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What is the probability that two cards drawn from a deck are both face cards and at least one is red?

Probability that two cards are black and one is redTwo cards are drawn without replacement from an ordinary deck, find the probability..Two cards are chosen from a deck of 52 cards without replacement.Probability of cards drawn from a deckProbability: Four cards face down, two are red, two are black. Aim is to find the red ones.Probability of face cards drawn from a deck.Two cards are drawn without replacement. Find a probability when Queens are drawn.What is the probability that if two cards are drawn from a standard deck without replacement that the first is red and the second is a heart?If five cards are drawn randomly from an ordinary deck, what is the probability of drawing exactly three face cards?Probability that two drawn cards are face cards

3

0

$begingroup$

Two cards are drawn from a deck of cards. What is the probability that they are both face cards and at least one is red? Assume that there are $52$ cards and without replacement.

I have two different methods, and they got different solutions. The second method below is the same as the teacher's answer: $0.0385$, but the first is not. What is wrong with my first solution?

Method 1:
$frac12cdot1152cdot51cdotfrac34=0.0373$. I got this because I calculated the probability that I get two face cards, and then I multiplied by $frac 34$ because there is $frac 34$ chance that I get at least $1$ red.

Method 2: $fracbinom122binom522-fracbinom62binom522=0.0385$. The first fraction is the probability that both are face cards. The second fraction is the probability that they are both face cards and both black.

probability card-games

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asked 1 hour ago

BadAtGeometryBadAtGeometry

291317

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add a comment | 

3

0

$begingroup$

Two cards are drawn from a deck of cards. What is the probability that they are both face cards and at least one is red? Assume that there are $52$ cards and without replacement.

I have two different methods, and they got different solutions. The second method below is the same as the teacher's answer: $0.0385$, but the first is not. What is wrong with my first solution?

Method 1:
$frac12cdot1152cdot51cdotfrac34=0.0373$. I got this because I calculated the probability that I get two face cards, and then I multiplied by $frac 34$ because there is $frac 34$ chance that I get at least $1$ red.

Method 2: $fracbinom122binom522-fracbinom62binom522=0.0385$. The first fraction is the probability that both are face cards. The second fraction is the probability that they are both face cards and both black.

probability card-games

share|cite|improve this question

asked 1 hour ago

BadAtGeometryBadAtGeometry

291317

$endgroup$

add a comment | 

$begingroup$

Two cards are drawn from a deck of cards. What is the probability that they are both face cards and at least one is red? Assume that there are $52$ cards and without replacement.

I have two different methods, and they got different solutions. The second method below is the same as the teacher's answer: $0.0385$, but the first is not. What is wrong with my first solution?

Method 1:
$frac12cdot1152cdot51cdotfrac34=0.0373$. I got this because I calculated the probability that I get two face cards, and then I multiplied by $frac 34$ because there is $frac 34$ chance that I get at least $1$ red.

Method 2: $fracbinom122binom522-fracbinom62binom522=0.0385$. The first fraction is the probability that both are face cards. The second fraction is the probability that they are both face cards and both black.

probability card-games

share|cite|improve this question

asked 1 hour ago

BadAtGeometryBadAtGeometry

291317

$endgroup$

share|cite|improve this question

asked 1 hour ago

BadAtGeometryBadAtGeometry

291317

share|cite|improve this question

asked 1 hour ago

BadAtGeometryBadAtGeometry

291317

asked 1 hour ago

BadAtGeometryBadAtGeometry

291317

asked 1 hour ago

BadAtGeometryBadAtGeometry

291317

2 Answers
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3

$begingroup$

Your mistake is in thinking that there is a $tfrac34$ chance to get at least $1$ red; this would be the probability when drawing with replacement. But without replacement, the color of the one card affects the possibilities for the color of the other card.

If you count the number of ways to draw two face cards, you will find there are
$tbinom122=66$ ways. The number of ways to draw no red cards is $binom62=15$, so the number of ways to draw at least one red card is $66-15=51$. This means the probability of at least one of the two drawn cards being red is $tfrac5166neqtfrac34$. Your second computation also reflects this.

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edited 1 hour ago

answered 1 hour ago

ServaesServaes

32.9k444102

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It's $binom 62=15$ ways of drawing only black.
  $endgroup$
  – Arthur
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur Of course, you are right. For some reason I was thinking of four suits in stead of two colors.
  $endgroup$
  – Servaes
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

$3/4$ is the probability that two cards drawn with replacement are not both black. However, because the cards are drawn without replacement, you cannot use this computation (the color of the first card affects the color probabilities for the second card).

share|cite|improve this answer

answered 1 hour ago

angryavianangryavian

43.9k33484

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3

$begingroup$

Your mistake is in thinking that there is a $tfrac34$ chance to get at least $1$ red; this would be the probability when drawing with replacement. But without replacement, the color of the one card affects the possibilities for the color of the other card.

If you count the number of ways to draw two face cards, you will find there are
$tbinom122=66$ ways. The number of ways to draw no red cards is $binom62=15$, so the number of ways to draw at least one red card is $66-15=51$. This means the probability of at least one of the two drawn cards being red is $tfrac5166neqtfrac34$. Your second computation also reflects this.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

ServaesServaes

32.9k444102

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It's $binom 62=15$ ways of drawing only black.
  $endgroup$
  – Arthur
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur Of course, you are right. For some reason I was thinking of four suits in stead of two colors.
  $endgroup$
  – Servaes
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Your mistake is in thinking that there is a $tfrac34$ chance to get at least $1$ red; this would be the probability when drawing with replacement. But without replacement, the color of the one card affects the possibilities for the color of the other card.

If you count the number of ways to draw two face cards, you will find there are
$tbinom122=66$ ways. The number of ways to draw no red cards is $binom62=15$, so the number of ways to draw at least one red card is $66-15=51$. This means the probability of at least one of the two drawn cards being red is $tfrac5166neqtfrac34$. Your second computation also reflects this.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

ServaesServaes

32.9k444102

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It's $binom 62=15$ ways of drawing only black.
  $endgroup$
  – Arthur
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur Of course, you are right. For some reason I was thinking of four suits in stead of two colors.
  $endgroup$
  – Servaes
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Your mistake is in thinking that there is a $tfrac34$ chance to get at least $1$ red; this would be the probability when drawing with replacement. But without replacement, the color of the one card affects the possibilities for the color of the other card.

If you count the number of ways to draw two face cards, you will find there are
$tbinom122=66$ ways. The number of ways to draw no red cards is $binom62=15$, so the number of ways to draw at least one red card is $66-15=51$. This means the probability of at least one of the two drawn cards being red is $tfrac5166neqtfrac34$. Your second computation also reflects this.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

ServaesServaes

32.9k444102

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

ServaesServaes

32.9k444102

answered 1 hour ago

ServaesServaes

32.9k444102

answered 1 hour ago

ServaesServaes

32.9k444102

3

$begingroup$

$3/4$ is the probability that two cards drawn with replacement are not both black. However, because the cards are drawn without replacement, you cannot use this computation (the color of the first card affects the color probabilities for the second card).

share|cite|improve this answer

answered 1 hour ago

angryavianangryavian

43.9k33484

$endgroup$

add a comment | 

3

$begingroup$

$3/4$ is the probability that two cards drawn with replacement are not both black. However, because the cards are drawn without replacement, you cannot use this computation (the color of the first card affects the color probabilities for the second card).

share|cite|improve this answer

answered 1 hour ago

angryavianangryavian

43.9k33484

$endgroup$

add a comment | 

$begingroup$

$3/4$ is the probability that two cards drawn with replacement are not both black. However, because the cards are drawn without replacement, you cannot use this computation (the color of the first card affects the color probabilities for the second card).

share|cite|improve this answer

answered 1 hour ago

angryavianangryavian

43.9k33484

$endgroup$

share|cite|improve this answer

answered 1 hour ago

angryavianangryavian

43.9k33484

answered 1 hour ago

angryavianangryavian

43.9k33484

answered 1 hour ago

angryavianangryavian

43.9k33484