what is the probability that the letters e, g, and k are all in the same subset?
$begingroup$
the 26 letters of the alphabet are to be randomly partitioned into subsets with 5,10, and 11 letters respectively. what is the probability that the letters e, g, and k are all in the same subset?
I am unsure about how to go about this. I have tried 3!/5!10!11! And I know that is not it and have also tried (5C3)(10C3)(11C3)/(26C3) and that isnt correst either. I just need a push in the right direction
probability
asked Dec 9 '18 at 14:42
CourtCourt
1
$endgroup$
add a comment |
$begingroup$
the 26 letters of the alphabet are to be randomly partitioned into subsets with 5,10, and 11 letters respectively. what is the probability that the letters e, g, and k are all in the same subset?
I am unsure about how to go about this. I have tried 3!/5!10!11! And I know that is not it and have also tried (5C3)(10C3)(11C3)/(26C3) and that isnt correst either. I just need a push in the right direction
probability
asked Dec 9 '18 at 14:42
CourtCourt
1
$endgroup$
$begingroup$
Do you know how to count the ways of splitting 26 letters into subsets of sizes 5,10, and 11?
$endgroup$
– Ben
Dec 9 '18 at 14:46
$begingroup$
Once you’ve done that you can break it into three cases: egk are in the subset of size 5,10, or 11. How many ways are there for each case to happen?
$endgroup$
– Ben
Dec 9 '18 at 14:49
$begingroup$
This helps a lot, thank you!
$endgroup$
– Court
Dec 9 '18 at 14:51
1
$begingroup$
I’m glad! You should write the answer yourself down below once you solve it :)
$endgroup$
– Ben
Dec 9 '18 at 15:01
add a comment |
$begingroup$
the 26 letters of the alphabet are to be randomly partitioned into subsets with 5,10, and 11 letters respectively. what is the probability that the letters e, g, and k are all in the same subset?
I am unsure about how to go about this. I have tried 3!/5!10!11! And I know that is not it and have also tried (5C3)(10C3)(11C3)/(26C3) and that isnt correst either. I just need a push in the right direction
probability
asked Dec 9 '18 at 14:42
CourtCourt
1
$endgroup$
the 26 letters of the alphabet are to be randomly partitioned into subsets with 5,10, and 11 letters respectively. what is the probability that the letters e, g, and k are all in the same subset?
I am unsure about how to go about this. I have tried 3!/5!10!11! And I know that is not it and have also tried (5C3)(10C3)(11C3)/(26C3) and that isnt correst either. I just need a push in the right direction
probability
probability
asked Dec 9 '18 at 14:42
CourtCourt
1
asked Dec 9 '18 at 14:42
CourtCourt
1
asked Dec 9 '18 at 14:42
CourtCourt
1
asked Dec 9 '18 at 14:42
CourtCourt
1
asked Dec 9 '18 at 14:42
CourtCourt
1
1
$begingroup$
Do you know how to count the ways of splitting 26 letters into subsets of sizes 5,10, and 11?
$endgroup$
– Ben
Dec 9 '18 at 14:46
$begingroup$
Once you’ve done that you can break it into three cases: egk are in the subset of size 5,10, or 11. How many ways are there for each case to happen?
$endgroup$
– Ben
Dec 9 '18 at 14:49
$begingroup$
This helps a lot, thank you!
$endgroup$
– Court
Dec 9 '18 at 14:51
1
$begingroup$
I’m glad! You should write the answer yourself down below once you solve it :)
$endgroup$
– Ben
Dec 9 '18 at 15:01
add a comment |
$begingroup$
Do you know how to count the ways of splitting 26 letters into subsets of sizes 5,10, and 11?
$endgroup$
– Ben
Dec 9 '18 at 14:46
$begingroup$
Once you’ve done that you can break it into three cases: egk are in the subset of size 5,10, or 11. How many ways are there for each case to happen?
$endgroup$
– Ben
Dec 9 '18 at 14:49
$begingroup$
This helps a lot, thank you!
$endgroup$
– Court
Dec 9 '18 at 14:51
1
$begingroup$
I’m glad! You should write the answer yourself down below once you solve it :)
$endgroup$
– Ben
Dec 9 '18 at 15:01
$begingroup$
Do you know how to count the ways of splitting 26 letters into subsets of sizes 5,10, and 11?
$endgroup$
– Ben
Dec 9 '18 at 14:46
$begingroup$
Do you know how to count the ways of splitting 26 letters into subsets of sizes 5,10, and 11?
$endgroup$
– Ben
Dec 9 '18 at 14:46
$begingroup$
Once you’ve done that you can break it into three cases: egk are in the subset of size 5,10, or 11. How many ways are there for each case to happen?
$endgroup$
– Ben
Dec 9 '18 at 14:49
$begingroup$
Once you’ve done that you can break it into three cases: egk are in the subset of size 5,10, or 11. How many ways are there for each case to happen?
$endgroup$
– Ben
Dec 9 '18 at 14:49
$begingroup$
This helps a lot, thank you!
$endgroup$
– Court
Dec 9 '18 at 14:51
$begingroup$
This helps a lot, thank you!
$endgroup$
– Court
Dec 9 '18 at 14:51
1
1
$begingroup$
I’m glad! You should write the answer yourself down below once you solve it :)
$endgroup$
– Ben
Dec 9 '18 at 15:01
$begingroup$
I’m glad! You should write the answer yourself down below once you solve it :)
$endgroup$
– Ben
Dec 9 '18 at 15:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
If $S$ is a randomly chosen subset of ${text{a,b,}cdotstext{,z}}$ having $n$ elements then:
$P(text{e}in Swedgetext{g}in Swedgetext{k}in S)=P(text{e}in S)P(text{g}in Smidtext{e}in S)P(text{k}in Smidtext{g}in Swedgetext{e}in S)=frac{n}{26}frac{n-1}{25}frac{n-2}{24}$
answered Dec 9 '18 at 15:14
drhabdrhab
101k545136
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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$begingroup$
Hint:
If $S$ is a randomly chosen subset of ${text{a,b,}cdotstext{,z}}$ having $n$ elements then:
$P(text{e}in Swedgetext{g}in Swedgetext{k}in S)=P(text{e}in S)P(text{g}in Smidtext{e}in S)P(text{k}in Smidtext{g}in Swedgetext{e}in S)=frac{n}{26}frac{n-1}{25}frac{n-2}{24}$
answered Dec 9 '18 at 15:14
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
Hint:
If $S$ is a randomly chosen subset of ${text{a,b,}cdotstext{,z}}$ having $n$ elements then:
$P(text{e}in Swedgetext{g}in Swedgetext{k}in S)=P(text{e}in S)P(text{g}in Smidtext{e}in S)P(text{k}in Smidtext{g}in Swedgetext{e}in S)=frac{n}{26}frac{n-1}{25}frac{n-2}{24}$
answered Dec 9 '18 at 15:14
drhabdrhab
101k545136
$endgroup$
add a comment |
$begingroup$
Hint:
If $S$ is a randomly chosen subset of ${text{a,b,}cdotstext{,z}}$ having $n$ elements then:
$P(text{e}in Swedgetext{g}in Swedgetext{k}in S)=P(text{e}in S)P(text{g}in Smidtext{e}in S)P(text{k}in Smidtext{g}in Swedgetext{e}in S)=frac{n}{26}frac{n-1}{25}frac{n-2}{24}$
answered Dec 9 '18 at 15:14
drhabdrhab
101k545136
$endgroup$
Hint:
If $S$ is a randomly chosen subset of ${text{a,b,}cdotstext{,z}}$ having $n$ elements then:
$P(text{e}in Swedgetext{g}in Swedgetext{k}in S)=P(text{e}in S)P(text{g}in Smidtext{e}in S)P(text{k}in Smidtext{g}in Swedgetext{e}in S)=frac{n}{26}frac{n-1}{25}frac{n-2}{24}$
answered Dec 9 '18 at 15:14
drhabdrhab
101k545136
answered Dec 9 '18 at 15:14
drhabdrhab
101k545136
answered Dec 9 '18 at 15:14
drhabdrhab
101k545136
answered Dec 9 '18 at 15:14
drhabdrhab
101k545136
101k545136
add a comment |
add a comment |
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$begingroup$
Do you know how to count the ways of splitting 26 letters into subsets of sizes 5,10, and 11?
$endgroup$
– Ben
Dec 9 '18 at 14:46
$begingroup$
Once you’ve done that you can break it into three cases: egk are in the subset of size 5,10, or 11. How many ways are there for each case to happen?
$endgroup$
– Ben
Dec 9 '18 at 14:49
$begingroup$
This helps a lot, thank you!
$endgroup$
– Court
Dec 9 '18 at 14:51
1
$begingroup$
I’m glad! You should write the answer yourself down below once you solve it :)
$endgroup$
– Ben
Dec 9 '18 at 15:01