Rolling three dice…am I doing this correctly?
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Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?
Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first
For 1) The first die, you have 6/6. The second die needs to be equal to the first, so you have probability of 1/6. Then the third die can't be equal to the first and second dice, so it's 5/6.
All together you get $1*frac{1}{6}*frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3*(1*frac{1}{6}*frac{5}{6})=frac{5}{12}$$
Did I do this correctly?
Thank you.
probability
asked Feb 12 '13 at 6:40
AltiAlti
1,18342246
$endgroup$
add a comment |
$begingroup$
Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?
Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first
For 1) The first die, you have 6/6. The second die needs to be equal to the first, so you have probability of 1/6. Then the third die can't be equal to the first and second dice, so it's 5/6.
All together you get $1*frac{1}{6}*frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3*(1*frac{1}{6}*frac{5}{6})=frac{5}{12}$$
Did I do this correctly?
Thank you.
probability
asked Feb 12 '13 at 6:40
AltiAlti
1,18342246
$endgroup$
1
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This is the answer I get and your reasoning seems sound.
$endgroup$
– user45150
Feb 12 '13 at 6:41
add a comment |
$begingroup$
Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?
Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first
For 1) The first die, you have 6/6. The second die needs to be equal to the first, so you have probability of 1/6. Then the third die can't be equal to the first and second dice, so it's 5/6.
All together you get $1*frac{1}{6}*frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3*(1*frac{1}{6}*frac{5}{6})=frac{5}{12}$$
Did I do this correctly?
Thank you.
probability
asked Feb 12 '13 at 6:40
AltiAlti
1,18342246
$endgroup$
Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?
Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first
For 1) The first die, you have 6/6. The second die needs to be equal to the first, so you have probability of 1/6. Then the third die can't be equal to the first and second dice, so it's 5/6.
All together you get $1*frac{1}{6}*frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3*(1*frac{1}{6}*frac{5}{6})=frac{5}{12}$$
Did I do this correctly?
Thank you.
probability
probability
asked Feb 12 '13 at 6:40
AltiAlti
1,18342246
asked Feb 12 '13 at 6:40
AltiAlti
1,18342246
asked Feb 12 '13 at 6:40
AltiAlti
1,18342246
asked Feb 12 '13 at 6:40
AltiAlti
1,18342246
asked Feb 12 '13 at 6:40
AltiAlti
1,18342246
1,18342246
1
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This is the answer I get and your reasoning seems sound.
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– user45150
Feb 12 '13 at 6:41
add a comment |
1
$begingroup$
This is the answer I get and your reasoning seems sound.
$endgroup$
– user45150
Feb 12 '13 at 6:41
1
1
$begingroup$
This is the answer I get and your reasoning seems sound.
$endgroup$
– user45150
Feb 12 '13 at 6:41
$begingroup$
This is the answer I get and your reasoning seems sound.
$endgroup$
– user45150
Feb 12 '13 at 6:41
add a comment |
3 Answers
3
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oldest
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Yes your answer is correct. $${3choose 2}cdotfrac{1}{6}cdotfrac{5}{6} = frac{5}{12}$$
Good job!
answered Feb 12 '13 at 6:44
MITjanitorMITjanitor
1,9631543
$endgroup$
1
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It's $1/5$ or $1/6$?
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– MonsieurGalois
Jan 17 '17 at 20:30
add a comment |
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As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3cdot6cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $frac{90}{216}=frac{5}{12}$.
answered Feb 12 '13 at 6:57
robjohn♦robjohn
271k27315643
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add a comment |
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One more way you can solve this is using complimentary counting. There are 6*$dbinom{6}{4}$=120 ways to get all numbers different. There are 6 ways to get all numbers same. So there are 216-126=90 ways that work. $frac{90}{216}=frac{5}{12}$
answered Dec 27 '18 at 16:15
BobBob
1
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
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active
oldest
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Yes your answer is correct. $${3choose 2}cdotfrac{1}{6}cdotfrac{5}{6} = frac{5}{12}$$
Good job!
answered Feb 12 '13 at 6:44
MITjanitorMITjanitor
1,9631543
$endgroup$
1
$begingroup$
It's $1/5$ or $1/6$?
$endgroup$
– MonsieurGalois
Jan 17 '17 at 20:30
add a comment |
$begingroup$
Yes your answer is correct. $${3choose 2}cdotfrac{1}{6}cdotfrac{5}{6} = frac{5}{12}$$
Good job!
answered Feb 12 '13 at 6:44
MITjanitorMITjanitor
1,9631543
$endgroup$
1
$begingroup$
It's $1/5$ or $1/6$?
$endgroup$
– MonsieurGalois
Jan 17 '17 at 20:30
add a comment |
$begingroup$
Yes your answer is correct. $${3choose 2}cdotfrac{1}{6}cdotfrac{5}{6} = frac{5}{12}$$
Good job!
answered Feb 12 '13 at 6:44
MITjanitorMITjanitor
1,9631543
$endgroup$
Yes your answer is correct. $${3choose 2}cdotfrac{1}{6}cdotfrac{5}{6} = frac{5}{12}$$
Good job!
answered Feb 12 '13 at 6:44
MITjanitorMITjanitor
1,9631543
edited Oct 10 '17 at 17:27
answered Feb 12 '13 at 6:44
MITjanitorMITjanitor
1,9631543
answered Feb 12 '13 at 6:44
MITjanitorMITjanitor
1,9631543
answered Feb 12 '13 at 6:44
MITjanitorMITjanitor
1,9631543
1,9631543
1
$begingroup$
It's $1/5$ or $1/6$?
$endgroup$
– MonsieurGalois
Jan 17 '17 at 20:30
add a comment |
1
$begingroup$
It's $1/5$ or $1/6$?
$endgroup$
– MonsieurGalois
Jan 17 '17 at 20:30
1
1
$begingroup$
It's $1/5$ or $1/6$?
$endgroup$
– MonsieurGalois
Jan 17 '17 at 20:30
$begingroup$
It's $1/5$ or $1/6$?
$endgroup$
– MonsieurGalois
Jan 17 '17 at 20:30
add a comment |
$begingroup$
As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3cdot6cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $frac{90}{216}=frac{5}{12}$.
answered Feb 12 '13 at 6:57
robjohn♦robjohn
271k27315643
$endgroup$
add a comment |
$begingroup$
As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3cdot6cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $frac{90}{216}=frac{5}{12}$.
answered Feb 12 '13 at 6:57
robjohn♦robjohn
271k27315643
$endgroup$
add a comment |
$begingroup$
As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3cdot6cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $frac{90}{216}=frac{5}{12}$.
answered Feb 12 '13 at 6:57
robjohn♦robjohn
271k27315643
$endgroup$
As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3cdot6cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $frac{90}{216}=frac{5}{12}$.
answered Feb 12 '13 at 6:57
robjohn♦robjohn
271k27315643
answered Feb 12 '13 at 6:57
robjohn♦robjohn
271k27315643
answered Feb 12 '13 at 6:57
robjohn♦robjohn
271k27315643
answered Feb 12 '13 at 6:57
robjohn♦robjohn
271k27315643
271k27315643
add a comment |
add a comment |
$begingroup$
One more way you can solve this is using complimentary counting. There are 6*$dbinom{6}{4}$=120 ways to get all numbers different. There are 6 ways to get all numbers same. So there are 216-126=90 ways that work. $frac{90}{216}=frac{5}{12}$
answered Dec 27 '18 at 16:15
BobBob
1
$endgroup$
add a comment |
$begingroup$
One more way you can solve this is using complimentary counting. There are 6*$dbinom{6}{4}$=120 ways to get all numbers different. There are 6 ways to get all numbers same. So there are 216-126=90 ways that work. $frac{90}{216}=frac{5}{12}$
answered Dec 27 '18 at 16:15
BobBob
1
$endgroup$
add a comment |
$begingroup$
One more way you can solve this is using complimentary counting. There are 6*$dbinom{6}{4}$=120 ways to get all numbers different. There are 6 ways to get all numbers same. So there are 216-126=90 ways that work. $frac{90}{216}=frac{5}{12}$
answered Dec 27 '18 at 16:15
BobBob
1
$endgroup$
One more way you can solve this is using complimentary counting. There are 6*$dbinom{6}{4}$=120 ways to get all numbers different. There are 6 ways to get all numbers same. So there are 216-126=90 ways that work. $frac{90}{216}=frac{5}{12}$
answered Dec 27 '18 at 16:15
BobBob
1
answered Dec 27 '18 at 16:15
BobBob
1
answered Dec 27 '18 at 16:15
BobBob
1
answered Dec 27 '18 at 16:15
BobBob
1
1
add a comment |
add a comment |
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This is the answer I get and your reasoning seems sound.
$endgroup$
– user45150
Feb 12 '13 at 6:41