Mixing
Probability that 4 single digit numbers whose sum is 28 contain at least two prime numbers
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What is the probability that a list of 4, not necessarily distinct, single-digit numbers, whose sum is 28, will contain at least two prime numbers?
There are not a whole lot of possible "desirable" cases ,but I was trying to see if there's a way to limit my search.
elementary-number-theory elementary-probability
asked Jan 25 at 19:05
The PianistThe Pianist
153
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|
show 2 more comments
$begingroup$
What is the probability that a list of 4, not necessarily distinct, single-digit numbers, whose sum is 28, will contain at least two prime numbers?
There are not a whole lot of possible "desirable" cases ,but I was trying to see if there's a way to limit my search.
elementary-number-theory elementary-probability
asked Jan 25 at 19:05
The PianistThe Pianist
153
$endgroup$
1
$begingroup$
why can you not compute this by straight enumeration in excel, say?
$endgroup$
– gt6989b
Jan 25 at 19:07
$begingroup$
Does $6,7,7,8$ count?
$endgroup$
– TonyK
Jan 25 at 19:08
3
$begingroup$
Notice that $9+9+5+5 = 28$ is the largest possible sum involving two primes both of which smaller than $7$. It follows that all sequences of four digits whose sum is $28$ and have at least two primes must include at least one $7$ such as $9+9+7+3, 9+7+7+5, 8+7+7+6, 7+7+7+7$. Convince yourself that we have found the full list and count the number of arrangements of the digits. Finally, divide by the total number of sequences of 4 digits whose sum is $28$ to get the probability.
$endgroup$
– JMoravitz
Jan 25 at 19:13
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@gt6989b I'm trying to see if anyone gets better way than bashing. Hence this is tagged "elementary"
$endgroup$
– The Pianist
Jan 25 at 19:13
1
$begingroup$
Welcome to MSE. You are much more likely to get a response if you show what you have tried and where you are stuck. Please show your working.
$endgroup$
– stuart stevenson
Jan 25 at 19:13
|
show 2 more comments
$begingroup$
What is the probability that a list of 4, not necessarily distinct, single-digit numbers, whose sum is 28, will contain at least two prime numbers?
There are not a whole lot of possible "desirable" cases ,but I was trying to see if there's a way to limit my search.
elementary-number-theory elementary-probability
asked Jan 25 at 19:05
The PianistThe Pianist
153
$endgroup$
What is the probability that a list of 4, not necessarily distinct, single-digit numbers, whose sum is 28, will contain at least two prime numbers?
There are not a whole lot of possible "desirable" cases ,but I was trying to see if there's a way to limit my search.
elementary-number-theory elementary-probability
elementary-number-theory elementary-probability
asked Jan 25 at 19:05
The PianistThe Pianist
153
asked Jan 25 at 19:05
The PianistThe Pianist
153
edited Jan 25 at 19:24
The Pianist
asked Jan 25 at 19:05
The PianistThe Pianist
153
asked Jan 25 at 19:05
The PianistThe Pianist
153
asked Jan 25 at 19:05
The PianistThe Pianist
153
153
1
$begingroup$
why can you not compute this by straight enumeration in excel, say?
$endgroup$
– gt6989b
Jan 25 at 19:07
$begingroup$
Does $6,7,7,8$ count?
$endgroup$
– TonyK
Jan 25 at 19:08
3
$begingroup$
Notice that $9+9+5+5 = 28$ is the largest possible sum involving two primes both of which smaller than $7$. It follows that all sequences of four digits whose sum is $28$ and have at least two primes must include at least one $7$ such as $9+9+7+3, 9+7+7+5, 8+7+7+6, 7+7+7+7$. Convince yourself that we have found the full list and count the number of arrangements of the digits. Finally, divide by the total number of sequences of 4 digits whose sum is $28$ to get the probability.
$endgroup$
– JMoravitz
Jan 25 at 19:13
$begingroup$
@gt6989b I'm trying to see if anyone gets better way than bashing. Hence this is tagged "elementary"
$endgroup$
– The Pianist
Jan 25 at 19:13
1
$begingroup$
Welcome to MSE. You are much more likely to get a response if you show what you have tried and where you are stuck. Please show your working.
$endgroup$
– stuart stevenson
Jan 25 at 19:13
|
show 2 more comments
1
$begingroup$
why can you not compute this by straight enumeration in excel, say?
$endgroup$
– gt6989b
Jan 25 at 19:07
$begingroup$
Does $6,7,7,8$ count?
$endgroup$
– TonyK
Jan 25 at 19:08
3
$begingroup$
Notice that $9+9+5+5 = 28$ is the largest possible sum involving two primes both of which smaller than $7$. It follows that all sequences of four digits whose sum is $28$ and have at least two primes must include at least one $7$ such as $9+9+7+3, 9+7+7+5, 8+7+7+6, 7+7+7+7$. Convince yourself that we have found the full list and count the number of arrangements of the digits. Finally, divide by the total number of sequences of 4 digits whose sum is $28$ to get the probability.
$endgroup$
– JMoravitz
Jan 25 at 19:13
$begingroup$
@gt6989b I'm trying to see if anyone gets better way than bashing. Hence this is tagged "elementary"
$endgroup$
– The Pianist
Jan 25 at 19:13
1
$begingroup$
Welcome to MSE. You are much more likely to get a response if you show what you have tried and where you are stuck. Please show your working.
$endgroup$
– stuart stevenson
Jan 25 at 19:13
1
1
$begingroup$
why can you not compute this by straight enumeration in excel, say?
$endgroup$
– gt6989b
Jan 25 at 19:07
$begingroup$
why can you not compute this by straight enumeration in excel, say?
$endgroup$
– gt6989b
Jan 25 at 19:07
$begingroup$
Does $6,7,7,8$ count?
$endgroup$
– TonyK
Jan 25 at 19:08
$begingroup$
Does $6,7,7,8$ count?
$endgroup$
– TonyK
Jan 25 at 19:08
3
3
$begingroup$
Notice that $9+9+5+5 = 28$ is the largest possible sum involving two primes both of which smaller than $7$. It follows that all sequences of four digits whose sum is $28$ and have at least two primes must include at least one $7$ such as $9+9+7+3, 9+7+7+5, 8+7+7+6, 7+7+7+7$. Convince yourself that we have found the full list and count the number of arrangements of the digits. Finally, divide by the total number of sequences of 4 digits whose sum is $28$ to get the probability.
$endgroup$
– JMoravitz
Jan 25 at 19:13
$begingroup$
Notice that $9+9+5+5 = 28$ is the largest possible sum involving two primes both of which smaller than $7$. It follows that all sequences of four digits whose sum is $28$ and have at least two primes must include at least one $7$ such as $9+9+7+3, 9+7+7+5, 8+7+7+6, 7+7+7+7$. Convince yourself that we have found the full list and count the number of arrangements of the digits. Finally, divide by the total number of sequences of 4 digits whose sum is $28$ to get the probability.
$endgroup$
– JMoravitz
Jan 25 at 19:13
$begingroup$
@gt6989b I'm trying to see if anyone gets better way than bashing. Hence this is tagged "elementary"
$endgroup$
– The Pianist
Jan 25 at 19:13
$begingroup$
@gt6989b I'm trying to see if anyone gets better way than bashing. Hence this is tagged "elementary"
$endgroup$
– The Pianist
Jan 25 at 19:13
1
1
$begingroup$
Welcome to MSE. You are much more likely to get a response if you show what you have tried and where you are stuck. Please show your working.
$endgroup$
– stuart stevenson
Jan 25 at 19:13
$begingroup$
Welcome to MSE. You are much more likely to get a response if you show what you have tried and where you are stuck. Please show your working.
$endgroup$
– stuart stevenson
Jan 25 at 19:13
|
show 2 more comments
1 Answer
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$begingroup$
There are only six cases that fulfill the requirements:
{{9, 9, 7, 3}, {9, 9, 5, 5}, {9, 7, 7, 5}, {8, 8, 7, 5}, {8, 7, 7, 6}, {7, 7, 7, 7}}
The total number of cases is 15, so the probability is $0.4$.
Start with the highest value in a set of four is $9$. If the next value is also $9$, then the following two values must be primes, giving $7$ and $3$ or $5$ and $5$.
It is simple to see that $9, 8$ has no two prime completions, to move to $9, 7$. The only completion is then $7, 5$. And we're done with $9$ as the largest digit.
Consider $8$ as the largest digit. The largest second digit is then $8$, which gives a completion of $7, 5$.
And so on....
answered Jan 25 at 20:12
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
1
$begingroup$
The only missing case seems to be {8,7,7,6}. Inspired by replies, I came to this approach: since 28/4=7, the 4 numbers should be relatively large. Let's start from the two primes, we have only four cases {7,7}, {7,5}, {5,5}, and {7,3}. For each case we can easily find the remaining two numbers. Total number of desirable cases is 6. If I'm not mistaken, the number of all cases with sum of 28 is 15 (9/9/9/1, all the way to 7/7/7/7). So the probability is 6/15=2/5.
$endgroup$
– The Pianist
Jan 25 at 21:10
add a comment |
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$begingroup$
There are only six cases that fulfill the requirements:
{{9, 9, 7, 3}, {9, 9, 5, 5}, {9, 7, 7, 5}, {8, 8, 7, 5}, {8, 7, 7, 6}, {7, 7, 7, 7}}
The total number of cases is 15, so the probability is $0.4$.
Start with the highest value in a set of four is $9$. If the next value is also $9$, then the following two values must be primes, giving $7$ and $3$ or $5$ and $5$.
It is simple to see that $9, 8$ has no two prime completions, to move to $9, 7$. The only completion is then $7, 5$. And we're done with $9$ as the largest digit.
Consider $8$ as the largest digit. The largest second digit is then $8$, which gives a completion of $7, 5$.
And so on....
answered Jan 25 at 20:12
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
1
$begingroup$
The only missing case seems to be {8,7,7,6}. Inspired by replies, I came to this approach: since 28/4=7, the 4 numbers should be relatively large. Let's start from the two primes, we have only four cases {7,7}, {7,5}, {5,5}, and {7,3}. For each case we can easily find the remaining two numbers. Total number of desirable cases is 6. If I'm not mistaken, the number of all cases with sum of 28 is 15 (9/9/9/1, all the way to 7/7/7/7). So the probability is 6/15=2/5.
$endgroup$
– The Pianist
Jan 25 at 21:10
add a comment |
$begingroup$
There are only six cases that fulfill the requirements:
{{9, 9, 7, 3}, {9, 9, 5, 5}, {9, 7, 7, 5}, {8, 8, 7, 5}, {8, 7, 7, 6}, {7, 7, 7, 7}}
The total number of cases is 15, so the probability is $0.4$.
Start with the highest value in a set of four is $9$. If the next value is also $9$, then the following two values must be primes, giving $7$ and $3$ or $5$ and $5$.
It is simple to see that $9, 8$ has no two prime completions, to move to $9, 7$. The only completion is then $7, 5$. And we're done with $9$ as the largest digit.
Consider $8$ as the largest digit. The largest second digit is then $8$, which gives a completion of $7, 5$.
And so on....
answered Jan 25 at 20:12
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
1
$begingroup$
The only missing case seems to be {8,7,7,6}. Inspired by replies, I came to this approach: since 28/4=7, the 4 numbers should be relatively large. Let's start from the two primes, we have only four cases {7,7}, {7,5}, {5,5}, and {7,3}. For each case we can easily find the remaining two numbers. Total number of desirable cases is 6. If I'm not mistaken, the number of all cases with sum of 28 is 15 (9/9/9/1, all the way to 7/7/7/7). So the probability is 6/15=2/5.
$endgroup$
– The Pianist
Jan 25 at 21:10
add a comment |
$begingroup$
There are only six cases that fulfill the requirements:
{{9, 9, 7, 3}, {9, 9, 5, 5}, {9, 7, 7, 5}, {8, 8, 7, 5}, {8, 7, 7, 6}, {7, 7, 7, 7}}
The total number of cases is 15, so the probability is $0.4$.
Start with the highest value in a set of four is $9$. If the next value is also $9$, then the following two values must be primes, giving $7$ and $3$ or $5$ and $5$.
It is simple to see that $9, 8$ has no two prime completions, to move to $9, 7$. The only completion is then $7, 5$. And we're done with $9$ as the largest digit.
Consider $8$ as the largest digit. The largest second digit is then $8$, which gives a completion of $7, 5$.
And so on....
answered Jan 25 at 20:12
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
There are only six cases that fulfill the requirements:
{{9, 9, 7, 3}, {9, 9, 5, 5}, {9, 7, 7, 5}, {8, 8, 7, 5}, {8, 7, 7, 6}, {7, 7, 7, 7}}
The total number of cases is 15, so the probability is $0.4$.
Start with the highest value in a set of four is $9$. If the next value is also $9$, then the following two values must be primes, giving $7$ and $3$ or $5$ and $5$.
It is simple to see that $9, 8$ has no two prime completions, to move to $9, 7$. The only completion is then $7, 5$. And we're done with $9$ as the largest digit.
Consider $8$ as the largest digit. The largest second digit is then $8$, which gives a completion of $7, 5$.
And so on....
answered Jan 25 at 20:12
David G. StorkDavid G. Stork
11.1k41432
edited Jan 25 at 21:32
answered Jan 25 at 20:12
David G. StorkDavid G. Stork
11.1k41432
answered Jan 25 at 20:12
David G. StorkDavid G. Stork
11.1k41432
answered Jan 25 at 20:12
David G. StorkDavid G. Stork
11.1k41432
11.1k41432
1
$begingroup$
The only missing case seems to be {8,7,7,6}. Inspired by replies, I came to this approach: since 28/4=7, the 4 numbers should be relatively large. Let's start from the two primes, we have only four cases {7,7}, {7,5}, {5,5}, and {7,3}. For each case we can easily find the remaining two numbers. Total number of desirable cases is 6. If I'm not mistaken, the number of all cases with sum of 28 is 15 (9/9/9/1, all the way to 7/7/7/7). So the probability is 6/15=2/5.
$endgroup$
– The Pianist
Jan 25 at 21:10
add a comment |
1
$begingroup$
The only missing case seems to be {8,7,7,6}. Inspired by replies, I came to this approach: since 28/4=7, the 4 numbers should be relatively large. Let's start from the two primes, we have only four cases {7,7}, {7,5}, {5,5}, and {7,3}. For each case we can easily find the remaining two numbers. Total number of desirable cases is 6. If I'm not mistaken, the number of all cases with sum of 28 is 15 (9/9/9/1, all the way to 7/7/7/7). So the probability is 6/15=2/5.
$endgroup$
– The Pianist
Jan 25 at 21:10
1
1
$begingroup$
The only missing case seems to be {8,7,7,6}. Inspired by replies, I came to this approach: since 28/4=7, the 4 numbers should be relatively large. Let's start from the two primes, we have only four cases {7,7}, {7,5}, {5,5}, and {7,3}. For each case we can easily find the remaining two numbers. Total number of desirable cases is 6. If I'm not mistaken, the number of all cases with sum of 28 is 15 (9/9/9/1, all the way to 7/7/7/7). So the probability is 6/15=2/5.
$endgroup$
– The Pianist
Jan 25 at 21:10
$begingroup$
The only missing case seems to be {8,7,7,6}. Inspired by replies, I came to this approach: since 28/4=7, the 4 numbers should be relatively large. Let's start from the two primes, we have only four cases {7,7}, {7,5}, {5,5}, and {7,3}. For each case we can easily find the remaining two numbers. Total number of desirable cases is 6. If I'm not mistaken, the number of all cases with sum of 28 is 15 (9/9/9/1, all the way to 7/7/7/7). So the probability is 6/15=2/5.
$endgroup$
– The Pianist
Jan 25 at 21:10
add a comment |
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1
$begingroup$
why can you not compute this by straight enumeration in excel, say?
$endgroup$
– gt6989b
Jan 25 at 19:07
$begingroup$
Does $6,7,7,8$ count?
$endgroup$
– TonyK
Jan 25 at 19:08
3
$begingroup$
Notice that $9+9+5+5 = 28$ is the largest possible sum involving two primes both of which smaller than $7$. It follows that all sequences of four digits whose sum is $28$ and have at least two primes must include at least one $7$ such as $9+9+7+3, 9+7+7+5, 8+7+7+6, 7+7+7+7$. Convince yourself that we have found the full list and count the number of arrangements of the digits. Finally, divide by the total number of sequences of 4 digits whose sum is $28$ to get the probability.
$endgroup$
– JMoravitz
Jan 25 at 19:13
$begingroup$
@gt6989b I'm trying to see if anyone gets better way than bashing. Hence this is tagged "elementary"
$endgroup$
– The Pianist
Jan 25 at 19:13
1
$begingroup$
Welcome to MSE. You are much more likely to get a response if you show what you have tried and where you are stuck. Please show your working.
$endgroup$
– stuart stevenson
Jan 25 at 19:13