Probability with a shrinking sample space
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If I have a set that contains some lowercase and uppercase letters, for example $S = {a,b,c,d,e,f,g,h,V,W,X,Y,Z}$. Say I choose 4 letters from this set in a uniformly random way; however, each time I choose an element I remove it. I.e, each element may only be chosen once.
My sample space is initially $S$, after I choose 4 elements the size of my sample space would then be $|S| - 4$.
I'm not too sure about how I would go about calculating the probabilities of various things, for example consider the event $B=text{"The 4th letter chosen is uppercase"}$, how would I calculate $Pr(B)$?
If there's anything I can do to improve my question I'm happy to get feedback :)
probability statistics random-variables
asked Nov 15 at 23:58
Netwinder
52
add a comment |
up vote
0
down vote
favorite
If I have a set that contains some lowercase and uppercase letters, for example $S = {a,b,c,d,e,f,g,h,V,W,X,Y,Z}$. Say I choose 4 letters from this set in a uniformly random way; however, each time I choose an element I remove it. I.e, each element may only be chosen once.
My sample space is initially $S$, after I choose 4 elements the size of my sample space would then be $|S| - 4$.
I'm not too sure about how I would go about calculating the probabilities of various things, for example consider the event $B=text{"The 4th letter chosen is uppercase"}$, how would I calculate $Pr(B)$?
If there's anything I can do to improve my question I'm happy to get feedback :)
probability statistics random-variables
asked Nov 15 at 23:58
Netwinder
52
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If I have a set that contains some lowercase and uppercase letters, for example $S = {a,b,c,d,e,f,g,h,V,W,X,Y,Z}$. Say I choose 4 letters from this set in a uniformly random way; however, each time I choose an element I remove it. I.e, each element may only be chosen once.
My sample space is initially $S$, after I choose 4 elements the size of my sample space would then be $|S| - 4$.
I'm not too sure about how I would go about calculating the probabilities of various things, for example consider the event $B=text{"The 4th letter chosen is uppercase"}$, how would I calculate $Pr(B)$?
If there's anything I can do to improve my question I'm happy to get feedback :)
probability statistics random-variables
asked Nov 15 at 23:58
Netwinder
52
If I have a set that contains some lowercase and uppercase letters, for example $S = {a,b,c,d,e,f,g,h,V,W,X,Y,Z}$. Say I choose 4 letters from this set in a uniformly random way; however, each time I choose an element I remove it. I.e, each element may only be chosen once.
My sample space is initially $S$, after I choose 4 elements the size of my sample space would then be $|S| - 4$.
I'm not too sure about how I would go about calculating the probabilities of various things, for example consider the event $B=text{"The 4th letter chosen is uppercase"}$, how would I calculate $Pr(B)$?
If there's anything I can do to improve my question I'm happy to get feedback :)
probability statistics random-variables
probability statistics random-variables
asked Nov 15 at 23:58
Netwinder
52
asked Nov 15 at 23:58
Netwinder
52
asked Nov 15 at 23:58
Netwinder
52
asked Nov 15 at 23:58
Netwinder
52
asked Nov 15 at 23:58
Netwinder
52
52
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
It can be treated as a permutation counting problem where:
$$Pr(B) = frac{text{permutations with $4$th letter UC}}{text{permutations of 4 from 13}}$$
$$Pr(B) = frac{5cdotfrac{12!}{9!}}{frac{13!}{9!}} = frac{5}{13}$$
answered Nov 16 at 2:35
Phil H
3,8782312
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
It can be treated as a permutation counting problem where:
$$Pr(B) = frac{text{permutations with $4$th letter UC}}{text{permutations of 4 from 13}}$$
$$Pr(B) = frac{5cdotfrac{12!}{9!}}{frac{13!}{9!}} = frac{5}{13}$$
answered Nov 16 at 2:35
Phil H
3,8782312
add a comment |
up vote
0
down vote
accepted
It can be treated as a permutation counting problem where:
$$Pr(B) = frac{text{permutations with $4$th letter UC}}{text{permutations of 4 from 13}}$$
$$Pr(B) = frac{5cdotfrac{12!}{9!}}{frac{13!}{9!}} = frac{5}{13}$$
answered Nov 16 at 2:35
Phil H
3,8782312
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
It can be treated as a permutation counting problem where:
$$Pr(B) = frac{text{permutations with $4$th letter UC}}{text{permutations of 4 from 13}}$$
$$Pr(B) = frac{5cdotfrac{12!}{9!}}{frac{13!}{9!}} = frac{5}{13}$$
answered Nov 16 at 2:35
Phil H
3,8782312
It can be treated as a permutation counting problem where:
$$Pr(B) = frac{text{permutations with $4$th letter UC}}{text{permutations of 4 from 13}}$$
$$Pr(B) = frac{5cdotfrac{12!}{9!}}{frac{13!}{9!}} = frac{5}{13}$$
answered Nov 16 at 2:35
Phil H
3,8782312
answered Nov 16 at 2:35
Phil H
3,8782312
answered Nov 16 at 2:35
Phil H
3,8782312
answered Nov 16 at 2:35
Phil H
3,8782312
3,8782312
add a comment |
add a comment |
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