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If you draw two cards in consecutively in a standard deck of 52 cards, what is the probability of getting black on the second draw?

Probability of getting two black ballsThe Deck of cards!Probability of cards drawn from a deckConditional Probability Problem (3 Cards, 1 red-red, 1 red-black, 1 black-black)Two Cards are drawn from the standard deck of 52 cards. What is the probability that the ff. occur?What is the amount of draws necessary to see all red cards from a standard deck of 52 cards if you draw 5 cards from the deck?Calculate the probability that the $i$-th draw returns a red ball.What is the probability that when 5 cards are picked out of a standard, 52-card deck, they alternate in color

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$begingroup$

This is my thought process:

$P(2^{nd} text{black}) = P(2^{nd} text{black} mid 1^{st} text{red}) P(1^{st} text{red}) + P(2^{nd} text{black} mid 1^{st} text{black}) P(1^{st} text{black}) $

$frac{26}{51}*frac{26}{52} + frac{25}{51}*frac{26}{52} = frac{1}{2} $

Is this correct?

probability

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edited 9 hours ago

cmk

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• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes you are right.
  $endgroup$
  – Mostafa Ayaz
  9 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

This is my thought process:

$P(2^{nd} text{black}) = P(2^{nd} text{black} mid 1^{st} text{red}) P(1^{st} text{red}) + P(2^{nd} text{black} mid 1^{st} text{black}) P(1^{st} text{black}) $

$frac{26}{51}*frac{26}{52} + frac{25}{51}*frac{26}{52} = frac{1}{2} $

Is this correct?

probability

share|cite|improve this question

edited 9 hours ago

cmk

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asked 9 hours ago

atnatn

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New contributor

atn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes you are right.
  $endgroup$
  – Mostafa Ayaz
  9 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This is my thought process:

$P(2^{nd} text{black}) = P(2^{nd} text{black} mid 1^{st} text{red}) P(1^{st} text{red}) + P(2^{nd} text{black} mid 1^{st} text{black}) P(1^{st} text{black}) $

$frac{26}{51}*frac{26}{52} + frac{25}{51}*frac{26}{52} = frac{1}{2} $

Is this correct?

probability

share|cite|improve this question

edited 9 hours ago

cmk

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asked 9 hours ago

atnatn

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New contributor

atn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

share|cite|improve this question

edited 9 hours ago

cmk

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asked 9 hours ago

atnatn

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share|cite|improve this question

edited 9 hours ago

cmk

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asked 9 hours ago

atnatn

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edited 9 hours ago

cmk

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edited 9 hours ago

cmk

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asked 9 hours ago

atnatn

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asked 9 hours ago

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3 Answers
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2

$begingroup$

Yes, it is correct.

Sanity check: By symmetry, the likelihood to get a black is equal to the probability og getting a red at any draw position.

share|cite|improve this answer

answered 9 hours ago

Siong Thye GohSiong Thye Goh

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "by symmetry"?
  $endgroup$
  – atn
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  propotion of black and red are equal. It is like two equally competent player are playing a fair game, they are equally likely to win the game.
  $endgroup$
  – Siong Thye Goh
  9 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Yes you are right.

The intuition behind why your answer is symmetrically ${1over 2}$ is that if you first draw a card and close your eyes over its color (e.g. throw it away), you are in a full ambiguity or lack of information about the color of the second one. Hence the probability is the same.

This is also the same case for any number of color types and drawing the second card.

share|cite|improve this answer

answered 9 hours ago

Mostafa AyazMostafa Ayaz

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$endgroup$

add a comment | 

1

$begingroup$

The "hard way":

To get a black card on the second draw you must get either "red, black" or "black, black".

1) "red, black": There are 52 cards in the deck, 26 or them red. The probability the first card drawn is red is 26/52= 1/2. Given that, there are now 51 cards in the deck, 26 of them black. The probability the second card drawn is black is 26/51. The probability of "red, black" is (1/2)(26/51)= 13/51.

2) "black, black: There are 52 cards in the deck, 26 or them black. The probability the first card drawn is red is 26/52= 1/2. Given that, there are now 51 cards in the deck, 25 of them black. The probability the second card drawn is black is 25/51. The probability of "red, black" is (1/2)(25/51)= 25/102.

The probability of either "red, black" or "black, black", since those are mutually exclusive, is 13/51+ 25/102= 26/102+ 25/102= 51/102= 1/2!

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answered 8 hours ago

user247327user247327

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2

$begingroup$

Yes, it is correct.

Sanity check: By symmetry, the likelihood to get a black is equal to the probability og getting a red at any draw position.

share|cite|improve this answer

answered 9 hours ago

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "by symmetry"?
  $endgroup$
  – atn
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  propotion of black and red are equal. It is like two equally competent player are playing a fair game, they are equally likely to win the game.
  $endgroup$
  – Siong Thye Goh
  9 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Yes, it is correct.

Sanity check: By symmetry, the likelihood to get a black is equal to the probability og getting a red at any draw position.

share|cite|improve this answer

answered 9 hours ago

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "by symmetry"?
  $endgroup$
  – atn
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  propotion of black and red are equal. It is like two equally competent player are playing a fair game, they are equally likely to win the game.
  $endgroup$
  – Siong Thye Goh
  9 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Yes, it is correct.

Sanity check: By symmetry, the likelihood to get a black is equal to the probability og getting a red at any draw position.

share|cite|improve this answer

answered 9 hours ago

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

answered 9 hours ago

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

answered 9 hours ago

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

2

$begingroup$

Yes you are right.

The intuition behind why your answer is symmetrically ${1over 2}$ is that if you first draw a card and close your eyes over its color (e.g. throw it away), you are in a full ambiguity or lack of information about the color of the second one. Hence the probability is the same.

This is also the same case for any number of color types and drawing the second card.

share|cite|improve this answer

answered 9 hours ago

Mostafa AyazMostafa Ayaz

20.4k33 gold badges1111 silver badges4646 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Yes you are right.

The intuition behind why your answer is symmetrically ${1over 2}$ is that if you first draw a card and close your eyes over its color (e.g. throw it away), you are in a full ambiguity or lack of information about the color of the second one. Hence the probability is the same.

This is also the same case for any number of color types and drawing the second card.

share|cite|improve this answer

answered 9 hours ago

Mostafa AyazMostafa Ayaz

20.4k33 gold badges1111 silver badges4646 bronze badges

$endgroup$

add a comment | 

$begingroup$

Yes you are right.

The intuition behind why your answer is symmetrically ${1over 2}$ is that if you first draw a card and close your eyes over its color (e.g. throw it away), you are in a full ambiguity or lack of information about the color of the second one. Hence the probability is the same.

This is also the same case for any number of color types and drawing the second card.

share|cite|improve this answer

answered 9 hours ago

Mostafa AyazMostafa Ayaz

20.4k33 gold badges1111 silver badges4646 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Mostafa AyazMostafa Ayaz

20.4k33 gold badges1111 silver badges4646 bronze badges

answered 9 hours ago

Mostafa AyazMostafa Ayaz

20.4k33 gold badges1111 silver badges4646 bronze badges

answered 9 hours ago

Mostafa AyazMostafa Ayaz

20.4k33 gold badges1111 silver badges4646 bronze badges

1

$begingroup$

The "hard way":

To get a black card on the second draw you must get either "red, black" or "black, black".

1) "red, black": There are 52 cards in the deck, 26 or them red. The probability the first card drawn is red is 26/52= 1/2. Given that, there are now 51 cards in the deck, 26 of them black. The probability the second card drawn is black is 26/51. The probability of "red, black" is (1/2)(26/51)= 13/51.

2) "black, black: There are 52 cards in the deck, 26 or them black. The probability the first card drawn is red is 26/52= 1/2. Given that, there are now 51 cards in the deck, 25 of them black. The probability the second card drawn is black is 25/51. The probability of "red, black" is (1/2)(25/51)= 25/102.

The probability of either "red, black" or "black, black", since those are mutually exclusive, is 13/51+ 25/102= 26/102+ 25/102= 51/102= 1/2!

share|cite|improve this answer

answered 8 hours ago

user247327user247327

12.7k22 gold badges77 silver badges1717 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

The "hard way":

To get a black card on the second draw you must get either "red, black" or "black, black".

1) "red, black": There are 52 cards in the deck, 26 or them red. The probability the first card drawn is red is 26/52= 1/2. Given that, there are now 51 cards in the deck, 26 of them black. The probability the second card drawn is black is 26/51. The probability of "red, black" is (1/2)(26/51)= 13/51.

2) "black, black: There are 52 cards in the deck, 26 or them black. The probability the first card drawn is red is 26/52= 1/2. Given that, there are now 51 cards in the deck, 25 of them black. The probability the second card drawn is black is 25/51. The probability of "red, black" is (1/2)(25/51)= 25/102.

The probability of either "red, black" or "black, black", since those are mutually exclusive, is 13/51+ 25/102= 26/102+ 25/102= 51/102= 1/2!

share|cite|improve this answer

answered 8 hours ago

user247327user247327

12.7k22 gold badges77 silver badges1717 bronze badges

$endgroup$

add a comment | 

$begingroup$

The "hard way":

To get a black card on the second draw you must get either "red, black" or "black, black".

1) "red, black": There are 52 cards in the deck, 26 or them red. The probability the first card drawn is red is 26/52= 1/2. Given that, there are now 51 cards in the deck, 26 of them black. The probability the second card drawn is black is 26/51. The probability of "red, black" is (1/2)(26/51)= 13/51.

2) "black, black: There are 52 cards in the deck, 26 or them black. The probability the first card drawn is red is 26/52= 1/2. Given that, there are now 51 cards in the deck, 25 of them black. The probability the second card drawn is black is 25/51. The probability of "red, black" is (1/2)(25/51)= 25/102.

The probability of either "red, black" or "black, black", since those are mutually exclusive, is 13/51+ 25/102= 26/102+ 25/102= 51/102= 1/2!

share|cite|improve this answer

answered 8 hours ago

user247327user247327

12.7k22 gold badges77 silver badges1717 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

user247327user247327

12.7k22 gold badges77 silver badges1717 bronze badges

answered 8 hours ago

user247327user247327

12.7k22 gold badges77 silver badges1717 bronze badges

answered 8 hours ago

user247327user247327

12.7k22 gold badges77 silver badges1717 bronze badges