Mixing
What is the probability that in a five card draw you get 0 aces and exactly 1 king?
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I have been stuck on this problem for hours and have absolutely no clue how to go about it. Any help would be amazing.
probability combinatorics
edited 1 hour ago
abc...
2,164529
asked 2 hours ago
Nathaniel Sexton
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up vote
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I have been stuck on this problem for hours and have absolutely no clue how to go about it. Any help would be amazing.
probability combinatorics
edited 1 hour ago
abc...
2,164529
asked 2 hours ago
Nathaniel Sexton
61
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have been stuck on this problem for hours and have absolutely no clue how to go about it. Any help would be amazing.
probability combinatorics
edited 1 hour ago
abc...
2,164529
asked 2 hours ago
Nathaniel Sexton
61
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have been stuck on this problem for hours and have absolutely no clue how to go about it. Any help would be amazing.
probability combinatorics
probability combinatorics
edited 1 hour ago
abc...
2,164529
asked 2 hours ago
Nathaniel Sexton
61
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
abc...
2,164529
asked 2 hours ago
Nathaniel Sexton
61
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
abc...
2,164529
edited 1 hour ago
abc...
2,164529
edited 1 hour ago
abc...
2,164529
2,164529
asked 2 hours ago
Nathaniel Sexton
61
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Nathaniel Sexton
61
asked 2 hours ago
Nathaniel Sexton
61
61
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
Set our sample space as the set of all unordered five-card hands possible.
Pick which $1$ king out of the $4$ possible are used. Pick which remaining $4$ non-king and non-ace cards out of the $44$ possible remaining cards are used in the hand.
Using binomial coefficients, applying multiplication principle, noting that our sample space is equiprobable, and taking the ratio we get our probability as:
$$dfracbinom41binom444binom525$$
answered 1 hour ago
JMoravitz
45k33582
what does that come out to as a decimal? my teacher never taught us that. thank you so much
– Nathaniel Sexton
1 hour ago
@NathanielSexton wolfram result
– JMoravitz
1 hour ago
You should have been introduced to binomial coefficients rather early on. They are an incredibly powerful tool for many counting problems.
– JMoravitz
1 hour ago
add a comment |Â
up vote
1
down vote
There are multiple ways you can attack this problem. One way is to count the hands you can get over all possible hands.
You know that all possible ordered hands are $52 times 51 times 50 times 49 times 48$
(make sure you understand why this is, if it's not obvious already)
If we take out the aces and just draw from this ace-less deck the possible (ordered) hands are $48 times 47 times 46 times 45 times 44$
Are these your possible hands? Not quite. You are counting some hands that are not allowed.
For example, what if you got 4 Kings, or 2 kings or no kings? These would not be permitted.
How many ordered hands are there with 4 Kings? There are $5 times 4 times 3 times 2$. Can you see why? It's like picking from 5 choices for the first position (either of the kings or no king), and then 4 choices for the second position, etc
3 or 2 kings are a bit trickier. It's doable but it gets a bit messy and we can make an error more easily.
Is there an another angle we can attack our problem?
Let's try the following: Let's take the kings out too. How many aces-less and king-less 5-card ordered hands can we have?
$44 times 43 times 42 times 41 times 40$
Now, for each of these hands, we want to replace one of its cards with a king. How many possible ways can we do it? If we replace the first card we have 4 possible ways (4 kings), if we replace the second another 4 ways, the third card the same, and so on. So we have $4+4+4+4+4 = 20$ ways to put a single king in every one of our ace-less and king-less choices.
So the answer is $$frac20 times (44 times 43 times 42 times 41 times 40)52 times 51 times 50 times 49 times 48 $$ (No, it's not! Read on.)
Update
As pointed out by JMoravitz, I am overcounting by several factors. This is because when I am replacing a card for a king I end up with a single hand. But this king might replace several possible cards in this position. So multiple hands produce a single hand.
That's why is good to validate your answer with multiple approaches. If I had carried out my initial approach I would have found the discrepancy on my own.
To correct the second approach we do something similar to Vasya's answer. Assume that we pick a king for the first position: 4 choices. Then we pick any non-ace and non-king card: 44 choices, then another non-ace and non-king card: 43 choices, etc. In total we have $4 times 44 times 43 times 42 times 41$ ordered hands. But remember these hands are with the restriction that the king is fixed in the first position. We can put the king in any of the 5 positions which means we multiply our possible hands by 5.
Thus the correct answer is $$frac5 times 4 times (44 times 43 times 42 times 41)52 times 51 times 50 times 49 times 48 $$
answered 1 hour ago
Thanassis
2,59511025
1
Your final approach at the bottom counts the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,6spadesuit)$ where we then replace the $6spadesuit$ with $Kspadesuit$ differently than the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,8spadesuit)$ where we then replace the $8spadesuit$ with $Kspadesuit$, etc...
– JMoravitz
1 hour ago
@JMoravitz ah you are right! The alternate solution is not so straightforward afterall.
– Thanassis
1 hour ago
1
You could easily fix it and it would effectively match the answer given by Vasya. Pick location of King. Pick suit of king. From left-to-right in remaining open positions pick which non-ace-non-king card is used. This gives our numerator as $5times 4times 44times 43times 42times 41$. You have the denominator correct for your approach. I still find this to be a great deal more writing and harder to see where the answer comes from than the approach with binomial coefficients I outlined in my answer.
– JMoravitz
1 hour ago
@JMoravitz yes, your approach is simpler and faster. But if a student is not comfortable with combinations (as it seems to be the case here), a more elaborate solution that is relying on simpler concepts might be more appropriate.
– Thanassis
1 hour ago
add a comment |Â
up vote
0
down vote
Hint: you need to break it down into cases: 1) first drawn card is a king; $colorredor$ 2) second drawn card is a king and so forth. Then you add probability for all cases. For case 1) probability to get a king first is $4/52$, probability to not get an ace or king ($7$ cards) for the second drawn card is $44/51$, for the third card is $43/50$, for the fourth card is $42/49$ and for the fifth card it's $41/48$. Thus the probability to get a king followed by four non-aces or kings is $frac4 cdot 44 cdot 43 cdot 42 cdot 4152 cdot 51 cdot 50 cdot 49 cdot 48$
answered 1 hour ago
Vasya
2,7181514
You can certainly break it into cases if you wanted to use the sample space as the set of ordered 5-card hands. It is far superior to use as our sample space the set of unordered 5-card hands as it removes all of the nasty case-work.
– JMoravitz
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Set our sample space as the set of all unordered five-card hands possible.
Pick which $1$ king out of the $4$ possible are used. Pick which remaining $4$ non-king and non-ace cards out of the $44$ possible remaining cards are used in the hand.
Using binomial coefficients, applying multiplication principle, noting that our sample space is equiprobable, and taking the ratio we get our probability as:
$$dfracbinom41binom444binom525$$
answered 1 hour ago
JMoravitz
45k33582
what does that come out to as a decimal? my teacher never taught us that. thank you so much
– Nathaniel Sexton
1 hour ago
@NathanielSexton wolfram result
– JMoravitz
1 hour ago
You should have been introduced to binomial coefficients rather early on. They are an incredibly powerful tool for many counting problems.
– JMoravitz
1 hour ago
add a comment |Â
up vote
5
down vote
Set our sample space as the set of all unordered five-card hands possible.
Pick which $1$ king out of the $4$ possible are used. Pick which remaining $4$ non-king and non-ace cards out of the $44$ possible remaining cards are used in the hand.
Using binomial coefficients, applying multiplication principle, noting that our sample space is equiprobable, and taking the ratio we get our probability as:
$$dfracbinom41binom444binom525$$
answered 1 hour ago
JMoravitz
45k33582
what does that come out to as a decimal? my teacher never taught us that. thank you so much
– Nathaniel Sexton
1 hour ago
@NathanielSexton wolfram result
– JMoravitz
1 hour ago
You should have been introduced to binomial coefficients rather early on. They are an incredibly powerful tool for many counting problems.
– JMoravitz
1 hour ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Set our sample space as the set of all unordered five-card hands possible.
Pick which $1$ king out of the $4$ possible are used. Pick which remaining $4$ non-king and non-ace cards out of the $44$ possible remaining cards are used in the hand.
Using binomial coefficients, applying multiplication principle, noting that our sample space is equiprobable, and taking the ratio we get our probability as:
$$dfracbinom41binom444binom525$$
answered 1 hour ago
JMoravitz
45k33582
Set our sample space as the set of all unordered five-card hands possible.
Pick which $1$ king out of the $4$ possible are used. Pick which remaining $4$ non-king and non-ace cards out of the $44$ possible remaining cards are used in the hand.
Using binomial coefficients, applying multiplication principle, noting that our sample space is equiprobable, and taking the ratio we get our probability as:
$$dfracbinom41binom444binom525$$
answered 1 hour ago
JMoravitz
45k33582
edited 1 hour ago
answered 1 hour ago
JMoravitz
45k33582
answered 1 hour ago
JMoravitz
45k33582
answered 1 hour ago
JMoravitz
45k33582
45k33582
what does that come out to as a decimal? my teacher never taught us that. thank you so much
– Nathaniel Sexton
1 hour ago
@NathanielSexton wolfram result
– JMoravitz
1 hour ago
You should have been introduced to binomial coefficients rather early on. They are an incredibly powerful tool for many counting problems.
– JMoravitz
1 hour ago
add a comment |Â
what does that come out to as a decimal? my teacher never taught us that. thank you so much
– Nathaniel Sexton
1 hour ago
@NathanielSexton wolfram result
– JMoravitz
1 hour ago
You should have been introduced to binomial coefficients rather early on. They are an incredibly powerful tool for many counting problems.
– JMoravitz
1 hour ago
what does that come out to as a decimal? my teacher never taught us that. thank you so much
– Nathaniel Sexton
1 hour ago
what does that come out to as a decimal? my teacher never taught us that. thank you so much
– Nathaniel Sexton
1 hour ago
@NathanielSexton wolfram result
– JMoravitz
1 hour ago
@NathanielSexton wolfram result
– JMoravitz
1 hour ago
You should have been introduced to binomial coefficients rather early on. They are an incredibly powerful tool for many counting problems.
– JMoravitz
1 hour ago
You should have been introduced to binomial coefficients rather early on. They are an incredibly powerful tool for many counting problems.
– JMoravitz
1 hour ago
add a comment |Â
up vote
1
down vote
There are multiple ways you can attack this problem. One way is to count the hands you can get over all possible hands.
You know that all possible ordered hands are $52 times 51 times 50 times 49 times 48$
(make sure you understand why this is, if it's not obvious already)
If we take out the aces and just draw from this ace-less deck the possible (ordered) hands are $48 times 47 times 46 times 45 times 44$
Are these your possible hands? Not quite. You are counting some hands that are not allowed.
For example, what if you got 4 Kings, or 2 kings or no kings? These would not be permitted.
How many ordered hands are there with 4 Kings? There are $5 times 4 times 3 times 2$. Can you see why? It's like picking from 5 choices for the first position (either of the kings or no king), and then 4 choices for the second position, etc
3 or 2 kings are a bit trickier. It's doable but it gets a bit messy and we can make an error more easily.
Is there an another angle we can attack our problem?
Let's try the following: Let's take the kings out too. How many aces-less and king-less 5-card ordered hands can we have?
$44 times 43 times 42 times 41 times 40$
Now, for each of these hands, we want to replace one of its cards with a king. How many possible ways can we do it? If we replace the first card we have 4 possible ways (4 kings), if we replace the second another 4 ways, the third card the same, and so on. So we have $4+4+4+4+4 = 20$ ways to put a single king in every one of our ace-less and king-less choices.
So the answer is $$frac20 times (44 times 43 times 42 times 41 times 40)52 times 51 times 50 times 49 times 48 $$ (No, it's not! Read on.)
Update
As pointed out by JMoravitz, I am overcounting by several factors. This is because when I am replacing a card for a king I end up with a single hand. But this king might replace several possible cards in this position. So multiple hands produce a single hand.
That's why is good to validate your answer with multiple approaches. If I had carried out my initial approach I would have found the discrepancy on my own.
To correct the second approach we do something similar to Vasya's answer. Assume that we pick a king for the first position: 4 choices. Then we pick any non-ace and non-king card: 44 choices, then another non-ace and non-king card: 43 choices, etc. In total we have $4 times 44 times 43 times 42 times 41$ ordered hands. But remember these hands are with the restriction that the king is fixed in the first position. We can put the king in any of the 5 positions which means we multiply our possible hands by 5.
Thus the correct answer is $$frac5 times 4 times (44 times 43 times 42 times 41)52 times 51 times 50 times 49 times 48 $$
answered 1 hour ago
Thanassis
2,59511025
1
Your final approach at the bottom counts the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,6spadesuit)$ where we then replace the $6spadesuit$ with $Kspadesuit$ differently than the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,8spadesuit)$ where we then replace the $8spadesuit$ with $Kspadesuit$, etc...
– JMoravitz
1 hour ago
@JMoravitz ah you are right! The alternate solution is not so straightforward afterall.
– Thanassis
1 hour ago
1
You could easily fix it and it would effectively match the answer given by Vasya. Pick location of King. Pick suit of king. From left-to-right in remaining open positions pick which non-ace-non-king card is used. This gives our numerator as $5times 4times 44times 43times 42times 41$. You have the denominator correct for your approach. I still find this to be a great deal more writing and harder to see where the answer comes from than the approach with binomial coefficients I outlined in my answer.
– JMoravitz
1 hour ago
@JMoravitz yes, your approach is simpler and faster. But if a student is not comfortable with combinations (as it seems to be the case here), a more elaborate solution that is relying on simpler concepts might be more appropriate.
– Thanassis
1 hour ago
add a comment |Â
up vote
1
down vote
There are multiple ways you can attack this problem. One way is to count the hands you can get over all possible hands.
You know that all possible ordered hands are $52 times 51 times 50 times 49 times 48$
(make sure you understand why this is, if it's not obvious already)
If we take out the aces and just draw from this ace-less deck the possible (ordered) hands are $48 times 47 times 46 times 45 times 44$
Are these your possible hands? Not quite. You are counting some hands that are not allowed.
For example, what if you got 4 Kings, or 2 kings or no kings? These would not be permitted.
How many ordered hands are there with 4 Kings? There are $5 times 4 times 3 times 2$. Can you see why? It's like picking from 5 choices for the first position (either of the kings or no king), and then 4 choices for the second position, etc
3 or 2 kings are a bit trickier. It's doable but it gets a bit messy and we can make an error more easily.
Is there an another angle we can attack our problem?
Let's try the following: Let's take the kings out too. How many aces-less and king-less 5-card ordered hands can we have?
$44 times 43 times 42 times 41 times 40$
Now, for each of these hands, we want to replace one of its cards with a king. How many possible ways can we do it? If we replace the first card we have 4 possible ways (4 kings), if we replace the second another 4 ways, the third card the same, and so on. So we have $4+4+4+4+4 = 20$ ways to put a single king in every one of our ace-less and king-less choices.
So the answer is $$frac20 times (44 times 43 times 42 times 41 times 40)52 times 51 times 50 times 49 times 48 $$ (No, it's not! Read on.)
Update
As pointed out by JMoravitz, I am overcounting by several factors. This is because when I am replacing a card for a king I end up with a single hand. But this king might replace several possible cards in this position. So multiple hands produce a single hand.
That's why is good to validate your answer with multiple approaches. If I had carried out my initial approach I would have found the discrepancy on my own.
To correct the second approach we do something similar to Vasya's answer. Assume that we pick a king for the first position: 4 choices. Then we pick any non-ace and non-king card: 44 choices, then another non-ace and non-king card: 43 choices, etc. In total we have $4 times 44 times 43 times 42 times 41$ ordered hands. But remember these hands are with the restriction that the king is fixed in the first position. We can put the king in any of the 5 positions which means we multiply our possible hands by 5.
Thus the correct answer is $$frac5 times 4 times (44 times 43 times 42 times 41)52 times 51 times 50 times 49 times 48 $$
answered 1 hour ago
Thanassis
2,59511025
1
Your final approach at the bottom counts the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,6spadesuit)$ where we then replace the $6spadesuit$ with $Kspadesuit$ differently than the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,8spadesuit)$ where we then replace the $8spadesuit$ with $Kspadesuit$, etc...
– JMoravitz
1 hour ago
@JMoravitz ah you are right! The alternate solution is not so straightforward afterall.
– Thanassis
1 hour ago
1
You could easily fix it and it would effectively match the answer given by Vasya. Pick location of King. Pick suit of king. From left-to-right in remaining open positions pick which non-ace-non-king card is used. This gives our numerator as $5times 4times 44times 43times 42times 41$. You have the denominator correct for your approach. I still find this to be a great deal more writing and harder to see where the answer comes from than the approach with binomial coefficients I outlined in my answer.
– JMoravitz
1 hour ago
@JMoravitz yes, your approach is simpler and faster. But if a student is not comfortable with combinations (as it seems to be the case here), a more elaborate solution that is relying on simpler concepts might be more appropriate.
– Thanassis
1 hour ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There are multiple ways you can attack this problem. One way is to count the hands you can get over all possible hands.
You know that all possible ordered hands are $52 times 51 times 50 times 49 times 48$
(make sure you understand why this is, if it's not obvious already)
If we take out the aces and just draw from this ace-less deck the possible (ordered) hands are $48 times 47 times 46 times 45 times 44$
Are these your possible hands? Not quite. You are counting some hands that are not allowed.
For example, what if you got 4 Kings, or 2 kings or no kings? These would not be permitted.
How many ordered hands are there with 4 Kings? There are $5 times 4 times 3 times 2$. Can you see why? It's like picking from 5 choices for the first position (either of the kings or no king), and then 4 choices for the second position, etc
3 or 2 kings are a bit trickier. It's doable but it gets a bit messy and we can make an error more easily.
Is there an another angle we can attack our problem?
Let's try the following: Let's take the kings out too. How many aces-less and king-less 5-card ordered hands can we have?
$44 times 43 times 42 times 41 times 40$
Now, for each of these hands, we want to replace one of its cards with a king. How many possible ways can we do it? If we replace the first card we have 4 possible ways (4 kings), if we replace the second another 4 ways, the third card the same, and so on. So we have $4+4+4+4+4 = 20$ ways to put a single king in every one of our ace-less and king-less choices.
So the answer is $$frac20 times (44 times 43 times 42 times 41 times 40)52 times 51 times 50 times 49 times 48 $$ (No, it's not! Read on.)
Update
As pointed out by JMoravitz, I am overcounting by several factors. This is because when I am replacing a card for a king I end up with a single hand. But this king might replace several possible cards in this position. So multiple hands produce a single hand.
That's why is good to validate your answer with multiple approaches. If I had carried out my initial approach I would have found the discrepancy on my own.
To correct the second approach we do something similar to Vasya's answer. Assume that we pick a king for the first position: 4 choices. Then we pick any non-ace and non-king card: 44 choices, then another non-ace and non-king card: 43 choices, etc. In total we have $4 times 44 times 43 times 42 times 41$ ordered hands. But remember these hands are with the restriction that the king is fixed in the first position. We can put the king in any of the 5 positions which means we multiply our possible hands by 5.
Thus the correct answer is $$frac5 times 4 times (44 times 43 times 42 times 41)52 times 51 times 50 times 49 times 48 $$
answered 1 hour ago
Thanassis
2,59511025
There are multiple ways you can attack this problem. One way is to count the hands you can get over all possible hands.
You know that all possible ordered hands are $52 times 51 times 50 times 49 times 48$
(make sure you understand why this is, if it's not obvious already)
If we take out the aces and just draw from this ace-less deck the possible (ordered) hands are $48 times 47 times 46 times 45 times 44$
Are these your possible hands? Not quite. You are counting some hands that are not allowed.
For example, what if you got 4 Kings, or 2 kings or no kings? These would not be permitted.
How many ordered hands are there with 4 Kings? There are $5 times 4 times 3 times 2$. Can you see why? It's like picking from 5 choices for the first position (either of the kings or no king), and then 4 choices for the second position, etc
3 or 2 kings are a bit trickier. It's doable but it gets a bit messy and we can make an error more easily.
Is there an another angle we can attack our problem?
Let's try the following: Let's take the kings out too. How many aces-less and king-less 5-card ordered hands can we have?
$44 times 43 times 42 times 41 times 40$
Now, for each of these hands, we want to replace one of its cards with a king. How many possible ways can we do it? If we replace the first card we have 4 possible ways (4 kings), if we replace the second another 4 ways, the third card the same, and so on. So we have $4+4+4+4+4 = 20$ ways to put a single king in every one of our ace-less and king-less choices.
So the answer is $$frac20 times (44 times 43 times 42 times 41 times 40)52 times 51 times 50 times 49 times 48 $$ (No, it's not! Read on.)
Update
As pointed out by JMoravitz, I am overcounting by several factors. This is because when I am replacing a card for a king I end up with a single hand. But this king might replace several possible cards in this position. So multiple hands produce a single hand.
That's why is good to validate your answer with multiple approaches. If I had carried out my initial approach I would have found the discrepancy on my own.
To correct the second approach we do something similar to Vasya's answer. Assume that we pick a king for the first position: 4 choices. Then we pick any non-ace and non-king card: 44 choices, then another non-ace and non-king card: 43 choices, etc. In total we have $4 times 44 times 43 times 42 times 41$ ordered hands. But remember these hands are with the restriction that the king is fixed in the first position. We can put the king in any of the 5 positions which means we multiply our possible hands by 5.
Thus the correct answer is $$frac5 times 4 times (44 times 43 times 42 times 41)52 times 51 times 50 times 49 times 48 $$
answered 1 hour ago
Thanassis
2,59511025
edited 1 hour ago
answered 1 hour ago
Thanassis
2,59511025
answered 1 hour ago
Thanassis
2,59511025
answered 1 hour ago
Thanassis
2,59511025
2,59511025
1
Your final approach at the bottom counts the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,6spadesuit)$ where we then replace the $6spadesuit$ with $Kspadesuit$ differently than the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,8spadesuit)$ where we then replace the $8spadesuit$ with $Kspadesuit$, etc...
– JMoravitz
1 hour ago
@JMoravitz ah you are right! The alternate solution is not so straightforward afterall.
– Thanassis
1 hour ago
1
You could easily fix it and it would effectively match the answer given by Vasya. Pick location of King. Pick suit of king. From left-to-right in remaining open positions pick which non-ace-non-king card is used. This gives our numerator as $5times 4times 44times 43times 42times 41$. You have the denominator correct for your approach. I still find this to be a great deal more writing and harder to see where the answer comes from than the approach with binomial coefficients I outlined in my answer.
– JMoravitz
1 hour ago
@JMoravitz yes, your approach is simpler and faster. But if a student is not comfortable with combinations (as it seems to be the case here), a more elaborate solution that is relying on simpler concepts might be more appropriate.
– Thanassis
1 hour ago
add a comment |Â
1
Your final approach at the bottom counts the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,6spadesuit)$ where we then replace the $6spadesuit$ with $Kspadesuit$ differently than the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,8spadesuit)$ where we then replace the $8spadesuit$ with $Kspadesuit$, etc...
– JMoravitz
1 hour ago
@JMoravitz ah you are right! The alternate solution is not so straightforward afterall.
– Thanassis
1 hour ago
1
You could easily fix it and it would effectively match the answer given by Vasya. Pick location of King. Pick suit of king. From left-to-right in remaining open positions pick which non-ace-non-king card is used. This gives our numerator as $5times 4times 44times 43times 42times 41$. You have the denominator correct for your approach. I still find this to be a great deal more writing and harder to see where the answer comes from than the approach with binomial coefficients I outlined in my answer.
– JMoravitz
1 hour ago
@JMoravitz yes, your approach is simpler and faster. But if a student is not comfortable with combinations (as it seems to be the case here), a more elaborate solution that is relying on simpler concepts might be more appropriate.
– Thanassis
1 hour ago
1
1
Your final approach at the bottom counts the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,6spadesuit)$ where we then replace the $6spadesuit$ with $Kspadesuit$ differently than the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,8spadesuit)$ where we then replace the $8spadesuit$ with $Kspadesuit$, etc...
– JMoravitz
1 hour ago
Your final approach at the bottom counts the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,6spadesuit)$ where we then replace the $6spadesuit$ with $Kspadesuit$ differently than the hand $(2spadesuit,3spadesuit,4spadesuit,5spadesuit,8spadesuit)$ where we then replace the $8spadesuit$ with $Kspadesuit$, etc...
– JMoravitz
1 hour ago
@JMoravitz ah you are right! The alternate solution is not so straightforward afterall.
– Thanassis
1 hour ago
@JMoravitz ah you are right! The alternate solution is not so straightforward afterall.
– Thanassis
1 hour ago
1
1
You could easily fix it and it would effectively match the answer given by Vasya. Pick location of King. Pick suit of king. From left-to-right in remaining open positions pick which non-ace-non-king card is used. This gives our numerator as $5times 4times 44times 43times 42times 41$. You have the denominator correct for your approach. I still find this to be a great deal more writing and harder to see where the answer comes from than the approach with binomial coefficients I outlined in my answer.
– JMoravitz
1 hour ago
You could easily fix it and it would effectively match the answer given by Vasya. Pick location of King. Pick suit of king. From left-to-right in remaining open positions pick which non-ace-non-king card is used. This gives our numerator as $5times 4times 44times 43times 42times 41$. You have the denominator correct for your approach. I still find this to be a great deal more writing and harder to see where the answer comes from than the approach with binomial coefficients I outlined in my answer.
– JMoravitz
1 hour ago
@JMoravitz yes, your approach is simpler and faster. But if a student is not comfortable with combinations (as it seems to be the case here), a more elaborate solution that is relying on simpler concepts might be more appropriate.
– Thanassis
1 hour ago
@JMoravitz yes, your approach is simpler and faster. But if a student is not comfortable with combinations (as it seems to be the case here), a more elaborate solution that is relying on simpler concepts might be more appropriate.
– Thanassis
1 hour ago
add a comment |Â
up vote
0
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Hint: you need to break it down into cases: 1) first drawn card is a king; $colorredor$ 2) second drawn card is a king and so forth. Then you add probability for all cases. For case 1) probability to get a king first is $4/52$, probability to not get an ace or king ($7$ cards) for the second drawn card is $44/51$, for the third card is $43/50$, for the fourth card is $42/49$ and for the fifth card it's $41/48$. Thus the probability to get a king followed by four non-aces or kings is $frac4 cdot 44 cdot 43 cdot 42 cdot 4152 cdot 51 cdot 50 cdot 49 cdot 48$
answered 1 hour ago
Vasya
2,7181514
You can certainly break it into cases if you wanted to use the sample space as the set of ordered 5-card hands. It is far superior to use as our sample space the set of unordered 5-card hands as it removes all of the nasty case-work.
– JMoravitz
1 hour ago
add a comment |Â
up vote
0
down vote
Hint: you need to break it down into cases: 1) first drawn card is a king; $colorredor$ 2) second drawn card is a king and so forth. Then you add probability for all cases. For case 1) probability to get a king first is $4/52$, probability to not get an ace or king ($7$ cards) for the second drawn card is $44/51$, for the third card is $43/50$, for the fourth card is $42/49$ and for the fifth card it's $41/48$. Thus the probability to get a king followed by four non-aces or kings is $frac4 cdot 44 cdot 43 cdot 42 cdot 4152 cdot 51 cdot 50 cdot 49 cdot 48$
answered 1 hour ago
Vasya
2,7181514
You can certainly break it into cases if you wanted to use the sample space as the set of ordered 5-card hands. It is far superior to use as our sample space the set of unordered 5-card hands as it removes all of the nasty case-work.
– JMoravitz
1 hour ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: you need to break it down into cases: 1) first drawn card is a king; $colorredor$ 2) second drawn card is a king and so forth. Then you add probability for all cases. For case 1) probability to get a king first is $4/52$, probability to not get an ace or king ($7$ cards) for the second drawn card is $44/51$, for the third card is $43/50$, for the fourth card is $42/49$ and for the fifth card it's $41/48$. Thus the probability to get a king followed by four non-aces or kings is $frac4 cdot 44 cdot 43 cdot 42 cdot 4152 cdot 51 cdot 50 cdot 49 cdot 48$
answered 1 hour ago
Vasya
2,7181514
Hint: you need to break it down into cases: 1) first drawn card is a king; $colorredor$ 2) second drawn card is a king and so forth. Then you add probability for all cases. For case 1) probability to get a king first is $4/52$, probability to not get an ace or king ($7$ cards) for the second drawn card is $44/51$, for the third card is $43/50$, for the fourth card is $42/49$ and for the fifth card it's $41/48$. Thus the probability to get a king followed by four non-aces or kings is $frac4 cdot 44 cdot 43 cdot 42 cdot 4152 cdot 51 cdot 50 cdot 49 cdot 48$
answered 1 hour ago
Vasya
2,7181514
edited 1 hour ago
answered 1 hour ago
Vasya
2,7181514
answered 1 hour ago
Vasya
2,7181514
answered 1 hour ago
Vasya
2,7181514
2,7181514
You can certainly break it into cases if you wanted to use the sample space as the set of ordered 5-card hands. It is far superior to use as our sample space the set of unordered 5-card hands as it removes all of the nasty case-work.
– JMoravitz
1 hour ago
add a comment |Â
You can certainly break it into cases if you wanted to use the sample space as the set of ordered 5-card hands. It is far superior to use as our sample space the set of unordered 5-card hands as it removes all of the nasty case-work.
– JMoravitz
1 hour ago
You can certainly break it into cases if you wanted to use the sample space as the set of ordered 5-card hands. It is far superior to use as our sample space the set of unordered 5-card hands as it removes all of the nasty case-work.
– JMoravitz
1 hour ago
You can certainly break it into cases if you wanted to use the sample space as the set of ordered 5-card hands. It is far superior to use as our sample space the set of unordered 5-card hands as it removes all of the nasty case-work.
– JMoravitz
1 hour ago
add a comment |Â
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