 15:20
$begingroup$
I hate the fact that I do not know any codes. Can you tell me how I can implement this myself?
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 15:25
|
show 3 more comments
1
$begingroup$
What is the correct answer? Unless you have mis-stated the question, this seems correct to me.
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 14:43
$begingroup$
I wrote a little simulation in Python and got approx. 44% chance. Despite this, the professor said that the solution is wrong.
$endgroup$
– Kranze
Jan 21 at 15:05
$begingroup$
Could you recheck the question. I don't see any way to claim that you are wrong.
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 15:11
$begingroup$
Here is my script: <!-- language: lang-py --> import random def throw(): '''coin toss''' coin = random.choice([5, 10]) die = random.choice([1,2,3,4,5,6]) return bool(coin+die==11) def main(): '''perform experiment''' t, f = 0, 0 # true , false counter for _ in range(1000000): throw1 = throw() throw2 = throw() if throw1 or throw2: t+=1 else: f+=1 print(t/f)
$endgroup$
– Kranze
Jan 21 at 15:20
$begingroup$
I hate the fact that I do not know any codes. Can you tell me how I can implement this myself?
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 15:25
1
1
$begingroup$
What is the correct answer? Unless you have mis-stated the question, this seems correct to me.
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 14:43
$begingroup$
What is the correct answer? Unless you have mis-stated the question, this seems correct to me.
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 14:43
$begingroup$
I wrote a little simulation in Python and got approx. 44% chance. Despite this, the professor said that the solution is wrong.
$endgroup$
– Kranze
Jan 21 at 15:05
$begingroup$
I wrote a little simulation in Python and got approx. 44% chance. Despite this, the professor said that the solution is wrong.
$endgroup$
– Kranze
Jan 21 at 15:05
$begingroup$
Could you recheck the question. I don't see any way to claim that you are wrong.
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 15:11
$begingroup$
Could you recheck the question. I don't see any way to claim that you are wrong.
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 15:11
$begingroup$
Here is my script: <!-- language: lang-py --> import random def throw(): '''coin toss''' coin = random.choice([5, 10]) die = random.choice([1,2,3,4,5,6]) return bool(coin+die==11) def main(): '''perform experiment''' t, f = 0, 0 # true , false counter for _ in range(1000000): throw1 = throw() throw2 = throw() if throw1 or throw2: t+=1 else: f+=1 print(t/f)
$endgroup$
– Kranze
Jan 21 at 15:20
$begingroup$
Here is my script: <!-- language: lang-py --> import random def throw(): '''coin toss''' coin = random.choice([5, 10]) die = random.choice([1,2,3,4,5,6]) return bool(coin+die==11) def main(): '''perform experiment''' t, f = 0, 0 # true , false counter for _ in range(1000000): throw1 = throw() throw2 = throw() if throw1 or throw2: t+=1 else: f+=1 print(t/f)
$endgroup$
– Kranze
Jan 21 at 15:20
$begingroup$
I hate the fact that I do not know any codes. Can you tell me how I can implement this myself?
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 15:25
$begingroup$
I hate the fact that I do not know any codes. Can you tell me how I can implement this myself?
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 15:25
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Your calculation is correct, the probability is $frac{11}{36}$. Your code (given in comments) is not giving the right answer because it outputs #successes / #failures instead of #successes / #attempts. Just change the last line to print(t/1000000).
answered Jan 21 at 15:41
Especially LimeEspecially Lime
22.3k22858
$endgroup$
$begingroup$
What a stupid bug. Thanks a lot for your answer
$endgroup$
– Kranze
Jan 21 at 15:49
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
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oldest
votes
$begingroup$
Your calculation is correct, the probability is $frac{11}{36}$. Your code (given in comments) is not giving the right answer because it outputs #successes / #failures instead of #successes / #attempts. Just change the last line to print(t/1000000).
answered Jan 21 at 15:41
Especially LimeEspecially Lime
22.3k22858
$endgroup$
$begingroup$
What a stupid bug. Thanks a lot for your answer
$endgroup$
– Kranze
Jan 21 at 15:49
add a comment |
$begingroup$
Your calculation is correct, the probability is $frac{11}{36}$. Your code (given in comments) is not giving the right answer because it outputs #successes / #failures instead of #successes / #attempts. Just change the last line to print(t/1000000).
answered Jan 21 at 15:41
Especially LimeEspecially Lime
22.3k22858
$endgroup$
$begingroup$
What a stupid bug. Thanks a lot for your answer
$endgroup$
– Kranze
Jan 21 at 15:49
add a comment |
$begingroup$
Your calculation is correct, the probability is $frac{11}{36}$. Your code (given in comments) is not giving the right answer because it outputs #successes / #failures instead of #successes / #attempts. Just change the last line to print(t/1000000).
answered Jan 21 at 15:41
Especially LimeEspecially Lime
22.3k22858
$endgroup$
Your calculation is correct, the probability is $frac{11}{36}$. Your code (given in comments) is not giving the right answer because it outputs #successes / #failures instead of #successes / #attempts. Just change the last line to print(t/1000000).
answered Jan 21 at 15:41
Especially LimeEspecially Lime
22.3k22858
answered Jan 21 at 15:41
Especially LimeEspecially Lime
22.3k22858
answered Jan 21 at 15:41
Especially LimeEspecially Lime
22.3k22858
answered Jan 21 at 15:41
Especially LimeEspecially Lime
22.3k22858
22.3k22858
$begingroup$
What a stupid bug. Thanks a lot for your answer
$endgroup$
– Kranze
Jan 21 at 15:49
add a comment |
$begingroup$
What a stupid bug. Thanks a lot for your answer
$endgroup$
– Kranze
Jan 21 at 15:49
$begingroup$
What a stupid bug. Thanks a lot for your answer
$endgroup$
– Kranze
Jan 21 at 15:49
$begingroup$
What a stupid bug. Thanks a lot for your answer
$endgroup$
– Kranze
Jan 21 at 15:49
add a comment |
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$begingroup$
What is the correct answer? Unless you have mis-stated the question, this seems correct to me.
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 14:43
$begingroup$
I wrote a little simulation in Python and got approx. 44% chance. Despite this, the professor said that the solution is wrong.
$endgroup$
– Kranze
Jan 21 at 15:05
$begingroup$
Could you recheck the question. I don't see any way to claim that you are wrong.
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 15:11
$begingroup$
Here is my script: <!-- language: lang-py --> import random def throw(): '''coin toss''' coin = random.choice([5, 10]) die = random.choice([1,2,3,4,5,6]) return bool(coin+die==11) def main(): '''perform experiment''' t, f = 0, 0 # true , false counter for _ in range(1000000): throw1 = throw() throw2 = throw() if throw1 or throw2: t+=1 else: f+=1 print(t/f)
$endgroup$
– Kranze
Jan 21 at 15:20
$begingroup$
I hate the fact that I do not know any codes. Can you tell me how I can implement this myself?
$endgroup$
– Mohammad Zuhair Khan
Jan 21 at 15:25