Find Expected Value of Random Variables with Indicator Variables
$begingroup$
Part a)
I calculated the Pr by ($10choose{1}$ * $5choose{1}$) / $10choose{5}$
Part b)
I defined X to be the 5 beer subset that Lindsay and Simon can pick. Multiplied each Probability by 2 for both Simon and Lindsay
E(X) = 1 * (($10choose{1}$ * $5choose{1}$) / $10choose{5}$)*2) +
2 * (($10choose{2}$ * $5choose{2}$) / $10choose{5}$)*2) +
3 * (($10choose{3}$ * $5choose{3}$) / $10choose{5}$)*2) +
4 * (($10choose{4}$ * $5choose{4}$) / $10choose{5}$)*2) +
5 * (($10choose{5}$ * $5choose{5}$) / $10choose{5}$)*2)
For Part c)
This was a tricky one, but here is my attempt on it:
I defined $X =1$ to be the number of days that $B_{ai}$ = $B_i$ and $0$ otherwise
I took $n =3$ and since Simon uses the uniform random permutations to select, it becomes:
{$B_1$,$B_2$,$B_3$}
{$B_3$,$B_2$,$B_1$}
{$B_2$,$B_1$,$B_3$}
{$B_1$,$B_3$,$B_2$}
{$B_1$,$B_3$,$B_2$}
{$B_3$,$B_1$,$B_2$}
And for Lindsay, it is simply {$B_1$,$B_2$,$B_3$}
Now for $3$ days, I compared when $B_i's$ are the same for both.
Day $1$: Simon Possibilities: $B_1$, $B_3$, $B_2$, $B_1$, $B_3$
Lindsay Possibilities: $B_1$
These are same only Pr = $frac{2}{5}$ times
Same thing with Day $2$ and Day $3$ = Pr = $frac{2}{5}$ times
E(X) = $1$ * $frac{2}{5}$ + $2$ * $frac{2}{5}$ + $3$ * $frac{2}{5}$ = $2.4$
This was for the scenario of taking $n=3$.
I am still learning on how to attempt these question so this was a wild attempt by me. I feel like my logic is still getting there so any help on these would be appreciated.
probability probability-theory discrete-mathematics random-variables expected-value
asked Dec 17 '18 at 19:59
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
Part a)
I calculated the Pr by ($10choose{1}$ * $5choose{1}$) / $10choose{5}$
Part b)
I defined X to be the 5 beer subset that Lindsay and Simon can pick. Multiplied each Probability by 2 for both Simon and Lindsay
E(X) = 1 * (($10choose{1}$ * $5choose{1}$) / $10choose{5}$)*2) +
2 * (($10choose{2}$ * $5choose{2}$) / $10choose{5}$)*2) +
3 * (($10choose{3}$ * $5choose{3}$) / $10choose{5}$)*2) +
4 * (($10choose{4}$ * $5choose{4}$) / $10choose{5}$)*2) +
5 * (($10choose{5}$ * $5choose{5}$) / $10choose{5}$)*2)
For Part c)
This was a tricky one, but here is my attempt on it:
I defined $X =1$ to be the number of days that $B_{ai}$ = $B_i$ and $0$ otherwise
I took $n =3$ and since Simon uses the uniform random permutations to select, it becomes:
{$B_1$,$B_2$,$B_3$}
{$B_3$,$B_2$,$B_1$}
{$B_2$,$B_1$,$B_3$}
{$B_1$,$B_3$,$B_2$}
{$B_1$,$B_3$,$B_2$}
{$B_3$,$B_1$,$B_2$}
And for Lindsay, it is simply {$B_1$,$B_2$,$B_3$}
Now for $3$ days, I compared when $B_i's$ are the same for both.
Day $1$: Simon Possibilities: $B_1$, $B_3$, $B_2$, $B_1$, $B_3$
Lindsay Possibilities: $B_1$
These are same only Pr = $frac{2}{5}$ times
Same thing with Day $2$ and Day $3$ = Pr = $frac{2}{5}$ times
E(X) = $1$ * $frac{2}{5}$ + $2$ * $frac{2}{5}$ + $3$ * $frac{2}{5}$ = $2.4$
This was for the scenario of taking $n=3$.
I am still learning on how to attempt these question so this was a wild attempt by me. I feel like my logic is still getting there so any help on these would be appreciated.
probability probability-theory discrete-mathematics random-variables expected-value
asked Dec 17 '18 at 19:59
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
Part a)
I calculated the Pr by ($10choose{1}$ * $5choose{1}$) / $10choose{5}$
Part b)
I defined X to be the 5 beer subset that Lindsay and Simon can pick. Multiplied each Probability by 2 for both Simon and Lindsay
E(X) = 1 * (($10choose{1}$ * $5choose{1}$) / $10choose{5}$)*2) +
2 * (($10choose{2}$ * $5choose{2}$) / $10choose{5}$)*2) +
3 * (($10choose{3}$ * $5choose{3}$) / $10choose{5}$)*2) +
4 * (($10choose{4}$ * $5choose{4}$) / $10choose{5}$)*2) +
5 * (($10choose{5}$ * $5choose{5}$) / $10choose{5}$)*2)
For Part c)
This was a tricky one, but here is my attempt on it:
I defined $X =1$ to be the number of days that $B_{ai}$ = $B_i$ and $0$ otherwise
I took $n =3$ and since Simon uses the uniform random permutations to select, it becomes:
{$B_1$,$B_2$,$B_3$}
{$B_3$,$B_2$,$B_1$}
{$B_2$,$B_1$,$B_3$}
{$B_1$,$B_3$,$B_2$}
{$B_1$,$B_3$,$B_2$}
{$B_3$,$B_1$,$B_2$}
And for Lindsay, it is simply {$B_1$,$B_2$,$B_3$}
Now for $3$ days, I compared when $B_i's$ are the same for both.
Day $1$: Simon Possibilities: $B_1$, $B_3$, $B_2$, $B_1$, $B_3$
Lindsay Possibilities: $B_1$
These are same only Pr = $frac{2}{5}$ times
Same thing with Day $2$ and Day $3$ = Pr = $frac{2}{5}$ times
E(X) = $1$ * $frac{2}{5}$ + $2$ * $frac{2}{5}$ + $3$ * $frac{2}{5}$ = $2.4$
This was for the scenario of taking $n=3$.
I am still learning on how to attempt these question so this was a wild attempt by me. I feel like my logic is still getting there so any help on these would be appreciated.
probability probability-theory discrete-mathematics random-variables expected-value
asked Dec 17 '18 at 19:59
TobyToby
1577
$endgroup$
Part a)
I calculated the Pr by ($10choose{1}$ * $5choose{1}$) / $10choose{5}$
Part b)
I defined X to be the 5 beer subset that Lindsay and Simon can pick. Multiplied each Probability by 2 for both Simon and Lindsay
E(X) = 1 * (($10choose{1}$ * $5choose{1}$) / $10choose{5}$)*2) +
2 * (($10choose{2}$ * $5choose{2}$) / $10choose{5}$)*2) +
3 * (($10choose{3}$ * $5choose{3}$) / $10choose{5}$)*2) +
4 * (($10choose{4}$ * $5choose{4}$) / $10choose{5}$)*2) +
5 * (($10choose{5}$ * $5choose{5}$) / $10choose{5}$)*2)
For Part c)
This was a tricky one, but here is my attempt on it:
I defined $X =1$ to be the number of days that $B_{ai}$ = $B_i$ and $0$ otherwise
I took $n =3$ and since Simon uses the uniform random permutations to select, it becomes:
{$B_1$,$B_2$,$B_3$}
{$B_3$,$B_2$,$B_1$}
{$B_2$,$B_1$,$B_3$}
{$B_1$,$B_3$,$B_2$}
{$B_1$,$B_3$,$B_2$}
{$B_3$,$B_1$,$B_2$}
And for Lindsay, it is simply {$B_1$,$B_2$,$B_3$}
Now for $3$ days, I compared when $B_i's$ are the same for both.
Day $1$: Simon Possibilities: $B_1$, $B_3$, $B_2$, $B_1$, $B_3$
Lindsay Possibilities: $B_1$
These are same only Pr = $frac{2}{5}$ times
Same thing with Day $2$ and Day $3$ = Pr = $frac{2}{5}$ times
E(X) = $1$ * $frac{2}{5}$ + $2$ * $frac{2}{5}$ + $3$ * $frac{2}{5}$ = $2.4$
This was for the scenario of taking $n=3$.
I am still learning on how to attempt these question so this was a wild attempt by me. I feel like my logic is still getting there so any help on these would be appreciated.
probability probability-theory discrete-mathematics random-variables expected-value
probability probability-theory discrete-mathematics random-variables expected-value
asked Dec 17 '18 at 19:59
TobyToby
1577
asked Dec 17 '18 at 19:59
TobyToby
1577
edited Dec 18 '18 at 0:13
Toby
asked Dec 17 '18 at 19:59
TobyToby
1577
asked Dec 17 '18 at 19:59
TobyToby
1577
asked Dec 17 '18 at 19:59
TobyToby
1577
1577
add a comment |
add a comment |
1 Answer
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$begingroup$
For part A, you get ${9 choose 4} / {10choose 5}$. The numerator is the number of ways that you can select 4 beers out of the remaining 9.
For part B, assume that Linday selects beers 1 through 5 (just relabel the beers if necessary), and let $X_i$ take the value 1 if Simon selects beer $i$ and 0 otherwise. The question asks for $E(sum_{i=1}^5 X_i)$. We have computed $EX_i$ in part A, so The answer is 5 times the answer for part A.
For part C, let $X_i$ take the value 1 if $a_i=i$, and 0 otherwise. The question asks for $E(sum_{i=1}^n X_i)$.
answered Dec 18 '18 at 0:54
LinAlgLinAlg
9,5791521
$endgroup$
add a comment |
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1 Answer
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$begingroup$
For part A, you get ${9 choose 4} / {10choose 5}$. The numerator is the number of ways that you can select 4 beers out of the remaining 9.
For part B, assume that Linday selects beers 1 through 5 (just relabel the beers if necessary), and let $X_i$ take the value 1 if Simon selects beer $i$ and 0 otherwise. The question asks for $E(sum_{i=1}^5 X_i)$. We have computed $EX_i$ in part A, so The answer is 5 times the answer for part A.
For part C, let $X_i$ take the value 1 if $a_i=i$, and 0 otherwise. The question asks for $E(sum_{i=1}^n X_i)$.
answered Dec 18 '18 at 0:54
LinAlgLinAlg
9,5791521
$endgroup$
add a comment |
$begingroup$
For part A, you get ${9 choose 4} / {10choose 5}$. The numerator is the number of ways that you can select 4 beers out of the remaining 9.
For part B, assume that Linday selects beers 1 through 5 (just relabel the beers if necessary), and let $X_i$ take the value 1 if Simon selects beer $i$ and 0 otherwise. The question asks for $E(sum_{i=1}^5 X_i)$. We have computed $EX_i$ in part A, so The answer is 5 times the answer for part A.
For part C, let $X_i$ take the value 1 if $a_i=i$, and 0 otherwise. The question asks for $E(sum_{i=1}^n X_i)$.
answered Dec 18 '18 at 0:54
LinAlgLinAlg
9,5791521
$endgroup$
add a comment |
$begingroup$
For part A, you get ${9 choose 4} / {10choose 5}$. The numerator is the number of ways that you can select 4 beers out of the remaining 9.
For part B, assume that Linday selects beers 1 through 5 (just relabel the beers if necessary), and let $X_i$ take the value 1 if Simon selects beer $i$ and 0 otherwise. The question asks for $E(sum_{i=1}^5 X_i)$. We have computed $EX_i$ in part A, so The answer is 5 times the answer for part A.
For part C, let $X_i$ take the value 1 if $a_i=i$, and 0 otherwise. The question asks for $E(sum_{i=1}^n X_i)$.
answered Dec 18 '18 at 0:54
LinAlgLinAlg
9,5791521
$endgroup$
For part A, you get ${9 choose 4} / {10choose 5}$. The numerator is the number of ways that you can select 4 beers out of the remaining 9.
For part B, assume that Linday selects beers 1 through 5 (just relabel the beers if necessary), and let $X_i$ take the value 1 if Simon selects beer $i$ and 0 otherwise. The question asks for $E(sum_{i=1}^5 X_i)$. We have computed $EX_i$ in part A, so The answer is 5 times the answer for part A.
For part C, let $X_i$ take the value 1 if $a_i=i$, and 0 otherwise. The question asks for $E(sum_{i=1}^n X_i)$.
answered Dec 18 '18 at 0:54
LinAlgLinAlg
9,5791521
answered Dec 18 '18 at 0:54
LinAlgLinAlg
9,5791521
answered Dec 18 '18 at 0:54
LinAlgLinAlg
9,5791521
answered Dec 18 '18 at 0:54
LinAlgLinAlg
9,5791521
9,5791521
add a comment |
add a comment |
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