Mixing
We throw $5$ dice: What is the probability to have $4$ different numbers?
$begingroup$
We throw $5$ dice: What is the probability to have $4$ different numbers?
I know it is $$frac{6cdot 5cdot 4cdot 3}{6^5}.$$
I wanted to use an other argument, but it look to not work : I take $binom{6}{4}$ numbers. Then I have $$frac{6cdot 5cdot 4cdot 3}{4!}$$
possibilities. Then I have to multiply this result by $4!$ and I don't understand why. Indeed, I would like to multiply by $5!$ since we can distribute the $5$ colors in e.g. $1;2;3;4;4$ in $5!$ different ways.
- If I want all dice different, this argument works: I take $binom{6}{5}$ number, then I can distribute the colors in $5!$ different way which give $frac{5cdot 5cdot 4cdot 3cdot 2}{5!}5!=6cdot 5cdot 4cdot 3cdot 2$ possibilities, that is the correct answer.
So why doesn't it work with the previous situation ?
probability combinatorics
edited Jan 11 at 9:49
N. F. Taussig
44.1k93356
asked Jan 11 at 8:58
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
We throw $5$ dice: What is the probability to have $4$ different numbers?
I know it is $$frac{6cdot 5cdot 4cdot 3}{6^5}.$$
I wanted to use an other argument, but it look to not work : I take $binom{6}{4}$ numbers. Then I have $$frac{6cdot 5cdot 4cdot 3}{4!}$$
possibilities. Then I have to multiply this result by $4!$ and I don't understand why. Indeed, I would like to multiply by $5!$ since we can distribute the $5$ colors in e.g. $1;2;3;4;4$ in $5!$ different ways.
- If I want all dice different, this argument works: I take $binom{6}{5}$ number, then I can distribute the colors in $5!$ different way which give $frac{5cdot 5cdot 4cdot 3cdot 2}{5!}5!=6cdot 5cdot 4cdot 3cdot 2$ possibilities, that is the correct answer.
So why doesn't it work with the previous situation ?
probability combinatorics
edited Jan 11 at 9:49
N. F. Taussig
44.1k93356
asked Jan 11 at 8:58
NewMathNewMath
4059
$endgroup$
1
$begingroup$
"I know it is..." - how?
$endgroup$
– zhoraster
Jan 11 at 9:34
add a comment |
$begingroup$
We throw $5$ dice: What is the probability to have $4$ different numbers?
I know it is $$frac{6cdot 5cdot 4cdot 3}{6^5}.$$
I wanted to use an other argument, but it look to not work : I take $binom{6}{4}$ numbers. Then I have $$frac{6cdot 5cdot 4cdot 3}{4!}$$
possibilities. Then I have to multiply this result by $4!$ and I don't understand why. Indeed, I would like to multiply by $5!$ since we can distribute the $5$ colors in e.g. $1;2;3;4;4$ in $5!$ different ways.
- If I want all dice different, this argument works: I take $binom{6}{5}$ number, then I can distribute the colors in $5!$ different way which give $frac{5cdot 5cdot 4cdot 3cdot 2}{5!}5!=6cdot 5cdot 4cdot 3cdot 2$ possibilities, that is the correct answer.
So why doesn't it work with the previous situation ?
probability combinatorics
edited Jan 11 at 9:49
N. F. Taussig
44.1k93356
asked Jan 11 at 8:58
NewMathNewMath
4059
$endgroup$
We throw $5$ dice: What is the probability to have $4$ different numbers?
I know it is $$frac{6cdot 5cdot 4cdot 3}{6^5}.$$
I wanted to use an other argument, but it look to not work : I take $binom{6}{4}$ numbers. Then I have $$frac{6cdot 5cdot 4cdot 3}{4!}$$
possibilities. Then I have to multiply this result by $4!$ and I don't understand why. Indeed, I would like to multiply by $5!$ since we can distribute the $5$ colors in e.g. $1;2;3;4;4$ in $5!$ different ways.
- If I want all dice different, this argument works: I take $binom{6}{5}$ number, then I can distribute the colors in $5!$ different way which give $frac{5cdot 5cdot 4cdot 3cdot 2}{5!}5!=6cdot 5cdot 4cdot 3cdot 2$ possibilities, that is the correct answer.
So why doesn't it work with the previous situation ?
probability combinatorics
probability combinatorics
edited Jan 11 at 9:49
N. F. Taussig
44.1k93356
asked Jan 11 at 8:58
NewMathNewMath
4059
edited Jan 11 at 9:49
N. F. Taussig
44.1k93356
asked Jan 11 at 8:58
NewMathNewMath
4059
edited Jan 11 at 9:49
N. F. Taussig
44.1k93356
edited Jan 11 at 9:49
N. F. Taussig
44.1k93356
edited Jan 11 at 9:49
N. F. Taussig
44.1k93356
44.1k93356
asked Jan 11 at 8:58
NewMathNewMath
4059
asked Jan 11 at 8:58
NewMathNewMath
4059
asked Jan 11 at 8:58
NewMathNewMath
4059
4059
1
$begingroup$
"I know it is..." - how?
$endgroup$
– zhoraster
Jan 11 at 9:34
add a comment |
1
$begingroup$
"I know it is..." - how?
$endgroup$
– zhoraster
Jan 11 at 9:34
1
1
$begingroup$
"I know it is..." - how?
$endgroup$
– zhoraster
Jan 11 at 9:34
$begingroup$
"I know it is..." - how?
$endgroup$
– zhoraster
Jan 11 at 9:34
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Notice that for a favorable outcome, exactly two of the five dice must display the same number, while each of the other three dice must display different numbers.
Suppose the dice are five different colors so that we can distinguish between them. There are $binom{5}{2}$ ways for two of the five dice to display the same number and $binom{6}{1} = 6$ possible numbers for those two dice to display. There are five possible numbers remaining for the other three dice to display. The number of possible ways those three dice can display the numbers is $5 cdot 4 cdot 3$ (where we list the outcomes as ordered triples in the order the colors appear in an alphabetical list). Hence, the number of favorable outcomes is
$$binom{5}{2} cdot 6 cdot 5 cdot 4 cdot 3$$
from which we obtain the probability
$$frac{binom{5}{2} cdot 6 cdot 5 cdot 4 cdot 3}{6^5}$$
that exactly four different numbers are displayed when five dice are rolled.
answered Jan 11 at 9:46
N. F. TaussigN. F. Taussig
44.1k93356
$endgroup$
add a comment |
$begingroup$
You can throw one by one $5$ dice.
Discern events $AXBCD$, $ABXCD$, $ABCXD$ and $ABCDX$ where e.g. $ABCXD$ stands for the event that the fourth throws give the same result as one of the former throws and there are exactly $4$ distinct results.
The events are mutually exclusive so the probability to get exactly $4$ different numbers is:$$P(AXBCD)+P(ABXCD)+P(ABCXD)+P(ABCDX)=$$$$frac66frac16frac56frac46frac36+frac66frac56frac26frac46frac36+frac66frac56frac46frac36frac36+frac66frac56frac46frac36frac46=frac56frac46frac36left(frac16+frac26+frac36+frac46right)$$
answered Jan 11 at 11:32
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
Suppose the output is in 5 boxes like [x][x][x][x][x]
For 4 definite different outcomes we have 6 x 5 x 4 x 3 (in any order)
For the remaining one, it can be any of the 6 digits so we have to choose 1 out of 5 boxes
and also the one box can contain any of the 6 digits i.e. 5c1 x 6
Therefore the probability should be (5c1 x 6 x 6 x 5 x 4 x 3)/6^5
answered Jan 11 at 12:46
cognitivecognitive
264
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notice that for a favorable outcome, exactly two of the five dice must display the same number, while each of the other three dice must display different numbers.
Suppose the dice are five different colors so that we can distinguish between them. There are $binom{5}{2}$ ways for two of the five dice to display the same number and $binom{6}{1} = 6$ possible numbers for those two dice to display. There are five possible numbers remaining for the other three dice to display. The number of possible ways those three dice can display the numbers is $5 cdot 4 cdot 3$ (where we list the outcomes as ordered triples in the order the colors appear in an alphabetical list). Hence, the number of favorable outcomes is
$$binom{5}{2} cdot 6 cdot 5 cdot 4 cdot 3$$
from which we obtain the probability
$$frac{binom{5}{2} cdot 6 cdot 5 cdot 4 cdot 3}{6^5}$$
that exactly four different numbers are displayed when five dice are rolled.
answered Jan 11 at 9:46
N. F. TaussigN. F. Taussig
44.1k93356
$endgroup$
add a comment |
$begingroup$
Notice that for a favorable outcome, exactly two of the five dice must display the same number, while each of the other three dice must display different numbers.
Suppose the dice are five different colors so that we can distinguish between them. There are $binom{5}{2}$ ways for two of the five dice to display the same number and $binom{6}{1} = 6$ possible numbers for those two dice to display. There are five possible numbers remaining for the other three dice to display. The number of possible ways those three dice can display the numbers is $5 cdot 4 cdot 3$ (where we list the outcomes as ordered triples in the order the colors appear in an alphabetical list). Hence, the number of favorable outcomes is
$$binom{5}{2} cdot 6 cdot 5 cdot 4 cdot 3$$
from which we obtain the probability
$$frac{binom{5}{2} cdot 6 cdot 5 cdot 4 cdot 3}{6^5}$$
that exactly four different numbers are displayed when five dice are rolled.
answered Jan 11 at 9:46
N. F. TaussigN. F. Taussig
44.1k93356
$endgroup$
add a comment |
$begingroup$
Notice that for a favorable outcome, exactly two of the five dice must display the same number, while each of the other three dice must display different numbers.
Suppose the dice are five different colors so that we can distinguish between them. There are $binom{5}{2}$ ways for two of the five dice to display the same number and $binom{6}{1} = 6$ possible numbers for those two dice to display. There are five possible numbers remaining for the other three dice to display. The number of possible ways those three dice can display the numbers is $5 cdot 4 cdot 3$ (where we list the outcomes as ordered triples in the order the colors appear in an alphabetical list). Hence, the number of favorable outcomes is
$$binom{5}{2} cdot 6 cdot 5 cdot 4 cdot 3$$
from which we obtain the probability
$$frac{binom{5}{2} cdot 6 cdot 5 cdot 4 cdot 3}{6^5}$$
that exactly four different numbers are displayed when five dice are rolled.
answered Jan 11 at 9:46
N. F. TaussigN. F. Taussig
44.1k93356
$endgroup$
Notice that for a favorable outcome, exactly two of the five dice must display the same number, while each of the other three dice must display different numbers.
Suppose the dice are five different colors so that we can distinguish between them. There are $binom{5}{2}$ ways for two of the five dice to display the same number and $binom{6}{1} = 6$ possible numbers for those two dice to display. There are five possible numbers remaining for the other three dice to display. The number of possible ways those three dice can display the numbers is $5 cdot 4 cdot 3$ (where we list the outcomes as ordered triples in the order the colors appear in an alphabetical list). Hence, the number of favorable outcomes is
$$binom{5}{2} cdot 6 cdot 5 cdot 4 cdot 3$$
from which we obtain the probability
$$frac{binom{5}{2} cdot 6 cdot 5 cdot 4 cdot 3}{6^5}$$
that exactly four different numbers are displayed when five dice are rolled.
answered Jan 11 at 9:46
N. F. TaussigN. F. Taussig
44.1k93356
answered Jan 11 at 9:46
N. F. TaussigN. F. Taussig
44.1k93356
answered Jan 11 at 9:46
N. F. TaussigN. F. Taussig
44.1k93356
answered Jan 11 at 9:46
N. F. TaussigN. F. Taussig
44.1k93356
44.1k93356
add a comment |
add a comment |
$begingroup$
You can throw one by one $5$ dice.
Discern events $AXBCD$, $ABXCD$, $ABCXD$ and $ABCDX$ where e.g. $ABCXD$ stands for the event that the fourth throws give the same result as one of the former throws and there are exactly $4$ distinct results.
The events are mutually exclusive so the probability to get exactly $4$ different numbers is:$$P(AXBCD)+P(ABXCD)+P(ABCXD)+P(ABCDX)=$$$$frac66frac16frac56frac46frac36+frac66frac56frac26frac46frac36+frac66frac56frac46frac36frac36+frac66frac56frac46frac36frac46=frac56frac46frac36left(frac16+frac26+frac36+frac46right)$$
answered Jan 11 at 11:32
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
You can throw one by one $5$ dice.
Discern events $AXBCD$, $ABXCD$, $ABCXD$ and $ABCDX$ where e.g. $ABCXD$ stands for the event that the fourth throws give the same result as one of the former throws and there are exactly $4$ distinct results.
The events are mutually exclusive so the probability to get exactly $4$ different numbers is:$$P(AXBCD)+P(ABXCD)+P(ABCXD)+P(ABCDX)=$$$$frac66frac16frac56frac46frac36+frac66frac56frac26frac46frac36+frac66frac56frac46frac36frac36+frac66frac56frac46frac36frac46=frac56frac46frac36left(frac16+frac26+frac36+frac46right)$$
answered Jan 11 at 11:32
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
You can throw one by one $5$ dice.
Discern events $AXBCD$, $ABXCD$, $ABCXD$ and $ABCDX$ where e.g. $ABCXD$ stands for the event that the fourth throws give the same result as one of the former throws and there are exactly $4$ distinct results.
The events are mutually exclusive so the probability to get exactly $4$ different numbers is:$$P(AXBCD)+P(ABXCD)+P(ABCXD)+P(ABCDX)=$$$$frac66frac16frac56frac46frac36+frac66frac56frac26frac46frac36+frac66frac56frac46frac36frac36+frac66frac56frac46frac36frac46=frac56frac46frac36left(frac16+frac26+frac36+frac46right)$$
answered Jan 11 at 11:32
drhabdrhab
101k544130
$endgroup$
You can throw one by one $5$ dice.
Discern events $AXBCD$, $ABXCD$, $ABCXD$ and $ABCDX$ where e.g. $ABCXD$ stands for the event that the fourth throws give the same result as one of the former throws and there are exactly $4$ distinct results.
The events are mutually exclusive so the probability to get exactly $4$ different numbers is:$$P(AXBCD)+P(ABXCD)+P(ABCXD)+P(ABCDX)=$$$$frac66frac16frac56frac46frac36+frac66frac56frac26frac46frac36+frac66frac56frac46frac36frac36+frac66frac56frac46frac36frac46=frac56frac46frac36left(frac16+frac26+frac36+frac46right)$$
answered Jan 11 at 11:32
drhabdrhab
101k544130
answered Jan 11 at 11:32
drhabdrhab
101k544130
answered Jan 11 at 11:32
drhabdrhab
101k544130
answered Jan 11 at 11:32
drhabdrhab
101k544130
101k544130
add a comment |
add a comment |
$begingroup$
Suppose the output is in 5 boxes like [x][x][x][x][x]
For 4 definite different outcomes we have 6 x 5 x 4 x 3 (in any order)
For the remaining one, it can be any of the 6 digits so we have to choose 1 out of 5 boxes
and also the one box can contain any of the 6 digits i.e. 5c1 x 6
Therefore the probability should be (5c1 x 6 x 6 x 5 x 4 x 3)/6^5
answered Jan 11 at 12:46
cognitivecognitive
264
$endgroup$
add a comment |
$begingroup$
Suppose the output is in 5 boxes like [x][x][x][x][x]
For 4 definite different outcomes we have 6 x 5 x 4 x 3 (in any order)
For the remaining one, it can be any of the 6 digits so we have to choose 1 out of 5 boxes
and also the one box can contain any of the 6 digits i.e. 5c1 x 6
Therefore the probability should be (5c1 x 6 x 6 x 5 x 4 x 3)/6^5
answered Jan 11 at 12:46
cognitivecognitive
264
$endgroup$
add a comment |
$begingroup$
Suppose the output is in 5 boxes like [x][x][x][x][x]
For 4 definite different outcomes we have 6 x 5 x 4 x 3 (in any order)
For the remaining one, it can be any of the 6 digits so we have to choose 1 out of 5 boxes
and also the one box can contain any of the 6 digits i.e. 5c1 x 6
Therefore the probability should be (5c1 x 6 x 6 x 5 x 4 x 3)/6^5
answered Jan 11 at 12:46
cognitivecognitive
264
$endgroup$
Suppose the output is in 5 boxes like [x][x][x][x][x]
For 4 definite different outcomes we have 6 x 5 x 4 x 3 (in any order)
For the remaining one, it can be any of the 6 digits so we have to choose 1 out of 5 boxes
and also the one box can contain any of the 6 digits i.e. 5c1 x 6
Therefore the probability should be (5c1 x 6 x 6 x 5 x 4 x 3)/6^5
answered Jan 11 at 12:46
cognitivecognitive
264
answered Jan 11 at 12:46
cognitivecognitive
264
answered Jan 11 at 12:46
cognitivecognitive
264
answered Jan 11 at 12:46
cognitivecognitive
264
264
add a comment |
add a comment |
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$begingroup$
"I know it is..." - how?
$endgroup$
– zhoraster
Jan 11 at 9:34