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Is my solution for this probability question correct?
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Four fair dice are rolled. What is the probability of getting two different pairs?
So I used the multiplication rule here basically:
$$frac(6 times 5 times 1 times 1) + (6 times 1 times 5 times 1) + (6 times 5 times 1 times 1)6^4 = .0694$$
Each number is the number of possible outcomes for each dice. The first dice always has 6 outcomes since it could be anything, the second one could be a different number or the same number as the first one. Lastly, the third one could be a number that matches the first dice or the second dice. Thus the following three are my possibilities: 6-5-1-1, 6-1-5-1, 6-5-1-1
probability
edited Jul 20 at 15:33
Ahmad Bazzi
2,6271417
asked Jul 20 at 15:19
David
603
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up vote
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down vote
favorite
Four fair dice are rolled. What is the probability of getting two different pairs?
So I used the multiplication rule here basically:
$$frac(6 times 5 times 1 times 1) + (6 times 1 times 5 times 1) + (6 times 5 times 1 times 1)6^4 = .0694$$
Each number is the number of possible outcomes for each dice. The first dice always has 6 outcomes since it could be anything, the second one could be a different number or the same number as the first one. Lastly, the third one could be a number that matches the first dice or the second dice. Thus the following three are my possibilities: 6-5-1-1, 6-1-5-1, 6-5-1-1
probability
edited Jul 20 at 15:33
Ahmad Bazzi
2,6271417
asked Jul 20 at 15:19
David
603
This is hard to follow. There are three ways to divide the four rolls into two groups of two (determined by the partner of the first roll, say). There are $6$ possible values for the pair containing the first roll and $5$ possible values for the other pair. Hence there are $3times 6times 5=90$ ways to throw two distinct pairs. As there are $6^4$ ways to throw the dice without restrictions...
– lulu
Jul 20 at 15:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Four fair dice are rolled. What is the probability of getting two different pairs?
So I used the multiplication rule here basically:
$$frac(6 times 5 times 1 times 1) + (6 times 1 times 5 times 1) + (6 times 5 times 1 times 1)6^4 = .0694$$
Each number is the number of possible outcomes for each dice. The first dice always has 6 outcomes since it could be anything, the second one could be a different number or the same number as the first one. Lastly, the third one could be a number that matches the first dice or the second dice. Thus the following three are my possibilities: 6-5-1-1, 6-1-5-1, 6-5-1-1
probability
edited Jul 20 at 15:33
Ahmad Bazzi
2,6271417
asked Jul 20 at 15:19
David
603
Four fair dice are rolled. What is the probability of getting two different pairs?
So I used the multiplication rule here basically:
$$frac(6 times 5 times 1 times 1) + (6 times 1 times 5 times 1) + (6 times 5 times 1 times 1)6^4 = .0694$$
Each number is the number of possible outcomes for each dice. The first dice always has 6 outcomes since it could be anything, the second one could be a different number or the same number as the first one. Lastly, the third one could be a number that matches the first dice or the second dice. Thus the following three are my possibilities: 6-5-1-1, 6-1-5-1, 6-5-1-1
probability
edited Jul 20 at 15:33
Ahmad Bazzi
2,6271417
asked Jul 20 at 15:19
David
603
edited Jul 20 at 15:33
Ahmad Bazzi
2,6271417
edited Jul 20 at 15:33
Ahmad Bazzi
2,6271417
edited Jul 20 at 15:33
Ahmad Bazzi
2,6271417
2,6271417
asked Jul 20 at 15:19
David
603
asked Jul 20 at 15:19
David
603
asked Jul 20 at 15:19
David
603
603
This is hard to follow. There are three ways to divide the four rolls into two groups of two (determined by the partner of the first roll, say). There are $6$ possible values for the pair containing the first roll and $5$ possible values for the other pair. Hence there are $3times 6times 5=90$ ways to throw two distinct pairs. As there are $6^4$ ways to throw the dice without restrictions...
– lulu
Jul 20 at 15:24
add a comment |Â
This is hard to follow. There are three ways to divide the four rolls into two groups of two (determined by the partner of the first roll, say). There are $6$ possible values for the pair containing the first roll and $5$ possible values for the other pair. Hence there are $3times 6times 5=90$ ways to throw two distinct pairs. As there are $6^4$ ways to throw the dice without restrictions...
– lulu
Jul 20 at 15:24
This is hard to follow. There are three ways to divide the four rolls into two groups of two (determined by the partner of the first roll, say). There are $6$ possible values for the pair containing the first roll and $5$ possible values for the other pair. Hence there are $3times 6times 5=90$ ways to throw two distinct pairs. As there are $6^4$ ways to throw the dice without restrictions...
– lulu
Jul 20 at 15:24
This is hard to follow. There are three ways to divide the four rolls into two groups of two (determined by the partner of the first roll, say). There are $6$ possible values for the pair containing the first roll and $5$ possible values for the other pair. Hence there are $3times 6times 5=90$ ways to throw two distinct pairs. As there are $6^4$ ways to throw the dice without restrictions...
– lulu
Jul 20 at 15:24
add a comment |Â
1 Answer
1
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Yes. Another way to think of it is 6*1 for one pair and 5*1 for a distinct pair then multiply by 4 choose 2 to select which dice make one pair and divide by 2 because you don't care which pair you selected. Same number, though: 6*5*1*1*6/2
answered Jul 20 at 15:25
L. Scott Johnson
1011
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes. Another way to think of it is 6*1 for one pair and 5*1 for a distinct pair then multiply by 4 choose 2 to select which dice make one pair and divide by 2 because you don't care which pair you selected. Same number, though: 6*5*1*1*6/2
answered Jul 20 at 15:25
L. Scott Johnson
1011
add a comment |Â
up vote
0
down vote
Yes. Another way to think of it is 6*1 for one pair and 5*1 for a distinct pair then multiply by 4 choose 2 to select which dice make one pair and divide by 2 because you don't care which pair you selected. Same number, though: 6*5*1*1*6/2
answered Jul 20 at 15:25
L. Scott Johnson
1011
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes. Another way to think of it is 6*1 for one pair and 5*1 for a distinct pair then multiply by 4 choose 2 to select which dice make one pair and divide by 2 because you don't care which pair you selected. Same number, though: 6*5*1*1*6/2
answered Jul 20 at 15:25
L. Scott Johnson
1011
Yes. Another way to think of it is 6*1 for one pair and 5*1 for a distinct pair then multiply by 4 choose 2 to select which dice make one pair and divide by 2 because you don't care which pair you selected. Same number, though: 6*5*1*1*6/2
answered Jul 20 at 15:25
L. Scott Johnson
1011
answered Jul 20 at 15:25
L. Scott Johnson
1011
answered Jul 20 at 15:25
L. Scott Johnson
1011
answered Jul 20 at 15:25
L. Scott Johnson
1011
1011
add a comment |Â
add a comment |Â
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This is hard to follow. There are three ways to divide the four rolls into two groups of two (determined by the partner of the first roll, say). There are $6$ possible values for the pair containing the first roll and $5$ possible values for the other pair. Hence there are $3times 6times 5=90$ ways to throw two distinct pairs. As there are $6^4$ ways to throw the dice without restrictions...
– lulu
Jul 20 at 15:24