A deck of cards is split into $2$ halves of $26$ cards each. What is the probability of one partition containing all $4$ aces?
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A deck of cards is split into $2$ halves of $26$ cards each. What is the probability of one partition containing all $4$ aces?
Find the probability corresponding to this. It deals with the concepts of partitions which explicitly added twist.
probability
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A deck of cards is split into $2$ halves of $26$ cards each. What is the probability of one partition containing all $4$ aces?
Find the probability corresponding to this. It deals with the concepts of partitions which explicitly added twist.
probability
Please, instead of uploading images, type your question and add what have you tried so far, in order to encourage us to help you.
â Taroccoesbrocco
Aug 22 at 9:23
Welcome to MathSE. Please type your question rather than posting a link since links may be deleted. This MathJax tutorial explains how to typeset mathematics on this site. Also, when you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have tried and explain where you are stuck.
â N. F. Taussig
Aug 22 at 9:24
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up vote
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down vote
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up vote
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down vote
favorite
A deck of cards is split into $2$ halves of $26$ cards each. What is the probability of one partition containing all $4$ aces?
Find the probability corresponding to this. It deals with the concepts of partitions which explicitly added twist.
probability
A deck of cards is split into $2$ halves of $26$ cards each. What is the probability of one partition containing all $4$ aces?
Find the probability corresponding to this. It deals with the concepts of partitions which explicitly added twist.
probability
edited Aug 22 at 9:29
N. F. Taussig
38.8k93153
38.8k93153
asked Aug 22 at 9:20
Sahil Arora
1
1
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â Taroccoesbrocco
Aug 22 at 9:23
Welcome to MathSE. Please type your question rather than posting a link since links may be deleted. This MathJax tutorial explains how to typeset mathematics on this site. Also, when you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have tried and explain where you are stuck.
â N. F. Taussig
Aug 22 at 9:24
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â Taroccoesbrocco
Aug 22 at 9:23
Welcome to MathSE. Please type your question rather than posting a link since links may be deleted. This MathJax tutorial explains how to typeset mathematics on this site. Also, when you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have tried and explain where you are stuck.
â N. F. Taussig
Aug 22 at 9:24
Please, instead of uploading images, type your question and add what have you tried so far, in order to encourage us to help you.
â Taroccoesbrocco
Aug 22 at 9:23
Please, instead of uploading images, type your question and add what have you tried so far, in order to encourage us to help you.
â Taroccoesbrocco
Aug 22 at 9:23
Welcome to MathSE. Please type your question rather than posting a link since links may be deleted. This MathJax tutorial explains how to typeset mathematics on this site. Also, when you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have tried and explain where you are stuck.
â N. F. Taussig
Aug 22 at 9:24
Welcome to MathSE. Please type your question rather than posting a link since links may be deleted. This MathJax tutorial explains how to typeset mathematics on this site. Also, when you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have tried and explain where you are stuck.
â N. F. Taussig
Aug 22 at 9:24
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2 Answers
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Hint:
Number the parts: $1,2$.
The probability that part $1$ has all $4$ aces is:
$$fracbinom264binom260binom524$$
It cannot happen that both parts have all $4$ aces.
How did you get that "Click to show Spoiler"?
â Mohammad Zuhair Khan
Aug 22 at 9:31
@MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
â drhab
Aug 22 at 9:34
we wont do $binom5226$ ?How do we split it into 2 parts?
â Damn1o1
Aug 22 at 9:50
@Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
â drhab
Aug 22 at 10:11
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We can divide the four aces over the two partitions, ignoring the other cards. For the first ace it is irrelevant in which partition it is placed, but the other aces must be in the same partition as the first. So the probability that all aces are in the same partition is given by $$1cdot frac2551cdotfrac2450cdotfrac2349=frac92833approx 0.11.$$
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint:
Number the parts: $1,2$.
The probability that part $1$ has all $4$ aces is:
$$fracbinom264binom260binom524$$
It cannot happen that both parts have all $4$ aces.
How did you get that "Click to show Spoiler"?
â Mohammad Zuhair Khan
Aug 22 at 9:31
@MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
â drhab
Aug 22 at 9:34
we wont do $binom5226$ ?How do we split it into 2 parts?
â Damn1o1
Aug 22 at 9:50
@Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
â drhab
Aug 22 at 10:11
add a comment |Â
up vote
2
down vote
Hint:
Number the parts: $1,2$.
The probability that part $1$ has all $4$ aces is:
$$fracbinom264binom260binom524$$
It cannot happen that both parts have all $4$ aces.
How did you get that "Click to show Spoiler"?
â Mohammad Zuhair Khan
Aug 22 at 9:31
@MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
â drhab
Aug 22 at 9:34
we wont do $binom5226$ ?How do we split it into 2 parts?
â Damn1o1
Aug 22 at 9:50
@Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
â drhab
Aug 22 at 10:11
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint:
Number the parts: $1,2$.
The probability that part $1$ has all $4$ aces is:
$$fracbinom264binom260binom524$$
It cannot happen that both parts have all $4$ aces.
Hint:
Number the parts: $1,2$.
The probability that part $1$ has all $4$ aces is:
$$fracbinom264binom260binom524$$
It cannot happen that both parts have all $4$ aces.
answered Aug 22 at 9:24
drhab
88k541120
88k541120
How did you get that "Click to show Spoiler"?
â Mohammad Zuhair Khan
Aug 22 at 9:31
@MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
â drhab
Aug 22 at 9:34
we wont do $binom5226$ ?How do we split it into 2 parts?
â Damn1o1
Aug 22 at 9:50
@Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
â drhab
Aug 22 at 10:11
add a comment |Â
How did you get that "Click to show Spoiler"?
â Mohammad Zuhair Khan
Aug 22 at 9:31
@MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
â drhab
Aug 22 at 9:34
we wont do $binom5226$ ?How do we split it into 2 parts?
â Damn1o1
Aug 22 at 9:50
@Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
â drhab
Aug 22 at 10:11
How did you get that "Click to show Spoiler"?
â Mohammad Zuhair Khan
Aug 22 at 9:31
How did you get that "Click to show Spoiler"?
â Mohammad Zuhair Khan
Aug 22 at 9:31
@MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
â drhab
Aug 22 at 9:34
@MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
â drhab
Aug 22 at 9:34
we wont do $binom5226$ ?How do we split it into 2 parts?
â Damn1o1
Aug 22 at 9:50
we wont do $binom5226$ ?How do we split it into 2 parts?
â Damn1o1
Aug 22 at 9:50
@Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
â drhab
Aug 22 at 10:11
@Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
â drhab
Aug 22 at 10:11
add a comment |Â
up vote
0
down vote
We can divide the four aces over the two partitions, ignoring the other cards. For the first ace it is irrelevant in which partition it is placed, but the other aces must be in the same partition as the first. So the probability that all aces are in the same partition is given by $$1cdot frac2551cdotfrac2450cdotfrac2349=frac92833approx 0.11.$$
add a comment |Â
up vote
0
down vote
We can divide the four aces over the two partitions, ignoring the other cards. For the first ace it is irrelevant in which partition it is placed, but the other aces must be in the same partition as the first. So the probability that all aces are in the same partition is given by $$1cdot frac2551cdotfrac2450cdotfrac2349=frac92833approx 0.11.$$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We can divide the four aces over the two partitions, ignoring the other cards. For the first ace it is irrelevant in which partition it is placed, but the other aces must be in the same partition as the first. So the probability that all aces are in the same partition is given by $$1cdot frac2551cdotfrac2450cdotfrac2349=frac92833approx 0.11.$$
We can divide the four aces over the two partitions, ignoring the other cards. For the first ace it is irrelevant in which partition it is placed, but the other aces must be in the same partition as the first. So the probability that all aces are in the same partition is given by $$1cdot frac2551cdotfrac2450cdotfrac2349=frac92833approx 0.11.$$
answered Aug 22 at 10:57
molarmass
1,327513
1,327513
add a comment |Â
add a comment |Â
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Please, instead of uploading images, type your question and add what have you tried so far, in order to encourage us to help you.
â Taroccoesbrocco
Aug 22 at 9:23
Welcome to MathSE. Please type your question rather than posting a link since links may be deleted. This MathJax tutorial explains how to typeset mathematics on this site. Also, when you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have tried and explain where you are stuck.
â N. F. Taussig
Aug 22 at 9:24