Mixing
Measurement of probability from nine consecutive integers
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Three distinct positive integers are selected at random from 9 consecutive positive integers. What is the probability that their average is also an integer?
Is there any way to find it efficiently? I'd taken example of $1,2,3,4,5,6,7,8,9$ then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D
Thank you.
probability combinatorics
edited Jul 15 at 8:42
Parcly Taxel
33.6k136588
asked Jul 15 at 8:25
Sagar Gautam
1064
add a comment |Â
up vote
0
down vote
favorite
Three distinct positive integers are selected at random from 9 consecutive positive integers. What is the probability that their average is also an integer?
Is there any way to find it efficiently? I'd taken example of $1,2,3,4,5,6,7,8,9$ then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D
Thank you.
probability combinatorics
edited Jul 15 at 8:42
Parcly Taxel
33.6k136588
asked Jul 15 at 8:25
Sagar Gautam
1064
Is this selection with or without replacement?
– Lord Shark the Unknown
Jul 15 at 8:26
@LordSharktheUnknown Yes, without replacement
– Sagar Gautam
Jul 15 at 8:27
I'd taken example of 1,2,3,4,5,6,7,8,9 then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D
– Sagar Gautam
Jul 15 at 8:37
This only depends on what your picks are modulo $3$.
– Lord Shark the Unknown
Jul 15 at 8:40
@LordSharktheUnknown can you describe it in detail please ?
– Sagar Gautam
Jul 15 at 8:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Three distinct positive integers are selected at random from 9 consecutive positive integers. What is the probability that their average is also an integer?
Is there any way to find it efficiently? I'd taken example of $1,2,3,4,5,6,7,8,9$ then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D
Thank you.
probability combinatorics
edited Jul 15 at 8:42
Parcly Taxel
33.6k136588
asked Jul 15 at 8:25
Sagar Gautam
1064
Three distinct positive integers are selected at random from 9 consecutive positive integers. What is the probability that their average is also an integer?
Is there any way to find it efficiently? I'd taken example of $1,2,3,4,5,6,7,8,9$ then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D
Thank you.
probability combinatorics
edited Jul 15 at 8:42
Parcly Taxel
33.6k136588
asked Jul 15 at 8:25
Sagar Gautam
1064
edited Jul 15 at 8:42
Parcly Taxel
33.6k136588
edited Jul 15 at 8:42
Parcly Taxel
33.6k136588
edited Jul 15 at 8:42
Parcly Taxel
33.6k136588
33.6k136588
asked Jul 15 at 8:25
Sagar Gautam
1064
asked Jul 15 at 8:25
Sagar Gautam
1064
asked Jul 15 at 8:25
Sagar Gautam
1064
1064
Is this selection with or without replacement?
– Lord Shark the Unknown
Jul 15 at 8:26
@LordSharktheUnknown Yes, without replacement
– Sagar Gautam
Jul 15 at 8:27
I'd taken example of 1,2,3,4,5,6,7,8,9 then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D
– Sagar Gautam
Jul 15 at 8:37
This only depends on what your picks are modulo $3$.
– Lord Shark the Unknown
Jul 15 at 8:40
@LordSharktheUnknown can you describe it in detail please ?
– Sagar Gautam
Jul 15 at 8:42
add a comment |Â
Is this selection with or without replacement?
– Lord Shark the Unknown
Jul 15 at 8:26
@LordSharktheUnknown Yes, without replacement
– Sagar Gautam
Jul 15 at 8:27
I'd taken example of 1,2,3,4,5,6,7,8,9 then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D
– Sagar Gautam
Jul 15 at 8:37
This only depends on what your picks are modulo $3$.
– Lord Shark the Unknown
Jul 15 at 8:40
@LordSharktheUnknown can you describe it in detail please ?
– Sagar Gautam
Jul 15 at 8:42
Is this selection with or without replacement?
– Lord Shark the Unknown
Jul 15 at 8:26
Is this selection with or without replacement?
– Lord Shark the Unknown
Jul 15 at 8:26
@LordSharktheUnknown Yes, without replacement
– Sagar Gautam
Jul 15 at 8:27
@LordSharktheUnknown Yes, without replacement
– Sagar Gautam
Jul 15 at 8:27
I'd taken example of 1,2,3,4,5,6,7,8,9 then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D
– Sagar Gautam
Jul 15 at 8:37
I'd taken example of 1,2,3,4,5,6,7,8,9 then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D
– Sagar Gautam
Jul 15 at 8:37
This only depends on what your picks are modulo $3$.
– Lord Shark the Unknown
Jul 15 at 8:40
This only depends on what your picks are modulo $3$.
– Lord Shark the Unknown
Jul 15 at 8:40
@LordSharktheUnknown can you describe it in detail please ?
– Sagar Gautam
Jul 15 at 8:42
@LordSharktheUnknown can you describe it in detail please ?
– Sagar Gautam
Jul 15 at 8:42
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
There will be three each of integers that are $0,1,2bmod3$ in the given range. The admissible selections sum to $0bmod3$, of which there are four ways to achieve it:
$$000 (1text selection)$$
$$111 (1)$$
$$222 (1)$$
$$012 (27)$$
This gives 30 admissible selections out of $binom93=84$ total selections, for a probability of $frac3084=frac514$.
answered Jul 15 at 8:50
Parcly Taxel
33.6k136588
Can you describe how you get 1,1,1 and 27 ? I've understand all other things
– Sagar Gautam
Jul 15 at 8:54
@SagarGautam concretely, using 1 to 9, there are exactly three numbers $0bmod3$ (3, 6, 9), so only one way to choose all of them in one selection (the 000 case). Same for 111 (1, 4, 7) and 222 (2, 5, 8). For 012, we are free to choose one of the $0bmod3$ numbers, and likewise for the 1 and 2 modulo 3 numbers, giving $3×3×3=27$ ways.
– Parcly Taxel
Jul 15 at 8:57
add a comment |Â
up vote
1
down vote
The average is an integer if the sum is divisible by $3$, which is the case if and only if the residues of the integers modulo $3$ are either all different or all the same. The probability for them to be all different is
$$
fracbinom31^3binom93=frac928;,
$$
and the probability for them to all be the same is
$$
frac3binom93=frac128;.
$$
Thus the probability for the average to be an integer is $frac1+928=frac514$.
answered Jul 15 at 8:51
joriki
165k10180328
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
There will be three each of integers that are $0,1,2bmod3$ in the given range. The admissible selections sum to $0bmod3$, of which there are four ways to achieve it:
$$000 (1text selection)$$
$$111 (1)$$
$$222 (1)$$
$$012 (27)$$
This gives 30 admissible selections out of $binom93=84$ total selections, for a probability of $frac3084=frac514$.
answered Jul 15 at 8:50
Parcly Taxel
33.6k136588
Can you describe how you get 1,1,1 and 27 ? I've understand all other things
– Sagar Gautam
Jul 15 at 8:54
@SagarGautam concretely, using 1 to 9, there are exactly three numbers $0bmod3$ (3, 6, 9), so only one way to choose all of them in one selection (the 000 case). Same for 111 (1, 4, 7) and 222 (2, 5, 8). For 012, we are free to choose one of the $0bmod3$ numbers, and likewise for the 1 and 2 modulo 3 numbers, giving $3×3×3=27$ ways.
– Parcly Taxel
Jul 15 at 8:57
add a comment |Â
up vote
2
down vote
accepted
There will be three each of integers that are $0,1,2bmod3$ in the given range. The admissible selections sum to $0bmod3$, of which there are four ways to achieve it:
$$000 (1text selection)$$
$$111 (1)$$
$$222 (1)$$
$$012 (27)$$
This gives 30 admissible selections out of $binom93=84$ total selections, for a probability of $frac3084=frac514$.
answered Jul 15 at 8:50
Parcly Taxel
33.6k136588
Can you describe how you get 1,1,1 and 27 ? I've understand all other things
– Sagar Gautam
Jul 15 at 8:54
@SagarGautam concretely, using 1 to 9, there are exactly three numbers $0bmod3$ (3, 6, 9), so only one way to choose all of them in one selection (the 000 case). Same for 111 (1, 4, 7) and 222 (2, 5, 8). For 012, we are free to choose one of the $0bmod3$ numbers, and likewise for the 1 and 2 modulo 3 numbers, giving $3×3×3=27$ ways.
– Parcly Taxel
Jul 15 at 8:57
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
There will be three each of integers that are $0,1,2bmod3$ in the given range. The admissible selections sum to $0bmod3$, of which there are four ways to achieve it:
$$000 (1text selection)$$
$$111 (1)$$
$$222 (1)$$
$$012 (27)$$
This gives 30 admissible selections out of $binom93=84$ total selections, for a probability of $frac3084=frac514$.
answered Jul 15 at 8:50
Parcly Taxel
33.6k136588
There will be three each of integers that are $0,1,2bmod3$ in the given range. The admissible selections sum to $0bmod3$, of which there are four ways to achieve it:
$$000 (1text selection)$$
$$111 (1)$$
$$222 (1)$$
$$012 (27)$$
This gives 30 admissible selections out of $binom93=84$ total selections, for a probability of $frac3084=frac514$.
answered Jul 15 at 8:50
Parcly Taxel
33.6k136588
answered Jul 15 at 8:50
Parcly Taxel
33.6k136588
answered Jul 15 at 8:50
Parcly Taxel
33.6k136588
answered Jul 15 at 8:50
Parcly Taxel
33.6k136588
33.6k136588
Can you describe how you get 1,1,1 and 27 ? I've understand all other things
– Sagar Gautam
Jul 15 at 8:54
@SagarGautam concretely, using 1 to 9, there are exactly three numbers $0bmod3$ (3, 6, 9), so only one way to choose all of them in one selection (the 000 case). Same for 111 (1, 4, 7) and 222 (2, 5, 8). For 012, we are free to choose one of the $0bmod3$ numbers, and likewise for the 1 and 2 modulo 3 numbers, giving $3×3×3=27$ ways.
– Parcly Taxel
Jul 15 at 8:57
add a comment |Â
Can you describe how you get 1,1,1 and 27 ? I've understand all other things
– Sagar Gautam
Jul 15 at 8:54
@SagarGautam concretely, using 1 to 9, there are exactly three numbers $0bmod3$ (3, 6, 9), so only one way to choose all of them in one selection (the 000 case). Same for 111 (1, 4, 7) and 222 (2, 5, 8). For 012, we are free to choose one of the $0bmod3$ numbers, and likewise for the 1 and 2 modulo 3 numbers, giving $3×3×3=27$ ways.
– Parcly Taxel
Jul 15 at 8:57
Can you describe how you get 1,1,1 and 27 ? I've understand all other things
– Sagar Gautam
Jul 15 at 8:54
Can you describe how you get 1,1,1 and 27 ? I've understand all other things
– Sagar Gautam
Jul 15 at 8:54
@SagarGautam concretely, using 1 to 9, there are exactly three numbers $0bmod3$ (3, 6, 9), so only one way to choose all of them in one selection (the 000 case). Same for 111 (1, 4, 7) and 222 (2, 5, 8). For 012, we are free to choose one of the $0bmod3$ numbers, and likewise for the 1 and 2 modulo 3 numbers, giving $3×3×3=27$ ways.
– Parcly Taxel
Jul 15 at 8:57
@SagarGautam concretely, using 1 to 9, there are exactly three numbers $0bmod3$ (3, 6, 9), so only one way to choose all of them in one selection (the 000 case). Same for 111 (1, 4, 7) and 222 (2, 5, 8). For 012, we are free to choose one of the $0bmod3$ numbers, and likewise for the 1 and 2 modulo 3 numbers, giving $3×3×3=27$ ways.
– Parcly Taxel
Jul 15 at 8:57
add a comment |Â
up vote
1
down vote
The average is an integer if the sum is divisible by $3$, which is the case if and only if the residues of the integers modulo $3$ are either all different or all the same. The probability for them to be all different is
$$
fracbinom31^3binom93=frac928;,
$$
and the probability for them to all be the same is
$$
frac3binom93=frac128;.
$$
Thus the probability for the average to be an integer is $frac1+928=frac514$.
answered Jul 15 at 8:51
joriki
165k10180328
add a comment |Â
up vote
1
down vote
The average is an integer if the sum is divisible by $3$, which is the case if and only if the residues of the integers modulo $3$ are either all different or all the same. The probability for them to be all different is
$$
fracbinom31^3binom93=frac928;,
$$
and the probability for them to all be the same is
$$
frac3binom93=frac128;.
$$
Thus the probability for the average to be an integer is $frac1+928=frac514$.
answered Jul 15 at 8:51
joriki
165k10180328
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The average is an integer if the sum is divisible by $3$, which is the case if and only if the residues of the integers modulo $3$ are either all different or all the same. The probability for them to be all different is
$$
fracbinom31^3binom93=frac928;,
$$
and the probability for them to all be the same is
$$
frac3binom93=frac128;.
$$
Thus the probability for the average to be an integer is $frac1+928=frac514$.
answered Jul 15 at 8:51
joriki
165k10180328
The average is an integer if the sum is divisible by $3$, which is the case if and only if the residues of the integers modulo $3$ are either all different or all the same. The probability for them to be all different is
$$
fracbinom31^3binom93=frac928;,
$$
and the probability for them to all be the same is
$$
frac3binom93=frac128;.
$$
Thus the probability for the average to be an integer is $frac1+928=frac514$.
answered Jul 15 at 8:51
joriki
165k10180328
answered Jul 15 at 8:51
joriki
165k10180328
answered Jul 15 at 8:51
joriki
165k10180328
answered Jul 15 at 8:51
joriki
165k10180328
165k10180328
add a comment |Â
add a comment |Â
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Is this selection with or without replacement?
– Lord Shark the Unknown
Jul 15 at 8:26
@LordSharktheUnknown Yes, without replacement
– Sagar Gautam
Jul 15 at 8:27
I'd taken example of 1,2,3,4,5,6,7,8,9 then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D
– Sagar Gautam
Jul 15 at 8:37
This only depends on what your picks are modulo $3$.
– Lord Shark the Unknown
Jul 15 at 8:40
@LordSharktheUnknown can you describe it in detail please ?
– Sagar Gautam
Jul 15 at 8:42