Suppose we roll two die, find the mass of the largest roll
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roll two dice and let $X$ be the largest value obtained and $Y$ the smallest. find pmf of $X$ and $Y$.
Im trying to find $P(X=i)$. I start with easy cases. For example, $P(X=1)$ this can only occur when we have $(1,1)$ so we have $frac{1}{36}$. For $P(X=2)$ we have now $(2,1), (1,2), (2,2)$ so now we have $frac{3}{36}$. Also $P(X=3)$ occurs when $(3,2),(3,1),(3,3), (3,1),(3,2)$ so this gives $P(X=3) = frac{5}{36}$ so the pattern is that we have odd numbers of event for $X$ bein largest and so we have
$$ P(X=i) = frac{2i-1}{36} $$
Isnt is $P(Y=j) = frac{2i-1}{26}$ the same?
Is my reasoning correct? How can I approach this problem without doing the easy cases first? Is there any general strategy to handle this type of problems?
probability
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up vote
1
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roll two dice and let $X$ be the largest value obtained and $Y$ the smallest. find pmf of $X$ and $Y$.
Im trying to find $P(X=i)$. I start with easy cases. For example, $P(X=1)$ this can only occur when we have $(1,1)$ so we have $frac{1}{36}$. For $P(X=2)$ we have now $(2,1), (1,2), (2,2)$ so now we have $frac{3}{36}$. Also $P(X=3)$ occurs when $(3,2),(3,1),(3,3), (3,1),(3,2)$ so this gives $P(X=3) = frac{5}{36}$ so the pattern is that we have odd numbers of event for $X$ bein largest and so we have
$$ P(X=i) = frac{2i-1}{36} $$
Isnt is $P(Y=j) = frac{2i-1}{26}$ the same?
Is my reasoning correct? How can I approach this problem without doing the easy cases first? Is there any general strategy to handle this type of problems?
probability
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
roll two dice and let $X$ be the largest value obtained and $Y$ the smallest. find pmf of $X$ and $Y$.
Im trying to find $P(X=i)$. I start with easy cases. For example, $P(X=1)$ this can only occur when we have $(1,1)$ so we have $frac{1}{36}$. For $P(X=2)$ we have now $(2,1), (1,2), (2,2)$ so now we have $frac{3}{36}$. Also $P(X=3)$ occurs when $(3,2),(3,1),(3,3), (3,1),(3,2)$ so this gives $P(X=3) = frac{5}{36}$ so the pattern is that we have odd numbers of event for $X$ bein largest and so we have
$$ P(X=i) = frac{2i-1}{36} $$
Isnt is $P(Y=j) = frac{2i-1}{26}$ the same?
Is my reasoning correct? How can I approach this problem without doing the easy cases first? Is there any general strategy to handle this type of problems?
probability
roll two dice and let $X$ be the largest value obtained and $Y$ the smallest. find pmf of $X$ and $Y$.
Im trying to find $P(X=i)$. I start with easy cases. For example, $P(X=1)$ this can only occur when we have $(1,1)$ so we have $frac{1}{36}$. For $P(X=2)$ we have now $(2,1), (1,2), (2,2)$ so now we have $frac{3}{36}$. Also $P(X=3)$ occurs when $(3,2),(3,1),(3,3), (3,1),(3,2)$ so this gives $P(X=3) = frac{5}{36}$ so the pattern is that we have odd numbers of event for $X$ bein largest and so we have
$$ P(X=i) = frac{2i-1}{36} $$
Isnt is $P(Y=j) = frac{2i-1}{26}$ the same?
Is my reasoning correct? How can I approach this problem without doing the easy cases first? Is there any general strategy to handle this type of problems?
probability
probability
asked yesterday
Neymar
36713
36713
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Your work for $X$ is correct, but the distribution for $Y$ is not the same, though it is the mirror image. You can calculate it in a similar way: $P(Y=6) = 1/36,$ $P(Y=5) = 3/36,ldots$ $P(Y=1) = 11/36.$
It is actually a pretty good approach in general to do the easy cases first. You do them and then think about how to generalize, like how you caught the pattern was $(2i-1)/36.$ There are plenty of methods and tricks you'll learn for dealing with combinatorics and probability problems, but I would still approach this problem exactly as you did here.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your work for $X$ is correct, but the distribution for $Y$ is not the same, though it is the mirror image. You can calculate it in a similar way: $P(Y=6) = 1/36,$ $P(Y=5) = 3/36,ldots$ $P(Y=1) = 11/36.$
It is actually a pretty good approach in general to do the easy cases first. You do them and then think about how to generalize, like how you caught the pattern was $(2i-1)/36.$ There are plenty of methods and tricks you'll learn for dealing with combinatorics and probability problems, but I would still approach this problem exactly as you did here.
add a comment |
up vote
2
down vote
accepted
Your work for $X$ is correct, but the distribution for $Y$ is not the same, though it is the mirror image. You can calculate it in a similar way: $P(Y=6) = 1/36,$ $P(Y=5) = 3/36,ldots$ $P(Y=1) = 11/36.$
It is actually a pretty good approach in general to do the easy cases first. You do them and then think about how to generalize, like how you caught the pattern was $(2i-1)/36.$ There are plenty of methods and tricks you'll learn for dealing with combinatorics and probability problems, but I would still approach this problem exactly as you did here.
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your work for $X$ is correct, but the distribution for $Y$ is not the same, though it is the mirror image. You can calculate it in a similar way: $P(Y=6) = 1/36,$ $P(Y=5) = 3/36,ldots$ $P(Y=1) = 11/36.$
It is actually a pretty good approach in general to do the easy cases first. You do them and then think about how to generalize, like how you caught the pattern was $(2i-1)/36.$ There are plenty of methods and tricks you'll learn for dealing with combinatorics and probability problems, but I would still approach this problem exactly as you did here.
Your work for $X$ is correct, but the distribution for $Y$ is not the same, though it is the mirror image. You can calculate it in a similar way: $P(Y=6) = 1/36,$ $P(Y=5) = 3/36,ldots$ $P(Y=1) = 11/36.$
It is actually a pretty good approach in general to do the easy cases first. You do them and then think about how to generalize, like how you caught the pattern was $(2i-1)/36.$ There are plenty of methods and tricks you'll learn for dealing with combinatorics and probability problems, but I would still approach this problem exactly as you did here.
edited yesterday
answered yesterday
spaceisdarkgreen
31.2k21552
31.2k21552
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