Mixing
What is the probability of selecting $2$ red queens when $5$ cards are selected from the $12$ face cards in a...
$begingroup$
I have $12$ face cards ($4$ kings, $4$ queens, $4$ jacks). What is the probability that I’ll pick $2$ red queens if $5$ cards are chosen?
My answer is $frac{binom{10}{3}}{binom{12}{5}} = frac{5}{33}$, since the number of ways to choose the other 3 cards once the 2 red queens are already chosen is $binom{10}{3}$, and the number of ways to choose the 5 cards is $binom{12}{5}$.
I seem to be missing out something here, particularly on the part when choosing the $2$ red queens.
Can anyone help me point out my mistake? Thank you!
probability combinatorics
edited Jan 5 at 9:54
N. F. Taussig
44k93355
asked Jan 5 at 8:35
Jonelle YuJonelle Yu
1776
$endgroup$
add a comment |
$begingroup$
I have $12$ face cards ($4$ kings, $4$ queens, $4$ jacks). What is the probability that I’ll pick $2$ red queens if $5$ cards are chosen?
My answer is $frac{binom{10}{3}}{binom{12}{5}} = frac{5}{33}$, since the number of ways to choose the other 3 cards once the 2 red queens are already chosen is $binom{10}{3}$, and the number of ways to choose the 5 cards is $binom{12}{5}$.
I seem to be missing out something here, particularly on the part when choosing the $2$ red queens.
Can anyone help me point out my mistake? Thank you!
probability combinatorics
edited Jan 5 at 9:54
N. F. Taussig
44k93355
asked Jan 5 at 8:35
Jonelle YuJonelle Yu
1776
$endgroup$
add a comment |
$begingroup$
I have $12$ face cards ($4$ kings, $4$ queens, $4$ jacks). What is the probability that I’ll pick $2$ red queens if $5$ cards are chosen?
My answer is $frac{binom{10}{3}}{binom{12}{5}} = frac{5}{33}$, since the number of ways to choose the other 3 cards once the 2 red queens are already chosen is $binom{10}{3}$, and the number of ways to choose the 5 cards is $binom{12}{5}$.
I seem to be missing out something here, particularly on the part when choosing the $2$ red queens.
Can anyone help me point out my mistake? Thank you!
probability combinatorics
edited Jan 5 at 9:54
N. F. Taussig
44k93355
asked Jan 5 at 8:35
Jonelle YuJonelle Yu
1776
$endgroup$
I have $12$ face cards ($4$ kings, $4$ queens, $4$ jacks). What is the probability that I’ll pick $2$ red queens if $5$ cards are chosen?
My answer is $frac{binom{10}{3}}{binom{12}{5}} = frac{5}{33}$, since the number of ways to choose the other 3 cards once the 2 red queens are already chosen is $binom{10}{3}$, and the number of ways to choose the 5 cards is $binom{12}{5}$.
I seem to be missing out something here, particularly on the part when choosing the $2$ red queens.
Can anyone help me point out my mistake? Thank you!
probability combinatorics
probability combinatorics
edited Jan 5 at 9:54
N. F. Taussig
44k93355
asked Jan 5 at 8:35
Jonelle YuJonelle Yu
1776
edited Jan 5 at 9:54
N. F. Taussig
44k93355
asked Jan 5 at 8:35
Jonelle YuJonelle Yu
1776
edited Jan 5 at 9:54
N. F. Taussig
44k93355
edited Jan 5 at 9:54
N. F. Taussig
44k93355
edited Jan 5 at 9:54
N. F. Taussig
44k93355
44k93355
asked Jan 5 at 8:35
Jonelle YuJonelle Yu
1776
asked Jan 5 at 8:35
Jonelle YuJonelle Yu
1776
asked Jan 5 at 8:35
Jonelle YuJonelle Yu
1776
1776
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your answer is okay and applies hypergeometric distribution:$$frac{binom22binom{10}3}{binom{12}5}$$
Maybe the expression $binom22$ is what you are missing in your own approach.
Fortunately that does not harm here because it equals $1$ hence is neutral wrt multiplication.
answered Jan 5 at 9:49
drhabdrhab
99.5k544130
$endgroup$
add a comment |
$begingroup$
Hint: Try to answer the following questions.
How many events you have?
How many of them are ''good''?
Yes, your answer is correct.
answered Jan 5 at 8:37
greedoidgreedoid
39.2k114797
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
Your answer is okay and applies hypergeometric distribution:$$frac{binom22binom{10}3}{binom{12}5}$$
Maybe the expression $binom22$ is what you are missing in your own approach.
Fortunately that does not harm here because it equals $1$ hence is neutral wrt multiplication.
answered Jan 5 at 9:49
drhabdrhab
99.5k544130
$endgroup$
add a comment |
$begingroup$
Your answer is okay and applies hypergeometric distribution:$$frac{binom22binom{10}3}{binom{12}5}$$
Maybe the expression $binom22$ is what you are missing in your own approach.
Fortunately that does not harm here because it equals $1$ hence is neutral wrt multiplication.
answered Jan 5 at 9:49
drhabdrhab
99.5k544130
$endgroup$
add a comment |
$begingroup$
Your answer is okay and applies hypergeometric distribution:$$frac{binom22binom{10}3}{binom{12}5}$$
Maybe the expression $binom22$ is what you are missing in your own approach.
Fortunately that does not harm here because it equals $1$ hence is neutral wrt multiplication.
answered Jan 5 at 9:49
drhabdrhab
99.5k544130
$endgroup$
Your answer is okay and applies hypergeometric distribution:$$frac{binom22binom{10}3}{binom{12}5}$$
Maybe the expression $binom22$ is what you are missing in your own approach.
Fortunately that does not harm here because it equals $1$ hence is neutral wrt multiplication.
answered Jan 5 at 9:49
drhabdrhab
99.5k544130
answered Jan 5 at 9:49
drhabdrhab
99.5k544130
answered Jan 5 at 9:49
drhabdrhab
99.5k544130
answered Jan 5 at 9:49
drhabdrhab
99.5k544130
99.5k544130
add a comment |
add a comment |
$begingroup$
Hint: Try to answer the following questions.
How many events you have?
How many of them are ''good''?
Yes, your answer is correct.
answered Jan 5 at 8:37
greedoidgreedoid
39.2k114797
$endgroup$
add a comment |
$begingroup$
Hint: Try to answer the following questions.
How many events you have?
How many of them are ''good''?
Yes, your answer is correct.
answered Jan 5 at 8:37
greedoidgreedoid
39.2k114797
$endgroup$
add a comment |
$begingroup$
Hint: Try to answer the following questions.
How many events you have?
How many of them are ''good''?
Yes, your answer is correct.
answered Jan 5 at 8:37
greedoidgreedoid
39.2k114797
$endgroup$
Hint: Try to answer the following questions.
How many events you have?
How many of them are ''good''?
Yes, your answer is correct.
answered Jan 5 at 8:37
greedoidgreedoid
39.2k114797
edited Jan 5 at 9:32
answered Jan 5 at 8:37
greedoidgreedoid
39.2k114797
answered Jan 5 at 8:37
greedoidgreedoid
39.2k114797
answered Jan 5 at 8:37
greedoidgreedoid
39.2k114797
39.2k114797
add a comment |
add a comment |
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