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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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 textblack) = P(2^nd textblack mid 1^st textred) P(1^st textred) + P(2^nd textblack mid 1^st textblack) P(1^st textblack) $
$frac2651*frac2652 + frac2551*frac2652 = frac12 $
Is this correct?
probability
edited 9 hours ago
cmk
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add a comment |
$begingroup$
This is my thought process:
$P(2^nd textblack) = P(2^nd textblack mid 1^st textred) P(1^st textred) + P(2^nd textblack mid 1^st textblack) P(1^st textblack) $
$frac2651*frac2652 + frac2551*frac2652 = frac12 $
Is this correct?
probability
edited 9 hours ago
cmk
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asked 9 hours ago
atnatn
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1
$begingroup$
Yes you are right.
$endgroup$
– Mostafa Ayaz
9 hours ago
add a comment |
$begingroup$
This is my thought process:
$P(2^nd textblack) = P(2^nd textblack mid 1^st textred) P(1^st textred) + P(2^nd textblack mid 1^st textblack) P(1^st textblack) $
$frac2651*frac2652 + frac2551*frac2652 = frac12 $
Is this correct?
probability
edited 9 hours ago
cmk
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asked 9 hours ago
atnatn
412 bronze badges
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atn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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This is my thought process:
$P(2^nd textblack) = P(2^nd textblack mid 1^st textred) P(1^st textred) + P(2^nd textblack mid 1^st textblack) P(1^st textblack) $
$frac2651*frac2652 + frac2551*frac2652 = frac12 $
Is this correct?
probability
probability
edited 9 hours ago
cmk
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edited 9 hours ago
cmk
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edited 9 hours ago
cmk
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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
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asked 9 hours ago
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1
$begingroup$
Yes you are right.
$endgroup$
– Mostafa Ayaz
9 hours ago
add a comment |
1
$begingroup$
Yes you are right.
$endgroup$
– Mostafa Ayaz
9 hours ago
1
1
$begingroup$
Yes you are right.
$endgroup$
– Mostafa Ayaz
9 hours ago
$begingroup$
Yes you are right.
$endgroup$
– Mostafa Ayaz
9 hours ago
add a comment |
4 Answers
4
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$begingroup$
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.
answered 9 hours ago
Siong Thye GohSiong Thye Goh
110k15 gold badges71 silver badges125 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 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.
answered 9 hours ago
Mostafa AyazMostafa Ayaz
20.4k3 gold badges11 silver badges46 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!
answered 8 hours ago
user247327user247327
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add a comment |
$begingroup$
You are correct... unless the colour of the first card is known, in which case it would be either:
$dfrac2551$ if the first card was black, or
$dfrac2651$ if the first card was red
edited 7 mins ago
Lee David Chung Lin
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 9 hours ago
Siong Thye GohSiong Thye Goh
110k15 gold badges71 silver badges125 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.
answered 9 hours ago
Siong Thye GohSiong Thye Goh
110k15 gold badges71 silver badges125 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.
answered 9 hours ago
Siong Thye GohSiong Thye Goh
110k15 gold badges71 silver badges125 bronze badges
$endgroup$
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.
answered 9 hours ago
Siong Thye GohSiong Thye Goh
110k15 gold badges71 silver badges125 bronze badges
answered 9 hours ago
Siong Thye GohSiong Thye Goh
110k15 gold badges71 silver badges125 bronze badges
answered 9 hours ago
Siong Thye GohSiong Thye Goh
110k15 gold badges71 silver badges125 bronze badges
answered 9 hours ago
Siong Thye GohSiong Thye Goh
110k15 gold badges71 silver badges125 bronze badges
110k15 gold badges71 silver badges125 bronze badges
$begingroup$
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 |
$begingroup$
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
$begingroup$
What do you mean by "by symmetry"?
$endgroup$
– atn
9 hours ago
$begingroup$
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
$begingroup$
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 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.
answered 9 hours ago
Mostafa AyazMostafa Ayaz
20.4k3 gold badges11 silver badges46 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.
answered 9 hours ago
Mostafa AyazMostafa Ayaz
20.4k3 gold badges11 silver badges46 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.
answered 9 hours ago
Mostafa AyazMostafa Ayaz
20.4k3 gold badges11 silver badges46 bronze badges
$endgroup$
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.
answered 9 hours ago
Mostafa AyazMostafa Ayaz
20.4k3 gold badges11 silver badges46 bronze badges
answered 9 hours ago
Mostafa AyazMostafa Ayaz
20.4k3 gold badges11 silver badges46 bronze badges
answered 9 hours ago
Mostafa AyazMostafa Ayaz
20.4k3 gold badges11 silver badges46 bronze badges
answered 9 hours ago
Mostafa AyazMostafa Ayaz
20.4k3 gold badges11 silver badges46 bronze badges
20.4k3 gold badges11 silver badges46 bronze badges
add a comment |
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!
answered 8 hours ago
user247327user247327
12.7k2 gold badges7 silver badges17 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!
answered 8 hours ago
user247327user247327
12.7k2 gold badges7 silver badges17 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!
answered 8 hours ago
user247327user247327
12.7k2 gold badges7 silver badges17 bronze badges
$endgroup$
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!
answered 8 hours ago
user247327user247327
12.7k2 gold badges7 silver badges17 bronze badges
answered 8 hours ago
user247327user247327
12.7k2 gold badges7 silver badges17 bronze badges
answered 8 hours ago
user247327user247327
12.7k2 gold badges7 silver badges17 bronze badges
answered 8 hours ago
user247327user247327
12.7k2 gold badges7 silver badges17 bronze badges
12.7k2 gold badges7 silver badges17 bronze badges
add a comment |
add a comment |
$begingroup$
You are correct... unless the colour of the first card is known, in which case it would be either:
$dfrac2551$ if the first card was black, or
$dfrac2651$ if the first card was red
edited 7 mins ago
Lee David Chung Lin
4,6315 gold badges15 silver badges43 bronze badges
answered 11 mins ago
spadelivesspadelives
1011 bronze badge
New contributor
spadelives is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
You are correct... unless the colour of the first card is known, in which case it would be either:
$dfrac2551$ if the first card was black, or
$dfrac2651$ if the first card was red
edited 7 mins ago
Lee David Chung Lin
4,6315 gold badges15 silver badges43 bronze badges
answered 11 mins ago
spadelivesspadelives
1011 bronze badge
New contributor
spadelives is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
You are correct... unless the colour of the first card is known, in which case it would be either:
$dfrac2551$ if the first card was black, or
$dfrac2651$ if the first card was red
edited 7 mins ago
Lee David Chung Lin
4,6315 gold badges15 silver badges43 bronze badges
answered 11 mins ago
spadelivesspadelives
1011 bronze badge
New contributor
spadelives is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You are correct... unless the colour of the first card is known, in which case it would be either:
$dfrac2551$ if the first card was black, or
$dfrac2651$ if the first card was red
edited 7 mins ago
Lee David Chung Lin
4,6315 gold badges15 silver badges43 bronze badges
answered 11 mins ago
spadelivesspadelives
1011 bronze badge
New contributor
spadelives is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 mins ago
Lee David Chung Lin
4,6315 gold badges15 silver badges43 bronze badges
edited 7 mins ago
Lee David Chung Lin
4,6315 gold badges15 silver badges43 bronze badges
edited 7 mins ago
Lee David Chung Lin
4,6315 gold badges15 silver badges43 bronze badges
4,6315 gold badges15 silver badges43 bronze badges
answered 11 mins ago
spadelivesspadelives
1011 bronze badge
New contributor
spadelives is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 11 mins ago
spadelivesspadelives
1011 bronze badge
answered 11 mins ago
spadelivesspadelives
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1011 bronze badge
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add a comment |
add a comment |
atn is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Yes you are right.
$endgroup$
– Mostafa Ayaz
9 hours ago