Determine the Expected Value of Uniformly random elements of sets
$begingroup$
Answer is D
The way I attempted this was that for X = MAX(a,b), the random variable X is equivalent to the max value of a and b. So, from the 2 sets, the probability of getting k from set {1,2...100} is $frac{k}{100}$ and for the second set we have to select $frac{k-1}{100}$ so that it is less than the first one.
E(X) would then just be the summation of $x*$$frac{{k*}{(k-1)}}{100}$, which is option B.
Don't understand how its D? Any explanations?
probability discrete-mathematics random-variables expected-value
asked Dec 12 '18 at 18:34
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
Answer is D
The way I attempted this was that for X = MAX(a,b), the random variable X is equivalent to the max value of a and b. So, from the 2 sets, the probability of getting k from set {1,2...100} is $frac{k}{100}$ and for the second set we have to select $frac{k-1}{100}$ so that it is less than the first one.
E(X) would then just be the summation of $x*$$frac{{k*}{(k-1)}}{100}$, which is option B.
Don't understand how its D? Any explanations?
probability discrete-mathematics random-variables expected-value
asked Dec 12 '18 at 18:34
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
Answer is D
The way I attempted this was that for X = MAX(a,b), the random variable X is equivalent to the max value of a and b. So, from the 2 sets, the probability of getting k from set {1,2...100} is $frac{k}{100}$ and for the second set we have to select $frac{k-1}{100}$ so that it is less than the first one.
E(X) would then just be the summation of $x*$$frac{{k*}{(k-1)}}{100}$, which is option B.
Don't understand how its D? Any explanations?
probability discrete-mathematics random-variables expected-value
asked Dec 12 '18 at 18:34
TobyToby
1577
$endgroup$
Answer is D
The way I attempted this was that for X = MAX(a,b), the random variable X is equivalent to the max value of a and b. So, from the 2 sets, the probability of getting k from set {1,2...100} is $frac{k}{100}$ and for the second set we have to select $frac{k-1}{100}$ so that it is less than the first one.
E(X) would then just be the summation of $x*$$frac{{k*}{(k-1)}}{100}$, which is option B.
Don't understand how its D? Any explanations?
probability discrete-mathematics random-variables expected-value
probability discrete-mathematics random-variables expected-value
asked Dec 12 '18 at 18:34
TobyToby
1577
asked Dec 12 '18 at 18:34
TobyToby
1577
asked Dec 12 '18 at 18:34
TobyToby
1577
asked Dec 12 '18 at 18:34
TobyToby
1577
asked Dec 12 '18 at 18:34
TobyToby
1577
1577
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your general logic is fine, although you calculate the probability incorrectly. Note that, if $max(a,b) = k$, then we have one of three cases:
$a = k$ and $b < k$.
$a < k$ and $b = k$.- $a = b = k$
It should be readily apparent that these cases are disjoint, so to find the probability $P(X = k)$, we can add the probabilities of these cases.
Since $a,b$ are uniform, it's really a matter of counting how many ways this can happen. There's one way for $a = k$ to happen and $k-1$ ways for $b < k$, so multiplying these gives $k-1$ ways; similarly, there's also $k-1$ ways to get $a < k$ and $b = k$. Finally, there's exactly one way to get $a = b = k$. Adding these gives you $1 + 2(k - 1)$; we divide by $100^2$ as the total number of outcomes for $a,b$. This give the probability $P(X = k) = frac{1 + 2(k-1)}{100^2}$, from which you can derive the expected value in the way you do above.
answered Dec 12 '18 at 22:03
plattyplatty
3,370320
$endgroup$
$begingroup$
I'm not familiar with the disjoint terminology. I'm assuming that means there is nothing in common with those 3 cases? If so, doesn't case 1 and 2 have a=k and b=k common with case 3? Does that make it not disjoint and you can't add probabilities (assuming again you can add probabilities only for disjoint?)
$endgroup$
– Toby
Dec 12 '18 at 22:42
1
$begingroup$
By disjoint, I mean that the cases have no overlap -- no outcome can be in multiple cases at once. You're right in noting disjointness is required to add probabilities. But it's impossible for case $3$ to overlap with either of the other two, because if $a=k$ and $b=k$, then one of the conditions has to break (because case 1 requires $b < k$ and case 2 requires $a < k$).
$endgroup$
– platty
Dec 12 '18 at 22:43
$begingroup$
Oh, I understand what you mean now! Cheers
$endgroup$
– Toby
Dec 12 '18 at 22:45
add a comment |
$begingroup$
You can use the following:
$$E(X)=sum_{k=1}^{100} P(Xge k)$$
$$P(Xge k)=1-P(Xle k-1)$$
$$P(Xle k)=P(ale ktext{ and }ble k)$$
answered Dec 12 '18 at 18:40
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
Your general logic is fine, although you calculate the probability incorrectly. Note that, if $max(a,b) = k$, then we have one of three cases:
$a = k$ and $b < k$.
$a < k$ and $b = k$.- $a = b = k$
It should be readily apparent that these cases are disjoint, so to find the probability $P(X = k)$, we can add the probabilities of these cases.
Since $a,b$ are uniform, it's really a matter of counting how many ways this can happen. There's one way for $a = k$ to happen and $k-1$ ways for $b < k$, so multiplying these gives $k-1$ ways; similarly, there's also $k-1$ ways to get $a < k$ and $b = k$. Finally, there's exactly one way to get $a = b = k$. Adding these gives you $1 + 2(k - 1)$; we divide by $100^2$ as the total number of outcomes for $a,b$. This give the probability $P(X = k) = frac{1 + 2(k-1)}{100^2}$, from which you can derive the expected value in the way you do above.
answered Dec 12 '18 at 22:03
plattyplatty
3,370320
$endgroup$
$begingroup$
I'm not familiar with the disjoint terminology. I'm assuming that means there is nothing in common with those 3 cases? If so, doesn't case 1 and 2 have a=k and b=k common with case 3? Does that make it not disjoint and you can't add probabilities (assuming again you can add probabilities only for disjoint?)
$endgroup$
– Toby
Dec 12 '18 at 22:42
1
$begingroup$
By disjoint, I mean that the cases have no overlap -- no outcome can be in multiple cases at once. You're right in noting disjointness is required to add probabilities. But it's impossible for case $3$ to overlap with either of the other two, because if $a=k$ and $b=k$, then one of the conditions has to break (because case 1 requires $b < k$ and case 2 requires $a < k$).
$endgroup$
– platty
Dec 12 '18 at 22:43
$begingroup$
Oh, I understand what you mean now! Cheers
$endgroup$
– Toby
Dec 12 '18 at 22:45
add a comment |
$begingroup$
Your general logic is fine, although you calculate the probability incorrectly. Note that, if $max(a,b) = k$, then we have one of three cases:
$a = k$ and $b < k$.
$a < k$ and $b = k$.- $a = b = k$
It should be readily apparent that these cases are disjoint, so to find the probability $P(X = k)$, we can add the probabilities of these cases.
Since $a,b$ are uniform, it's really a matter of counting how many ways this can happen. There's one way for $a = k$ to happen and $k-1$ ways for $b < k$, so multiplying these gives $k-1$ ways; similarly, there's also $k-1$ ways to get $a < k$ and $b = k$. Finally, there's exactly one way to get $a = b = k$. Adding these gives you $1 + 2(k - 1)$; we divide by $100^2$ as the total number of outcomes for $a,b$. This give the probability $P(X = k) = frac{1 + 2(k-1)}{100^2}$, from which you can derive the expected value in the way you do above.
answered Dec 12 '18 at 22:03
plattyplatty
3,370320
$endgroup$
$begingroup$
I'm not familiar with the disjoint terminology. I'm assuming that means there is nothing in common with those 3 cases? If so, doesn't case 1 and 2 have a=k and b=k common with case 3? Does that make it not disjoint and you can't add probabilities (assuming again you can add probabilities only for disjoint?)
$endgroup$
– Toby
Dec 12 '18 at 22:42
1
$begingroup$
By disjoint, I mean that the cases have no overlap -- no outcome can be in multiple cases at once. You're right in noting disjointness is required to add probabilities. But it's impossible for case $3$ to overlap with either of the other two, because if $a=k$ and $b=k$, then one of the conditions has to break (because case 1 requires $b < k$ and case 2 requires $a < k$).
$endgroup$
– platty
Dec 12 '18 at 22:43
$begingroup$
Oh, I understand what you mean now! Cheers
$endgroup$
– Toby
Dec 12 '18 at 22:45
add a comment |
$begingroup$
Your general logic is fine, although you calculate the probability incorrectly. Note that, if $max(a,b) = k$, then we have one of three cases:
$a = k$ and $b < k$.
$a < k$ and $b = k$.- $a = b = k$
It should be readily apparent that these cases are disjoint, so to find the probability $P(X = k)$, we can add the probabilities of these cases.
Since $a,b$ are uniform, it's really a matter of counting how many ways this can happen. There's one way for $a = k$ to happen and $k-1$ ways for $b < k$, so multiplying these gives $k-1$ ways; similarly, there's also $k-1$ ways to get $a < k$ and $b = k$. Finally, there's exactly one way to get $a = b = k$. Adding these gives you $1 + 2(k - 1)$; we divide by $100^2$ as the total number of outcomes for $a,b$. This give the probability $P(X = k) = frac{1 + 2(k-1)}{100^2}$, from which you can derive the expected value in the way you do above.
answered Dec 12 '18 at 22:03
plattyplatty
3,370320
$endgroup$
Your general logic is fine, although you calculate the probability incorrectly. Note that, if $max(a,b) = k$, then we have one of three cases:
$a = k$ and $b < k$.
$a < k$ and $b = k$.- $a = b = k$
It should be readily apparent that these cases are disjoint, so to find the probability $P(X = k)$, we can add the probabilities of these cases.
Since $a,b$ are uniform, it's really a matter of counting how many ways this can happen. There's one way for $a = k$ to happen and $k-1$ ways for $b < k$, so multiplying these gives $k-1$ ways; similarly, there's also $k-1$ ways to get $a < k$ and $b = k$. Finally, there's exactly one way to get $a = b = k$. Adding these gives you $1 + 2(k - 1)$; we divide by $100^2$ as the total number of outcomes for $a,b$. This give the probability $P(X = k) = frac{1 + 2(k-1)}{100^2}$, from which you can derive the expected value in the way you do above.
answered Dec 12 '18 at 22:03
plattyplatty
3,370320
answered Dec 12 '18 at 22:03
plattyplatty
3,370320
answered Dec 12 '18 at 22:03
plattyplatty
3,370320
answered Dec 12 '18 at 22:03
plattyplatty
3,370320
3,370320
$begingroup$
I'm not familiar with the disjoint terminology. I'm assuming that means there is nothing in common with those 3 cases? If so, doesn't case 1 and 2 have a=k and b=k common with case 3? Does that make it not disjoint and you can't add probabilities (assuming again you can add probabilities only for disjoint?)
$endgroup$
– Toby
Dec 12 '18 at 22:42
1
$begingroup$
By disjoint, I mean that the cases have no overlap -- no outcome can be in multiple cases at once. You're right in noting disjointness is required to add probabilities. But it's impossible for case $3$ to overlap with either of the other two, because if $a=k$ and $b=k$, then one of the conditions has to break (because case 1 requires $b < k$ and case 2 requires $a < k$).
$endgroup$
– platty
Dec 12 '18 at 22:43
$begingroup$
Oh, I understand what you mean now! Cheers
$endgroup$
– Toby
Dec 12 '18 at 22:45
add a comment |
$begingroup$
I'm not familiar with the disjoint terminology. I'm assuming that means there is nothing in common with those 3 cases? If so, doesn't case 1 and 2 have a=k and b=k common with case 3? Does that make it not disjoint and you can't add probabilities (assuming again you can add probabilities only for disjoint?)
$endgroup$
– Toby
Dec 12 '18 at 22:42
1
$begingroup$
By disjoint, I mean that the cases have no overlap -- no outcome can be in multiple cases at once. You're right in noting disjointness is required to add probabilities. But it's impossible for case $3$ to overlap with either of the other two, because if $a=k$ and $b=k$, then one of the conditions has to break (because case 1 requires $b < k$ and case 2 requires $a < k$).
$endgroup$
– platty
Dec 12 '18 at 22:43
$begingroup$
Oh, I understand what you mean now! Cheers
$endgroup$
– Toby
Dec 12 '18 at 22:45
$begingroup$
I'm not familiar with the disjoint terminology. I'm assuming that means there is nothing in common with those 3 cases? If so, doesn't case 1 and 2 have a=k and b=k common with case 3? Does that make it not disjoint and you can't add probabilities (assuming again you can add probabilities only for disjoint?)
$endgroup$
– Toby
Dec 12 '18 at 22:42
$begingroup$
I'm not familiar with the disjoint terminology. I'm assuming that means there is nothing in common with those 3 cases? If so, doesn't case 1 and 2 have a=k and b=k common with case 3? Does that make it not disjoint and you can't add probabilities (assuming again you can add probabilities only for disjoint?)
$endgroup$
– Toby
Dec 12 '18 at 22:42
1
1
$begingroup$
By disjoint, I mean that the cases have no overlap -- no outcome can be in multiple cases at once. You're right in noting disjointness is required to add probabilities. But it's impossible for case $3$ to overlap with either of the other two, because if $a=k$ and $b=k$, then one of the conditions has to break (because case 1 requires $b < k$ and case 2 requires $a < k$).
$endgroup$
– platty
Dec 12 '18 at 22:43
$begingroup$
By disjoint, I mean that the cases have no overlap -- no outcome can be in multiple cases at once. You're right in noting disjointness is required to add probabilities. But it's impossible for case $3$ to overlap with either of the other two, because if $a=k$ and $b=k$, then one of the conditions has to break (because case 1 requires $b < k$ and case 2 requires $a < k$).
$endgroup$
– platty
Dec 12 '18 at 22:43
$begingroup$
Oh, I understand what you mean now! Cheers
$endgroup$
– Toby
Dec 12 '18 at 22:45
$begingroup$
Oh, I understand what you mean now! Cheers
$endgroup$
– Toby
Dec 12 '18 at 22:45
add a comment |
$begingroup$
You can use the following:
$$E(X)=sum_{k=1}^{100} P(Xge k)$$
$$P(Xge k)=1-P(Xle k-1)$$
$$P(Xle k)=P(ale ktext{ and }ble k)$$
answered Dec 12 '18 at 18:40
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
$endgroup$
add a comment |
$begingroup$
You can use the following:
$$E(X)=sum_{k=1}^{100} P(Xge k)$$
$$P(Xge k)=1-P(Xle k-1)$$
$$P(Xle k)=P(ale ktext{ and }ble k)$$
answered Dec 12 '18 at 18:40
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
$endgroup$
add a comment |
$begingroup$
You can use the following:
$$E(X)=sum_{k=1}^{100} P(Xge k)$$
$$P(Xge k)=1-P(Xle k-1)$$
$$P(Xle k)=P(ale ktext{ and }ble k)$$
answered Dec 12 '18 at 18:40
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
$endgroup$
You can use the following:
$$E(X)=sum_{k=1}^{100} P(Xge k)$$
$$P(Xge k)=1-P(Xle k-1)$$
$$P(Xle k)=P(ale ktext{ and }ble k)$$
answered Dec 12 '18 at 18:40
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
answered Dec 12 '18 at 18:40
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
answered Dec 12 '18 at 18:40
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
answered Dec 12 '18 at 18:40
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
2,086918
add a comment |
add a comment |
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