Expectation and RIP question
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Hi I am trying to solve the following question but couldnt figure out if i am moving in the right direction or not. The question is as follows.
I have a large number of candies. One day i decided to distribute them to my friends. Suppose 10 of my friends have a collection of
1000 candies each, the other 15 have a collection of 1300 candies each while the rest of your 5 friends have collected 600
candies each.
What is the average number of candies your friends have?
If you mix all of these candies and randomly pick one out, what is the expected size of the collection that candies's
owner has?
I am trying to approach this problem Through Random Incidence Paradox. Am i doing correct can any one help me out here.
probability probability-theory probability-distributions expected-value
asked Nov 17 at 18:48
chuffles
203
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0
down vote
favorite
Hi I am trying to solve the following question but couldnt figure out if i am moving in the right direction or not. The question is as follows.
I have a large number of candies. One day i decided to distribute them to my friends. Suppose 10 of my friends have a collection of
1000 candies each, the other 15 have a collection of 1300 candies each while the rest of your 5 friends have collected 600
candies each.
What is the average number of candies your friends have?
If you mix all of these candies and randomly pick one out, what is the expected size of the collection that candies's
owner has?
I am trying to approach this problem Through Random Incidence Paradox. Am i doing correct can any one help me out here.
probability probability-theory probability-distributions expected-value
asked Nov 17 at 18:48
chuffles
203
1
Don´t forget to accept one of the given answers. Check your previous questions. Hint: At 1. you just have to calculate the arithmetic mean.
– callculus
Nov 17 at 18:50
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Hi I am trying to solve the following question but couldnt figure out if i am moving in the right direction or not. The question is as follows.
I have a large number of candies. One day i decided to distribute them to my friends. Suppose 10 of my friends have a collection of
1000 candies each, the other 15 have a collection of 1300 candies each while the rest of your 5 friends have collected 600
candies each.
What is the average number of candies your friends have?
If you mix all of these candies and randomly pick one out, what is the expected size of the collection that candies's
owner has?
I am trying to approach this problem Through Random Incidence Paradox. Am i doing correct can any one help me out here.
probability probability-theory probability-distributions expected-value
asked Nov 17 at 18:48
chuffles
203
Hi I am trying to solve the following question but couldnt figure out if i am moving in the right direction or not. The question is as follows.
I have a large number of candies. One day i decided to distribute them to my friends. Suppose 10 of my friends have a collection of
1000 candies each, the other 15 have a collection of 1300 candies each while the rest of your 5 friends have collected 600
candies each.
What is the average number of candies your friends have?
If you mix all of these candies and randomly pick one out, what is the expected size of the collection that candies's
owner has?
I am trying to approach this problem Through Random Incidence Paradox. Am i doing correct can any one help me out here.
probability probability-theory probability-distributions expected-value
probability probability-theory probability-distributions expected-value
asked Nov 17 at 18:48
chuffles
203
asked Nov 17 at 18:48
chuffles
203
asked Nov 17 at 18:48
chuffles
203
asked Nov 17 at 18:48
chuffles
203
asked Nov 17 at 18:48
chuffles
203
203
1
Don´t forget to accept one of the given answers. Check your previous questions. Hint: At 1. you just have to calculate the arithmetic mean.
– callculus
Nov 17 at 18:50
add a comment |
1
Don´t forget to accept one of the given answers. Check your previous questions. Hint: At 1. you just have to calculate the arithmetic mean.
– callculus
Nov 17 at 18:50
1
1
Don´t forget to accept one of the given answers. Check your previous questions. Hint: At 1. you just have to calculate the arithmetic mean.
– callculus
Nov 17 at 18:50
Don´t forget to accept one of the given answers. Check your previous questions. Hint: At 1. you just have to calculate the arithmetic mean.
– callculus
Nov 17 at 18:50
add a comment |
1 Answer
1
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Simply the weighted average: $frac{10 times 1000 + 15*1300 + 5*600}{10+15+5} = 1083.33$
Group 1 brings $10 times 1000 = 10000$ candies to the mix, group 2 brings $15 times 1300 = 19500$ candies to the mix, and group 3 brings $5 times 600 = 3000$ candies to the mix. There are a total of 32500 candies. You will pick a candy from an friend from group 1 with probability $10000/32500 = 0.3076$, group 2 with probability 0.6, and group 3 with probability 0.0923. Thus, the owner will have a candy collection of 1000 w.p. 0.3076, 1300 w.p. 0.6 and 600 w.p. 0.0923. You can now compute the expected value as $0.3076 times 1000 + 0.6 times 1300 + .0923 times 600 = 1143.08$, which is not the same as part 1.
The random incidence paradox is a good framework to understand the details of this problem.
answered Nov 18 at 3:41
Aditya Dua
5958
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Simply the weighted average: $frac{10 times 1000 + 15*1300 + 5*600}{10+15+5} = 1083.33$
Group 1 brings $10 times 1000 = 10000$ candies to the mix, group 2 brings $15 times 1300 = 19500$ candies to the mix, and group 3 brings $5 times 600 = 3000$ candies to the mix. There are a total of 32500 candies. You will pick a candy from an friend from group 1 with probability $10000/32500 = 0.3076$, group 2 with probability 0.6, and group 3 with probability 0.0923. Thus, the owner will have a candy collection of 1000 w.p. 0.3076, 1300 w.p. 0.6 and 600 w.p. 0.0923. You can now compute the expected value as $0.3076 times 1000 + 0.6 times 1300 + .0923 times 600 = 1143.08$, which is not the same as part 1.
The random incidence paradox is a good framework to understand the details of this problem.
answered Nov 18 at 3:41
Aditya Dua
5958
add a comment |
up vote
1
down vote
Simply the weighted average: $frac{10 times 1000 + 15*1300 + 5*600}{10+15+5} = 1083.33$
Group 1 brings $10 times 1000 = 10000$ candies to the mix, group 2 brings $15 times 1300 = 19500$ candies to the mix, and group 3 brings $5 times 600 = 3000$ candies to the mix. There are a total of 32500 candies. You will pick a candy from an friend from group 1 with probability $10000/32500 = 0.3076$, group 2 with probability 0.6, and group 3 with probability 0.0923. Thus, the owner will have a candy collection of 1000 w.p. 0.3076, 1300 w.p. 0.6 and 600 w.p. 0.0923. You can now compute the expected value as $0.3076 times 1000 + 0.6 times 1300 + .0923 times 600 = 1143.08$, which is not the same as part 1.
The random incidence paradox is a good framework to understand the details of this problem.
answered Nov 18 at 3:41
Aditya Dua
5958
add a comment |
up vote
1
down vote
up vote
1
down vote
Simply the weighted average: $frac{10 times 1000 + 15*1300 + 5*600}{10+15+5} = 1083.33$
Group 1 brings $10 times 1000 = 10000$ candies to the mix, group 2 brings $15 times 1300 = 19500$ candies to the mix, and group 3 brings $5 times 600 = 3000$ candies to the mix. There are a total of 32500 candies. You will pick a candy from an friend from group 1 with probability $10000/32500 = 0.3076$, group 2 with probability 0.6, and group 3 with probability 0.0923. Thus, the owner will have a candy collection of 1000 w.p. 0.3076, 1300 w.p. 0.6 and 600 w.p. 0.0923. You can now compute the expected value as $0.3076 times 1000 + 0.6 times 1300 + .0923 times 600 = 1143.08$, which is not the same as part 1.
The random incidence paradox is a good framework to understand the details of this problem.
answered Nov 18 at 3:41
Aditya Dua
5958
Simply the weighted average: $frac{10 times 1000 + 15*1300 + 5*600}{10+15+5} = 1083.33$
Group 1 brings $10 times 1000 = 10000$ candies to the mix, group 2 brings $15 times 1300 = 19500$ candies to the mix, and group 3 brings $5 times 600 = 3000$ candies to the mix. There are a total of 32500 candies. You will pick a candy from an friend from group 1 with probability $10000/32500 = 0.3076$, group 2 with probability 0.6, and group 3 with probability 0.0923. Thus, the owner will have a candy collection of 1000 w.p. 0.3076, 1300 w.p. 0.6 and 600 w.p. 0.0923. You can now compute the expected value as $0.3076 times 1000 + 0.6 times 1300 + .0923 times 600 = 1143.08$, which is not the same as part 1.
The random incidence paradox is a good framework to understand the details of this problem.
answered Nov 18 at 3:41
Aditya Dua
5958
answered Nov 18 at 3:41
Aditya Dua
5958
answered Nov 18 at 3:41
Aditya Dua
5958
answered Nov 18 at 3:41
Aditya Dua
5958
5958
add a comment |
add a comment |
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Don´t forget to accept one of the given answers. Check your previous questions. Hint: At 1. you just have to calculate the arithmetic mean.
– callculus
Nov 17 at 18:50