Probability of sending same postcards to 10 friends out of 15 types of postcards
$begingroup$
In a shop there are 15 types of postcards, and you want to send one postcard to each of 10 friends. To save precious vacation time, you decide to select each postcard independently at random from the 15 types.
a)What is the probability that you manage to send everyone the same type of postcard?
My solution for (a) is: The total possible ways to send the postcards is ${15} choose {10}$ which is 3003 ways. I know that to discern the probability of sending everyone the same postcard, I would have to find $x$ whereby $frac{x}{3003}$. However, I am having trouble finding $x$. Would $x$ be $frac{10}{15}$ since there are 10 out of 15 ways to send the same postcard, whereby each postcard would be sent out 10 times?
Could anyone help clarify this and if I am wrong or not? thanks.
probability discrete-mathematics
edited Dec 9 '18 at 15:02
mm-crj
423213
asked Dec 9 '18 at 14:53
MunchiesOatsMunchiesOats
373
$endgroup$
add a comment |
$begingroup$
In a shop there are 15 types of postcards, and you want to send one postcard to each of 10 friends. To save precious vacation time, you decide to select each postcard independently at random from the 15 types.
a)What is the probability that you manage to send everyone the same type of postcard?
My solution for (a) is: The total possible ways to send the postcards is ${15} choose {10}$ which is 3003 ways. I know that to discern the probability of sending everyone the same postcard, I would have to find $x$ whereby $frac{x}{3003}$. However, I am having trouble finding $x$. Would $x$ be $frac{10}{15}$ since there are 10 out of 15 ways to send the same postcard, whereby each postcard would be sent out 10 times?
Could anyone help clarify this and if I am wrong or not? thanks.
probability discrete-mathematics
edited Dec 9 '18 at 15:02
mm-crj
423213
asked Dec 9 '18 at 14:53
MunchiesOatsMunchiesOats
373
$endgroup$
$begingroup$
You can choose 1 out of 15 cards in 15 different ways. And for all the next persons you have to send the same card. So $x=15$
$endgroup$
– mm-crj
Dec 9 '18 at 15:05
add a comment |
$begingroup$
In a shop there are 15 types of postcards, and you want to send one postcard to each of 10 friends. To save precious vacation time, you decide to select each postcard independently at random from the 15 types.
a)What is the probability that you manage to send everyone the same type of postcard?
My solution for (a) is: The total possible ways to send the postcards is ${15} choose {10}$ which is 3003 ways. I know that to discern the probability of sending everyone the same postcard, I would have to find $x$ whereby $frac{x}{3003}$. However, I am having trouble finding $x$. Would $x$ be $frac{10}{15}$ since there are 10 out of 15 ways to send the same postcard, whereby each postcard would be sent out 10 times?
Could anyone help clarify this and if I am wrong or not? thanks.
probability discrete-mathematics
edited Dec 9 '18 at 15:02
mm-crj
423213
asked Dec 9 '18 at 14:53
MunchiesOatsMunchiesOats
373
$endgroup$
In a shop there are 15 types of postcards, and you want to send one postcard to each of 10 friends. To save precious vacation time, you decide to select each postcard independently at random from the 15 types.
a)What is the probability that you manage to send everyone the same type of postcard?
My solution for (a) is: The total possible ways to send the postcards is ${15} choose {10}$ which is 3003 ways. I know that to discern the probability of sending everyone the same postcard, I would have to find $x$ whereby $frac{x}{3003}$. However, I am having trouble finding $x$. Would $x$ be $frac{10}{15}$ since there are 10 out of 15 ways to send the same postcard, whereby each postcard would be sent out 10 times?
Could anyone help clarify this and if I am wrong or not? thanks.
probability discrete-mathematics
probability discrete-mathematics
edited Dec 9 '18 at 15:02
mm-crj
423213
asked Dec 9 '18 at 14:53
MunchiesOatsMunchiesOats
373
edited Dec 9 '18 at 15:02
mm-crj
423213
asked Dec 9 '18 at 14:53
MunchiesOatsMunchiesOats
373
edited Dec 9 '18 at 15:02
mm-crj
423213
edited Dec 9 '18 at 15:02
mm-crj
423213
edited Dec 9 '18 at 15:02
mm-crj
423213
423213
asked Dec 9 '18 at 14:53
MunchiesOatsMunchiesOats
373
asked Dec 9 '18 at 14:53
MunchiesOatsMunchiesOats
373
asked Dec 9 '18 at 14:53
MunchiesOatsMunchiesOats
373
373
$begingroup$
You can choose 1 out of 15 cards in 15 different ways. And for all the next persons you have to send the same card. So $x=15$
$endgroup$
– mm-crj
Dec 9 '18 at 15:05
add a comment |
$begingroup$
You can choose 1 out of 15 cards in 15 different ways. And for all the next persons you have to send the same card. So $x=15$
$endgroup$
– mm-crj
Dec 9 '18 at 15:05
$begingroup$
You can choose 1 out of 15 cards in 15 different ways. And for all the next persons you have to send the same card. So $x=15$
$endgroup$
– mm-crj
Dec 9 '18 at 15:05
$begingroup$
You can choose 1 out of 15 cards in 15 different ways. And for all the next persons you have to send the same card. So $x=15$
$endgroup$
– mm-crj
Dec 9 '18 at 15:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$15 choose 10$ would be the number of unordered ways to select $10$ different postcards out of the $15$. You did not specify that the postcards are different and in fact want them to all be the same.
You are interested in the number of ordered ways to select $15$ postcards with replacement. You can imagine putting your friends in order, then buying a card for the first, the second, and so on. There are $15$ choices for the first friend, $15$ for the second, so $15^2$ so far. There are $15^{10}$ choices for the whole list.
To send everybody the same type of card, you can choose any card for the first friend, which has $15$ choices. You then have to choose the same card for all the rest, so there are only $15$ ways to send everybody the same card.
answered Dec 9 '18 at 15:10
Ross MillikanRoss Millikan
300k24200374
$endgroup$
$begingroup$
Ooh now I see that if it's without replacement there is no way everybody gets the same card!
$endgroup$
– mm-crj
Dec 9 '18 at 15:13
add a comment |
$begingroup$
You're on the right track, although there is an easier way to think about this. Remember that the store has more than one postcard of each type in stock!
Consider the list of $10$ postcards all lined up. How many possible types of postcard are there for that one? Well, it could be any one of $15$. What about the second? Also one of 15. continuing down the line, each post card could be any of the $15$ possible types, so there are $15^{10}$ possible ways to send the postcards.
Now, say you get postcard type $1$ for the first postcard. We need the other $9$ to match, and only $1$ of the $15$ types is type $1$. So there is exactly one way to get all the other postcards to match, namely, if each postcard was type $1$. The same is true if you had chosen type $2$ first, etc, so there is one way to succeed for each of the $15$ possibilities for your first choice.
This gives that the probability in question is $dfrac{15}{15^{10}}=dfrac{1}{15^9}$.
answered Dec 9 '18 at 15:15
Adam CartisanoAdam Cartisano
1764
$endgroup$
add a comment |
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2 Answers
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$begingroup$
$15 choose 10$ would be the number of unordered ways to select $10$ different postcards out of the $15$. You did not specify that the postcards are different and in fact want them to all be the same.
You are interested in the number of ordered ways to select $15$ postcards with replacement. You can imagine putting your friends in order, then buying a card for the first, the second, and so on. There are $15$ choices for the first friend, $15$ for the second, so $15^2$ so far. There are $15^{10}$ choices for the whole list.
To send everybody the same type of card, you can choose any card for the first friend, which has $15$ choices. You then have to choose the same card for all the rest, so there are only $15$ ways to send everybody the same card.
answered Dec 9 '18 at 15:10
Ross MillikanRoss Millikan
300k24200374
$endgroup$
$begingroup$
Ooh now I see that if it's without replacement there is no way everybody gets the same card!
$endgroup$
– mm-crj
Dec 9 '18 at 15:13
add a comment |
$begingroup$
$15 choose 10$ would be the number of unordered ways to select $10$ different postcards out of the $15$. You did not specify that the postcards are different and in fact want them to all be the same.
You are interested in the number of ordered ways to select $15$ postcards with replacement. You can imagine putting your friends in order, then buying a card for the first, the second, and so on. There are $15$ choices for the first friend, $15$ for the second, so $15^2$ so far. There are $15^{10}$ choices for the whole list.
To send everybody the same type of card, you can choose any card for the first friend, which has $15$ choices. You then have to choose the same card for all the rest, so there are only $15$ ways to send everybody the same card.
answered Dec 9 '18 at 15:10
Ross MillikanRoss Millikan
300k24200374
$endgroup$
$begingroup$
Ooh now I see that if it's without replacement there is no way everybody gets the same card!
$endgroup$
– mm-crj
Dec 9 '18 at 15:13
add a comment |
$begingroup$
$15 choose 10$ would be the number of unordered ways to select $10$ different postcards out of the $15$. You did not specify that the postcards are different and in fact want them to all be the same.
You are interested in the number of ordered ways to select $15$ postcards with replacement. You can imagine putting your friends in order, then buying a card for the first, the second, and so on. There are $15$ choices for the first friend, $15$ for the second, so $15^2$ so far. There are $15^{10}$ choices for the whole list.
To send everybody the same type of card, you can choose any card for the first friend, which has $15$ choices. You then have to choose the same card for all the rest, so there are only $15$ ways to send everybody the same card.
answered Dec 9 '18 at 15:10
Ross MillikanRoss Millikan
300k24200374
$endgroup$
$15 choose 10$ would be the number of unordered ways to select $10$ different postcards out of the $15$. You did not specify that the postcards are different and in fact want them to all be the same.
You are interested in the number of ordered ways to select $15$ postcards with replacement. You can imagine putting your friends in order, then buying a card for the first, the second, and so on. There are $15$ choices for the first friend, $15$ for the second, so $15^2$ so far. There are $15^{10}$ choices for the whole list.
To send everybody the same type of card, you can choose any card for the first friend, which has $15$ choices. You then have to choose the same card for all the rest, so there are only $15$ ways to send everybody the same card.
answered Dec 9 '18 at 15:10
Ross MillikanRoss Millikan
300k24200374
answered Dec 9 '18 at 15:10
Ross MillikanRoss Millikan
300k24200374
answered Dec 9 '18 at 15:10
Ross MillikanRoss Millikan
300k24200374
answered Dec 9 '18 at 15:10
Ross MillikanRoss Millikan
300k24200374
300k24200374
$begingroup$
Ooh now I see that if it's without replacement there is no way everybody gets the same card!
$endgroup$
– mm-crj
Dec 9 '18 at 15:13
add a comment |
$begingroup$
Ooh now I see that if it's without replacement there is no way everybody gets the same card!
$endgroup$
– mm-crj
Dec 9 '18 at 15:13
$begingroup$
Ooh now I see that if it's without replacement there is no way everybody gets the same card!
$endgroup$
– mm-crj
Dec 9 '18 at 15:13
$begingroup$
Ooh now I see that if it's without replacement there is no way everybody gets the same card!
$endgroup$
– mm-crj
Dec 9 '18 at 15:13
add a comment |
$begingroup$
You're on the right track, although there is an easier way to think about this. Remember that the store has more than one postcard of each type in stock!
Consider the list of $10$ postcards all lined up. How many possible types of postcard are there for that one? Well, it could be any one of $15$. What about the second? Also one of 15. continuing down the line, each post card could be any of the $15$ possible types, so there are $15^{10}$ possible ways to send the postcards.
Now, say you get postcard type $1$ for the first postcard. We need the other $9$ to match, and only $1$ of the $15$ types is type $1$. So there is exactly one way to get all the other postcards to match, namely, if each postcard was type $1$. The same is true if you had chosen type $2$ first, etc, so there is one way to succeed for each of the $15$ possibilities for your first choice.
This gives that the probability in question is $dfrac{15}{15^{10}}=dfrac{1}{15^9}$.
answered Dec 9 '18 at 15:15
Adam CartisanoAdam Cartisano
1764
$endgroup$
add a comment |
$begingroup$
You're on the right track, although there is an easier way to think about this. Remember that the store has more than one postcard of each type in stock!
Consider the list of $10$ postcards all lined up. How many possible types of postcard are there for that one? Well, it could be any one of $15$. What about the second? Also one of 15. continuing down the line, each post card could be any of the $15$ possible types, so there are $15^{10}$ possible ways to send the postcards.
Now, say you get postcard type $1$ for the first postcard. We need the other $9$ to match, and only $1$ of the $15$ types is type $1$. So there is exactly one way to get all the other postcards to match, namely, if each postcard was type $1$. The same is true if you had chosen type $2$ first, etc, so there is one way to succeed for each of the $15$ possibilities for your first choice.
This gives that the probability in question is $dfrac{15}{15^{10}}=dfrac{1}{15^9}$.
answered Dec 9 '18 at 15:15
Adam CartisanoAdam Cartisano
1764
$endgroup$
add a comment |
$begingroup$
You're on the right track, although there is an easier way to think about this. Remember that the store has more than one postcard of each type in stock!
Consider the list of $10$ postcards all lined up. How many possible types of postcard are there for that one? Well, it could be any one of $15$. What about the second? Also one of 15. continuing down the line, each post card could be any of the $15$ possible types, so there are $15^{10}$ possible ways to send the postcards.
Now, say you get postcard type $1$ for the first postcard. We need the other $9$ to match, and only $1$ of the $15$ types is type $1$. So there is exactly one way to get all the other postcards to match, namely, if each postcard was type $1$. The same is true if you had chosen type $2$ first, etc, so there is one way to succeed for each of the $15$ possibilities for your first choice.
This gives that the probability in question is $dfrac{15}{15^{10}}=dfrac{1}{15^9}$.
answered Dec 9 '18 at 15:15
Adam CartisanoAdam Cartisano
1764
$endgroup$
You're on the right track, although there is an easier way to think about this. Remember that the store has more than one postcard of each type in stock!
Consider the list of $10$ postcards all lined up. How many possible types of postcard are there for that one? Well, it could be any one of $15$. What about the second? Also one of 15. continuing down the line, each post card could be any of the $15$ possible types, so there are $15^{10}$ possible ways to send the postcards.
Now, say you get postcard type $1$ for the first postcard. We need the other $9$ to match, and only $1$ of the $15$ types is type $1$. So there is exactly one way to get all the other postcards to match, namely, if each postcard was type $1$. The same is true if you had chosen type $2$ first, etc, so there is one way to succeed for each of the $15$ possibilities for your first choice.
This gives that the probability in question is $dfrac{15}{15^{10}}=dfrac{1}{15^9}$.
answered Dec 9 '18 at 15:15
Adam CartisanoAdam Cartisano
1764
answered Dec 9 '18 at 15:15
Adam CartisanoAdam Cartisano
1764
answered Dec 9 '18 at 15:15
Adam CartisanoAdam Cartisano
1764
answered Dec 9 '18 at 15:15
Adam CartisanoAdam Cartisano
1764
1764
add a comment |
add a comment |
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$begingroup$
You can choose 1 out of 15 cards in 15 different ways. And for all the next persons you have to send the same card. So $x=15$
$endgroup$
– mm-crj
Dec 9 '18 at 15:05