Random Variables and Expected Values Problem
$begingroup$
Let $ bgeq 1$,$cgeq 1$ and $hgeq 1$ be integers, and let $n=b+c+h$.
You have $b$ balls, denoted $B_1, B_2, dots , B_B$, $c$ cubes $C_1, C_2, dots , C_c$, and $h$ hexagons $H_1, H_2, dots , H_h$. Let $mgeq 1$ be an integer with $mleq b$ and $mleq n-b$.
All $n$ objects are in a box, from this box you choose a uniformly random subset consisting of $m$ objects.
Define the random variables:
- X = the number of balls in the chosen subset
- Y = the number of cubes in the chosen subset
- Z = the number of hexagons in a chosen subset
Two questions about this:
- How would I start, or go about proving that $Pr(X=k) = frac{binom{b}{k} binom{n-b}{m-k}}{binom{n}{m}}$
- How would I prove that $sum_{k=0}^m frac{kbinom{b}{k} binom{n-b}{m-k}}{binom{n}{m}} = frac{bm}{n}$
I'm not too sure where to start for these, so any help is appreciated. If there's anything I can do to improve my question I'm happy to get input :)
discrete-mathematics random-variables expected-value
asked Dec 1 '18 at 0:37
NetwinderNetwinder
52
$endgroup$
add a comment |
$begingroup$
Let $ bgeq 1$,$cgeq 1$ and $hgeq 1$ be integers, and let $n=b+c+h$.
You have $b$ balls, denoted $B_1, B_2, dots , B_B$, $c$ cubes $C_1, C_2, dots , C_c$, and $h$ hexagons $H_1, H_2, dots , H_h$. Let $mgeq 1$ be an integer with $mleq b$ and $mleq n-b$.
All $n$ objects are in a box, from this box you choose a uniformly random subset consisting of $m$ objects.
Define the random variables:
- X = the number of balls in the chosen subset
- Y = the number of cubes in the chosen subset
- Z = the number of hexagons in a chosen subset
Two questions about this:
- How would I start, or go about proving that $Pr(X=k) = frac{binom{b}{k} binom{n-b}{m-k}}{binom{n}{m}}$
- How would I prove that $sum_{k=0}^m frac{kbinom{b}{k} binom{n-b}{m-k}}{binom{n}{m}} = frac{bm}{n}$
I'm not too sure where to start for these, so any help is appreciated. If there's anything I can do to improve my question I'm happy to get input :)
discrete-mathematics random-variables expected-value
asked Dec 1 '18 at 0:37
NetwinderNetwinder
52
$endgroup$
add a comment |
$begingroup$
Let $ bgeq 1$,$cgeq 1$ and $hgeq 1$ be integers, and let $n=b+c+h$.
You have $b$ balls, denoted $B_1, B_2, dots , B_B$, $c$ cubes $C_1, C_2, dots , C_c$, and $h$ hexagons $H_1, H_2, dots , H_h$. Let $mgeq 1$ be an integer with $mleq b$ and $mleq n-b$.
All $n$ objects are in a box, from this box you choose a uniformly random subset consisting of $m$ objects.
Define the random variables:
- X = the number of balls in the chosen subset
- Y = the number of cubes in the chosen subset
- Z = the number of hexagons in a chosen subset
Two questions about this:
- How would I start, or go about proving that $Pr(X=k) = frac{binom{b}{k} binom{n-b}{m-k}}{binom{n}{m}}$
- How would I prove that $sum_{k=0}^m frac{kbinom{b}{k} binom{n-b}{m-k}}{binom{n}{m}} = frac{bm}{n}$
I'm not too sure where to start for these, so any help is appreciated. If there's anything I can do to improve my question I'm happy to get input :)
discrete-mathematics random-variables expected-value
asked Dec 1 '18 at 0:37
NetwinderNetwinder
52
$endgroup$
Let $ bgeq 1$,$cgeq 1$ and $hgeq 1$ be integers, and let $n=b+c+h$.
You have $b$ balls, denoted $B_1, B_2, dots , B_B$, $c$ cubes $C_1, C_2, dots , C_c$, and $h$ hexagons $H_1, H_2, dots , H_h$. Let $mgeq 1$ be an integer with $mleq b$ and $mleq n-b$.
All $n$ objects are in a box, from this box you choose a uniformly random subset consisting of $m$ objects.
Define the random variables:
- X = the number of balls in the chosen subset
- Y = the number of cubes in the chosen subset
- Z = the number of hexagons in a chosen subset
Two questions about this:
- How would I start, or go about proving that $Pr(X=k) = frac{binom{b}{k} binom{n-b}{m-k}}{binom{n}{m}}$
- How would I prove that $sum_{k=0}^m frac{kbinom{b}{k} binom{n-b}{m-k}}{binom{n}{m}} = frac{bm}{n}$
I'm not too sure where to start for these, so any help is appreciated. If there's anything I can do to improve my question I'm happy to get input :)
discrete-mathematics random-variables expected-value
discrete-mathematics random-variables expected-value
asked Dec 1 '18 at 0:37
NetwinderNetwinder
52
asked Dec 1 '18 at 0:37
NetwinderNetwinder
52
asked Dec 1 '18 at 0:37
NetwinderNetwinder
52
asked Dec 1 '18 at 0:37
NetwinderNetwinder
52
asked Dec 1 '18 at 0:37
NetwinderNetwinder
52
52
add a comment |
add a comment |
1 Answer
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$begingroup$
- This problem can be solved by a counting argument. $X = k$ is the event that we choose exactly $k$ balls (and therefore, $m-k$ other objects) when picking $m$ objects out of all $n$. You can count the number of ways to do this by multiplying $binom{b}{k}$ ways to choose $k$ balls and $binom{n-b}{m-k}$ ways to choose the $m-k$ non-balls. This is divided by the total number of ways to choose $m$ objects out of $n$, $binom{n}{m}$.
- Hint: Rewrite the sum as $sum_{k=0}^m k Pr(X = k)$. But this is just the expectation of $X$; you can solve for it using linearity of expectation. Namely, let $X_i$ be the indicator random variable for the $i^{th}$ object being a ball; $E[X_i] = Pr(X_i = 1) = frac{b}{n}$ (each object is equally likely to be chosen in any spot). Summing over $1 leq i leq m$ gives:
$$
E[X] = Eleft[ sum_{i=1}^m X_i right] = sum_{i=1}^m E[X_i] = frac{bm}{n}
$$
answered Dec 1 '18 at 2:00
plattyplatty
3,370320
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$begingroup$
- This problem can be solved by a counting argument. $X = k$ is the event that we choose exactly $k$ balls (and therefore, $m-k$ other objects) when picking $m$ objects out of all $n$. You can count the number of ways to do this by multiplying $binom{b}{k}$ ways to choose $k$ balls and $binom{n-b}{m-k}$ ways to choose the $m-k$ non-balls. This is divided by the total number of ways to choose $m$ objects out of $n$, $binom{n}{m}$.
- Hint: Rewrite the sum as $sum_{k=0}^m k Pr(X = k)$. But this is just the expectation of $X$; you can solve for it using linearity of expectation. Namely, let $X_i$ be the indicator random variable for the $i^{th}$ object being a ball; $E[X_i] = Pr(X_i = 1) = frac{b}{n}$ (each object is equally likely to be chosen in any spot). Summing over $1 leq i leq m$ gives:
$$
E[X] = Eleft[ sum_{i=1}^m X_i right] = sum_{i=1}^m E[X_i] = frac{bm}{n}
$$
answered Dec 1 '18 at 2:00
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
- This problem can be solved by a counting argument. $X = k$ is the event that we choose exactly $k$ balls (and therefore, $m-k$ other objects) when picking $m$ objects out of all $n$. You can count the number of ways to do this by multiplying $binom{b}{k}$ ways to choose $k$ balls and $binom{n-b}{m-k}$ ways to choose the $m-k$ non-balls. This is divided by the total number of ways to choose $m$ objects out of $n$, $binom{n}{m}$.
- Hint: Rewrite the sum as $sum_{k=0}^m k Pr(X = k)$. But this is just the expectation of $X$; you can solve for it using linearity of expectation. Namely, let $X_i$ be the indicator random variable for the $i^{th}$ object being a ball; $E[X_i] = Pr(X_i = 1) = frac{b}{n}$ (each object is equally likely to be chosen in any spot). Summing over $1 leq i leq m$ gives:
$$
E[X] = Eleft[ sum_{i=1}^m X_i right] = sum_{i=1}^m E[X_i] = frac{bm}{n}
$$
answered Dec 1 '18 at 2:00
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
- This problem can be solved by a counting argument. $X = k$ is the event that we choose exactly $k$ balls (and therefore, $m-k$ other objects) when picking $m$ objects out of all $n$. You can count the number of ways to do this by multiplying $binom{b}{k}$ ways to choose $k$ balls and $binom{n-b}{m-k}$ ways to choose the $m-k$ non-balls. This is divided by the total number of ways to choose $m$ objects out of $n$, $binom{n}{m}$.
- Hint: Rewrite the sum as $sum_{k=0}^m k Pr(X = k)$. But this is just the expectation of $X$; you can solve for it using linearity of expectation. Namely, let $X_i$ be the indicator random variable for the $i^{th}$ object being a ball; $E[X_i] = Pr(X_i = 1) = frac{b}{n}$ (each object is equally likely to be chosen in any spot). Summing over $1 leq i leq m$ gives:
$$
E[X] = Eleft[ sum_{i=1}^m X_i right] = sum_{i=1}^m E[X_i] = frac{bm}{n}
$$
answered Dec 1 '18 at 2:00
plattyplatty
3,370320
$endgroup$
- This problem can be solved by a counting argument. $X = k$ is the event that we choose exactly $k$ balls (and therefore, $m-k$ other objects) when picking $m$ objects out of all $n$. You can count the number of ways to do this by multiplying $binom{b}{k}$ ways to choose $k$ balls and $binom{n-b}{m-k}$ ways to choose the $m-k$ non-balls. This is divided by the total number of ways to choose $m$ objects out of $n$, $binom{n}{m}$.
- Hint: Rewrite the sum as $sum_{k=0}^m k Pr(X = k)$. But this is just the expectation of $X$; you can solve for it using linearity of expectation. Namely, let $X_i$ be the indicator random variable for the $i^{th}$ object being a ball; $E[X_i] = Pr(X_i = 1) = frac{b}{n}$ (each object is equally likely to be chosen in any spot). Summing over $1 leq i leq m$ gives:
$$
E[X] = Eleft[ sum_{i=1}^m X_i right] = sum_{i=1}^m E[X_i] = frac{bm}{n}
$$
answered Dec 1 '18 at 2:00
plattyplatty
3,370320
answered Dec 1 '18 at 2:00
plattyplatty
3,370320
answered Dec 1 '18 at 2:00
plattyplatty
3,370320
answered Dec 1 '18 at 2:00
plattyplatty
3,370320
3,370320
add a comment |
add a comment |
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