52 cards are equally given to 4 players Find probability that one of them has 3 spades out of remaining 5.
$begingroup$
52 cards are equally given to 4 players $A,B,C,D$.
Together, $A$ and $B$ have a total of 8 spades among them,
and we have to find probability that $C$ has 3 spades out if remaining 5.
In this question, the sample space reduces to 44 cards after we eliminate 8 spades from the cards, so now we have 5 spades left from which 3 will be given to $C$, and the remaining 10 cards can be chosen from the remaining 39 cards so I get the answer as
$$ frac{binom{5}{3}binom{39}{10}}{binom{44}{13}},$$
is this approach correct?
probability
edited Dec 26 '15 at 4:21
Alex Provost
15.5k22350
asked Dec 26 '15 at 4:06
radhikaradhika
141313
$endgroup$
add a comment |
$begingroup$
52 cards are equally given to 4 players $A,B,C,D$.
Together, $A$ and $B$ have a total of 8 spades among them,
and we have to find probability that $C$ has 3 spades out if remaining 5.
In this question, the sample space reduces to 44 cards after we eliminate 8 spades from the cards, so now we have 5 spades left from which 3 will be given to $C$, and the remaining 10 cards can be chosen from the remaining 39 cards so I get the answer as
$$ frac{binom{5}{3}binom{39}{10}}{binom{44}{13}},$$
is this approach correct?
probability
edited Dec 26 '15 at 4:21
Alex Provost
15.5k22350
asked Dec 26 '15 at 4:06
radhikaradhika
141313
$endgroup$
$begingroup$
Do you mean exactly three or at least three? (if the latter, the answer is $frac 12$ as either $C$ or $D$, but not both, must have at least three).
$endgroup$
– lulu
Dec 26 '15 at 4:11
$begingroup$
Aside from the issue of "exactly" vs "at least", your method is flawed. There aren't $44$ cards left to choose from, there are only $26$ of which $5$ are Spades and $21$ are Other. Also, having chosen your three Spades, you must then choose the remaining $10$ cards from the Other (else you count the same combinations multiple times).
$endgroup$
– lulu
Dec 26 '15 at 4:19
add a comment |
$begingroup$
52 cards are equally given to 4 players $A,B,C,D$.
Together, $A$ and $B$ have a total of 8 spades among them,
and we have to find probability that $C$ has 3 spades out if remaining 5.
In this question, the sample space reduces to 44 cards after we eliminate 8 spades from the cards, so now we have 5 spades left from which 3 will be given to $C$, and the remaining 10 cards can be chosen from the remaining 39 cards so I get the answer as
$$ frac{binom{5}{3}binom{39}{10}}{binom{44}{13}},$$
is this approach correct?
probability
edited Dec 26 '15 at 4:21
Alex Provost
15.5k22350
asked Dec 26 '15 at 4:06
radhikaradhika
141313
$endgroup$
52 cards are equally given to 4 players $A,B,C,D$.
Together, $A$ and $B$ have a total of 8 spades among them,
and we have to find probability that $C$ has 3 spades out if remaining 5.
In this question, the sample space reduces to 44 cards after we eliminate 8 spades from the cards, so now we have 5 spades left from which 3 will be given to $C$, and the remaining 10 cards can be chosen from the remaining 39 cards so I get the answer as
$$ frac{binom{5}{3}binom{39}{10}}{binom{44}{13}},$$
is this approach correct?
probability
probability
edited Dec 26 '15 at 4:21
Alex Provost
15.5k22350
asked Dec 26 '15 at 4:06
radhikaradhika
141313
edited Dec 26 '15 at 4:21
Alex Provost
15.5k22350
asked Dec 26 '15 at 4:06
radhikaradhika
141313
edited Dec 26 '15 at 4:21
Alex Provost
15.5k22350
edited Dec 26 '15 at 4:21
Alex Provost
15.5k22350
edited Dec 26 '15 at 4:21
Alex Provost
15.5k22350
15.5k22350
asked Dec 26 '15 at 4:06
radhikaradhika
141313
asked Dec 26 '15 at 4:06
radhikaradhika
141313
asked Dec 26 '15 at 4:06
radhikaradhika
141313
141313
$begingroup$
Do you mean exactly three or at least three? (if the latter, the answer is $frac 12$ as either $C$ or $D$, but not both, must have at least three).
$endgroup$
– lulu
Dec 26 '15 at 4:11
$begingroup$
Aside from the issue of "exactly" vs "at least", your method is flawed. There aren't $44$ cards left to choose from, there are only $26$ of which $5$ are Spades and $21$ are Other. Also, having chosen your three Spades, you must then choose the remaining $10$ cards from the Other (else you count the same combinations multiple times).
$endgroup$
– lulu
Dec 26 '15 at 4:19
add a comment |
$begingroup$
Do you mean exactly three or at least three? (if the latter, the answer is $frac 12$ as either $C$ or $D$, but not both, must have at least three).
$endgroup$
– lulu
Dec 26 '15 at 4:11
$begingroup$
Aside from the issue of "exactly" vs "at least", your method is flawed. There aren't $44$ cards left to choose from, there are only $26$ of which $5$ are Spades and $21$ are Other. Also, having chosen your three Spades, you must then choose the remaining $10$ cards from the Other (else you count the same combinations multiple times).
$endgroup$
– lulu
Dec 26 '15 at 4:19
$begingroup$
Do you mean exactly three or at least three? (if the latter, the answer is $frac 12$ as either $C$ or $D$, but not both, must have at least three).
$endgroup$
– lulu
Dec 26 '15 at 4:11
$begingroup$
Do you mean exactly three or at least three? (if the latter, the answer is $frac 12$ as either $C$ or $D$, but not both, must have at least three).
$endgroup$
– lulu
Dec 26 '15 at 4:11
$begingroup$
Aside from the issue of "exactly" vs "at least", your method is flawed. There aren't $44$ cards left to choose from, there are only $26$ of which $5$ are Spades and $21$ are Other. Also, having chosen your three Spades, you must then choose the remaining $10$ cards from the Other (else you count the same combinations multiple times).
$endgroup$
– lulu
Dec 26 '15 at 4:19
$begingroup$
Aside from the issue of "exactly" vs "at least", your method is flawed. There aren't $44$ cards left to choose from, there are only $26$ of which $5$ are Spades and $21$ are Other. Also, having chosen your three Spades, you must then choose the remaining $10$ cards from the Other (else you count the same combinations multiple times).
$endgroup$
– lulu
Dec 26 '15 at 4:19
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There are only 26 cards left in the deck. 5 are spades, 21 are not. We use the $frac{Number- of -favorable -outcomes}{Number- of- possible- outcomes}$ formula. Thus we get $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$.
answered Dec 26 '15 at 4:35
MaffredMaffred
2,680625
$endgroup$
add a comment |
$begingroup$
Question: In the card game bridge, the 52 cards are dealt out equally to 4 players - called East, West, North and South. If North and South have total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?
My Approach: My answer to the above question was:
East can get 3 spades out of remaining 5 spades in $binom{5}{3}$ ways.
All possible outcomes are $binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}$. i.e., East gets no spades, or 1 spade or... and so on.
Therefore, the desired probability = $frac{binom{5}{3}}{binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}}= frac{10}{32} = .3125$
However, as per Sheldon Ross' book, the answer is 0.339. I understand how he arrived at the solution. But, I do not see what is wrong in the above approach. Why don't the answers match?
answered Jul 9 '16 at 20:11
Venkatesh VinayakaraoVenkatesh Vinayakarao
4316
$endgroup$
$begingroup$
Quote: "3 of the remaining 5 spades" That seems wrong. It should be "3 of the remaining 8 spades", I think. So your calculation shall use 8 instead of 5
$endgroup$
– 4386427
Sep 15 '18 at 8:24
add a comment |
$begingroup$
It is given that A and B have a total of 8 spades among 26 cards.
∴ In the remaining 26 cards, there are exactly 5 spades.
These 26 cards are distributed equally among C and D [13 each]
P(C has 3 of 5 spades) = $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$
= 0.339
answered Dec 23 '18 at 17:45
shivam guptashivam gupta
1
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are only 26 cards left in the deck. 5 are spades, 21 are not. We use the $frac{Number- of -favorable -outcomes}{Number- of- possible- outcomes}$ formula. Thus we get $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$.
answered Dec 26 '15 at 4:35
MaffredMaffred
2,680625
$endgroup$
add a comment |
$begingroup$
There are only 26 cards left in the deck. 5 are spades, 21 are not. We use the $frac{Number- of -favorable -outcomes}{Number- of- possible- outcomes}$ formula. Thus we get $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$.
answered Dec 26 '15 at 4:35
MaffredMaffred
2,680625
$endgroup$
add a comment |
$begingroup$
There are only 26 cards left in the deck. 5 are spades, 21 are not. We use the $frac{Number- of -favorable -outcomes}{Number- of- possible- outcomes}$ formula. Thus we get $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$.
answered Dec 26 '15 at 4:35
MaffredMaffred
2,680625
$endgroup$
There are only 26 cards left in the deck. 5 are spades, 21 are not. We use the $frac{Number- of -favorable -outcomes}{Number- of- possible- outcomes}$ formula. Thus we get $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$.
answered Dec 26 '15 at 4:35
MaffredMaffred
2,680625
edited Dec 26 '15 at 5:14
answered Dec 26 '15 at 4:35
MaffredMaffred
2,680625
answered Dec 26 '15 at 4:35
MaffredMaffred
2,680625
answered Dec 26 '15 at 4:35
MaffredMaffred
2,680625
2,680625
add a comment |
add a comment |
$begingroup$
Question: In the card game bridge, the 52 cards are dealt out equally to 4 players - called East, West, North and South. If North and South have total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?
My Approach: My answer to the above question was:
East can get 3 spades out of remaining 5 spades in $binom{5}{3}$ ways.
All possible outcomes are $binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}$. i.e., East gets no spades, or 1 spade or... and so on.
Therefore, the desired probability = $frac{binom{5}{3}}{binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}}= frac{10}{32} = .3125$
However, as per Sheldon Ross' book, the answer is 0.339. I understand how he arrived at the solution. But, I do not see what is wrong in the above approach. Why don't the answers match?
answered Jul 9 '16 at 20:11
Venkatesh VinayakaraoVenkatesh Vinayakarao
4316
$endgroup$
$begingroup$
Quote: "3 of the remaining 5 spades" That seems wrong. It should be "3 of the remaining 8 spades", I think. So your calculation shall use 8 instead of 5
$endgroup$
– 4386427
Sep 15 '18 at 8:24
add a comment |
$begingroup$
Question: In the card game bridge, the 52 cards are dealt out equally to 4 players - called East, West, North and South. If North and South have total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?
My Approach: My answer to the above question was:
East can get 3 spades out of remaining 5 spades in $binom{5}{3}$ ways.
All possible outcomes are $binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}$. i.e., East gets no spades, or 1 spade or... and so on.
Therefore, the desired probability = $frac{binom{5}{3}}{binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}}= frac{10}{32} = .3125$
However, as per Sheldon Ross' book, the answer is 0.339. I understand how he arrived at the solution. But, I do not see what is wrong in the above approach. Why don't the answers match?
answered Jul 9 '16 at 20:11
Venkatesh VinayakaraoVenkatesh Vinayakarao
4316
$endgroup$
$begingroup$
Quote: "3 of the remaining 5 spades" That seems wrong. It should be "3 of the remaining 8 spades", I think. So your calculation shall use 8 instead of 5
$endgroup$
– 4386427
Sep 15 '18 at 8:24
add a comment |
$begingroup$
Question: In the card game bridge, the 52 cards are dealt out equally to 4 players - called East, West, North and South. If North and South have total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?
My Approach: My answer to the above question was:
East can get 3 spades out of remaining 5 spades in $binom{5}{3}$ ways.
All possible outcomes are $binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}$. i.e., East gets no spades, or 1 spade or... and so on.
Therefore, the desired probability = $frac{binom{5}{3}}{binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}}= frac{10}{32} = .3125$
However, as per Sheldon Ross' book, the answer is 0.339. I understand how he arrived at the solution. But, I do not see what is wrong in the above approach. Why don't the answers match?
answered Jul 9 '16 at 20:11
Venkatesh VinayakaraoVenkatesh Vinayakarao
4316
$endgroup$
Question: In the card game bridge, the 52 cards are dealt out equally to 4 players - called East, West, North and South. If North and South have total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?
My Approach: My answer to the above question was:
East can get 3 spades out of remaining 5 spades in $binom{5}{3}$ ways.
All possible outcomes are $binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}$. i.e., East gets no spades, or 1 spade or... and so on.
Therefore, the desired probability = $frac{binom{5}{3}}{binom{5}{0}+binom{5}{1}+binom{5}{2}+binom{5}{3}+binom{5}{4}+binom{5}{5}}= frac{10}{32} = .3125$
However, as per Sheldon Ross' book, the answer is 0.339. I understand how he arrived at the solution. But, I do not see what is wrong in the above approach. Why don't the answers match?
answered Jul 9 '16 at 20:11
Venkatesh VinayakaraoVenkatesh Vinayakarao
4316
answered Jul 9 '16 at 20:11
Venkatesh VinayakaraoVenkatesh Vinayakarao
4316
answered Jul 9 '16 at 20:11
Venkatesh VinayakaraoVenkatesh Vinayakarao
4316
answered Jul 9 '16 at 20:11
Venkatesh VinayakaraoVenkatesh Vinayakarao
4316
4316
$begingroup$
Quote: "3 of the remaining 5 spades" That seems wrong. It should be "3 of the remaining 8 spades", I think. So your calculation shall use 8 instead of 5
$endgroup$
– 4386427
Sep 15 '18 at 8:24
add a comment |
$begingroup$
Quote: "3 of the remaining 5 spades" That seems wrong. It should be "3 of the remaining 8 spades", I think. So your calculation shall use 8 instead of 5
$endgroup$
– 4386427
Sep 15 '18 at 8:24
$begingroup$
Quote: "3 of the remaining 5 spades" That seems wrong. It should be "3 of the remaining 8 spades", I think. So your calculation shall use 8 instead of 5
$endgroup$
– 4386427
Sep 15 '18 at 8:24
$begingroup$
Quote: "3 of the remaining 5 spades" That seems wrong. It should be "3 of the remaining 8 spades", I think. So your calculation shall use 8 instead of 5
$endgroup$
– 4386427
Sep 15 '18 at 8:24
add a comment |
$begingroup$
It is given that A and B have a total of 8 spades among 26 cards.
∴ In the remaining 26 cards, there are exactly 5 spades.
These 26 cards are distributed equally among C and D [13 each]
P(C has 3 of 5 spades) = $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$
= 0.339
answered Dec 23 '18 at 17:45
shivam guptashivam gupta
1
$endgroup$
add a comment |
$begingroup$
It is given that A and B have a total of 8 spades among 26 cards.
∴ In the remaining 26 cards, there are exactly 5 spades.
These 26 cards are distributed equally among C and D [13 each]
P(C has 3 of 5 spades) = $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$
= 0.339
answered Dec 23 '18 at 17:45
shivam guptashivam gupta
1
$endgroup$
add a comment |
$begingroup$
It is given that A and B have a total of 8 spades among 26 cards.
∴ In the remaining 26 cards, there are exactly 5 spades.
These 26 cards are distributed equally among C and D [13 each]
P(C has 3 of 5 spades) = $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$
= 0.339
answered Dec 23 '18 at 17:45
shivam guptashivam gupta
1
$endgroup$
It is given that A and B have a total of 8 spades among 26 cards.
∴ In the remaining 26 cards, there are exactly 5 spades.
These 26 cards are distributed equally among C and D [13 each]
P(C has 3 of 5 spades) = $frac{{5 choose 3}{21 choose 10}}{{26 choose 13}}$
= 0.339
answered Dec 23 '18 at 17:45
shivam guptashivam gupta
1
answered Dec 23 '18 at 17:45
shivam guptashivam gupta
1
answered Dec 23 '18 at 17:45
shivam guptashivam gupta
1
answered Dec 23 '18 at 17:45
shivam guptashivam gupta
1
1
add a comment |
add a comment |
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$begingroup$
Do you mean exactly three or at least three? (if the latter, the answer is $frac 12$ as either $C$ or $D$, but not both, must have at least three).
$endgroup$
– lulu
Dec 26 '15 at 4:11
$begingroup$
Aside from the issue of "exactly" vs "at least", your method is flawed. There aren't $44$ cards left to choose from, there are only $26$ of which $5$ are Spades and $21$ are Other. Also, having chosen your three Spades, you must then choose the remaining $10$ cards from the Other (else you count the same combinations multiple times).
$endgroup$
– lulu
Dec 26 '15 at 4:19