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
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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
$begingroup$
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
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
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
asked 1 hour ago
BadAtGeometryBadAtGeometry
291317
$endgroup$
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
probability card-games
asked 1 hour ago
BadAtGeometryBadAtGeometry
291317
asked 1 hour ago
BadAtGeometryBadAtGeometry
291317
asked 1 hour ago
BadAtGeometryBadAtGeometry
291317
asked 1 hour ago
BadAtGeometryBadAtGeometry
291317
asked 1 hour ago
BadAtGeometryBadAtGeometry
291317
291317
add a comment |
add a comment |
2 Answers
2
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$begingroup$
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.
answered 58 mins ago
ServaesServaes
32.9k444102
$endgroup$
2
$begingroup$
It's $binom 62=15$ ways of drawing only black.
$endgroup$
– Arthur
56 mins ago
$begingroup$
@Arthur Of course, you are right. For some reason I was thinking of four suits in stead of two colors.
$endgroup$
– Servaes
19 mins ago
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).
answered 1 hour ago
angryavianangryavian
43.9k33484
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 58 mins ago
ServaesServaes
32.9k444102
$endgroup$
2
$begingroup$
It's $binom 62=15$ ways of drawing only black.
$endgroup$
– Arthur
56 mins ago
$begingroup$
@Arthur Of course, you are right. For some reason I was thinking of four suits in stead of two colors.
$endgroup$
– Servaes
19 mins 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.
answered 58 mins ago
ServaesServaes
32.9k444102
$endgroup$
2
$begingroup$
It's $binom 62=15$ ways of drawing only black.
$endgroup$
– Arthur
56 mins ago
$begingroup$
@Arthur Of course, you are right. For some reason I was thinking of four suits in stead of two colors.
$endgroup$
– Servaes
19 mins 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.
answered 58 mins ago
ServaesServaes
32.9k444102
$endgroup$
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.
answered 58 mins ago
ServaesServaes
32.9k444102
edited 20 mins ago
answered 58 mins ago
ServaesServaes
32.9k444102
answered 58 mins ago
ServaesServaes
32.9k444102
answered 58 mins ago
ServaesServaes
32.9k444102
32.9k444102
2
$begingroup$
It's $binom 62=15$ ways of drawing only black.
$endgroup$
– Arthur
56 mins ago
$begingroup$
@Arthur Of course, you are right. For some reason I was thinking of four suits in stead of two colors.
$endgroup$
– Servaes
19 mins ago
add a comment |
2
$begingroup$
It's $binom 62=15$ ways of drawing only black.
$endgroup$
– Arthur
56 mins ago
$begingroup$
@Arthur Of course, you are right. For some reason I was thinking of four suits in stead of two colors.
$endgroup$
– Servaes
19 mins ago
2
2
$begingroup$
It's $binom 62=15$ ways of drawing only black.
$endgroup$
– Arthur
56 mins ago
$begingroup$
It's $binom 62=15$ ways of drawing only black.
$endgroup$
– Arthur
56 mins ago
$begingroup$
@Arthur Of course, you are right. For some reason I was thinking of four suits in stead of two colors.
$endgroup$
– Servaes
19 mins ago
$begingroup$
@Arthur Of course, you are right. For some reason I was thinking of four suits in stead of two colors.
$endgroup$
– Servaes
19 mins ago
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).
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).
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).
answered 1 hour ago
angryavianangryavian
43.9k33484
$endgroup$
$3/4$ 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).
answered 1 hour ago
angryavianangryavian
43.9k33484
answered 1 hour ago
angryavianangryavian
43.9k33484
answered 1 hour ago
angryavianangryavian
43.9k33484
answered 1 hour ago
angryavianangryavian
43.9k33484
43.9k33484
add a comment |
add a comment |
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