Given that the first two cards drawn are kings, what is the probability that the third is also a king?
$begingroup$
Four cards are randomly selected from a pack of 52 cards. If the first two cards are kings, what is the probability that the third card is a king?
The given answer is 2/50.
I was thinking 4 cards can be selected in 52C4 ways, out of which first three kings can be selected in 4P3 ways i.e. 24 ways and last 1 card can be choose by 49 ways. Hence result should be equal to (24*49)/(52C4)
Where am I going wrong?
probability
edited Dec 10 '18 at 17:48
Brian Tung
25.8k32554
asked Dec 10 '18 at 17:15
swapnilswapnil
335
$endgroup$
add a comment |
$begingroup$
Four cards are randomly selected from a pack of 52 cards. If the first two cards are kings, what is the probability that the third card is a king?
The given answer is 2/50.
I was thinking 4 cards can be selected in 52C4 ways, out of which first three kings can be selected in 4P3 ways i.e. 24 ways and last 1 card can be choose by 49 ways. Hence result should be equal to (24*49)/(52C4)
Where am I going wrong?
probability
edited Dec 10 '18 at 17:48
Brian Tung
25.8k32554
asked Dec 10 '18 at 17:15
swapnilswapnil
335
$endgroup$
$begingroup$
4P3 doesn't look correct as you're not using the fact that there were 2 kings already.
$endgroup$
– rsadhvika
Dec 10 '18 at 17:37
2
$begingroup$
The given answer is correct because you have $2$ kings left in a deck of $50$ cards. You seem to be trying to calculate the probability that the first three cards were kings. This problem is different because you are given that the first two cards are kings.
$endgroup$
– John Douma
Dec 10 '18 at 17:39
$begingroup$
Edited the title to match the body. (Sorry, I accidentally typed return and missed out on the opportunity to comment on the edit.)
$endgroup$
– Brian Tung
Dec 10 '18 at 17:49
add a comment |
$begingroup$
Four cards are randomly selected from a pack of 52 cards. If the first two cards are kings, what is the probability that the third card is a king?
The given answer is 2/50.
I was thinking 4 cards can be selected in 52C4 ways, out of which first three kings can be selected in 4P3 ways i.e. 24 ways and last 1 card can be choose by 49 ways. Hence result should be equal to (24*49)/(52C4)
Where am I going wrong?
probability
edited Dec 10 '18 at 17:48
Brian Tung
25.8k32554
asked Dec 10 '18 at 17:15
swapnilswapnil
335
$endgroup$
Four cards are randomly selected from a pack of 52 cards. If the first two cards are kings, what is the probability that the third card is a king?
The given answer is 2/50.
I was thinking 4 cards can be selected in 52C4 ways, out of which first three kings can be selected in 4P3 ways i.e. 24 ways and last 1 card can be choose by 49 ways. Hence result should be equal to (24*49)/(52C4)
Where am I going wrong?
probability
probability
edited Dec 10 '18 at 17:48
Brian Tung
25.8k32554
asked Dec 10 '18 at 17:15
swapnilswapnil
335
edited Dec 10 '18 at 17:48
Brian Tung
25.8k32554
asked Dec 10 '18 at 17:15
swapnilswapnil
335
edited Dec 10 '18 at 17:48
Brian Tung
25.8k32554
edited Dec 10 '18 at 17:48
Brian Tung
25.8k32554
edited Dec 10 '18 at 17:48
Brian Tung
25.8k32554
25.8k32554
asked Dec 10 '18 at 17:15
swapnilswapnil
335
asked Dec 10 '18 at 17:15
swapnilswapnil
335
asked Dec 10 '18 at 17:15
swapnilswapnil
335
335
$begingroup$
4P3 doesn't look correct as you're not using the fact that there were 2 kings already.
$endgroup$
– rsadhvika
Dec 10 '18 at 17:37
2
$begingroup$
The given answer is correct because you have $2$ kings left in a deck of $50$ cards. You seem to be trying to calculate the probability that the first three cards were kings. This problem is different because you are given that the first two cards are kings.
$endgroup$
– John Douma
Dec 10 '18 at 17:39
$begingroup$
Edited the title to match the body. (Sorry, I accidentally typed return and missed out on the opportunity to comment on the edit.)
$endgroup$
– Brian Tung
Dec 10 '18 at 17:49
add a comment |
$begingroup$
4P3 doesn't look correct as you're not using the fact that there were 2 kings already.
$endgroup$
– rsadhvika
Dec 10 '18 at 17:37
2
$begingroup$
The given answer is correct because you have $2$ kings left in a deck of $50$ cards. You seem to be trying to calculate the probability that the first three cards were kings. This problem is different because you are given that the first two cards are kings.
$endgroup$
– John Douma
Dec 10 '18 at 17:39
$begingroup$
Edited the title to match the body. (Sorry, I accidentally typed return and missed out on the opportunity to comment on the edit.)
$endgroup$
– Brian Tung
Dec 10 '18 at 17:49
$begingroup$
4P3 doesn't look correct as you're not using the fact that there were 2 kings already.
$endgroup$
– rsadhvika
Dec 10 '18 at 17:37
$begingroup$
4P3 doesn't look correct as you're not using the fact that there were 2 kings already.
$endgroup$
– rsadhvika
Dec 10 '18 at 17:37
2
2
$begingroup$
The given answer is correct because you have $2$ kings left in a deck of $50$ cards. You seem to be trying to calculate the probability that the first three cards were kings. This problem is different because you are given that the first two cards are kings.
$endgroup$
– John Douma
Dec 10 '18 at 17:39
$begingroup$
The given answer is correct because you have $2$ kings left in a deck of $50$ cards. You seem to be trying to calculate the probability that the first three cards were kings. This problem is different because you are given that the first two cards are kings.
$endgroup$
– John Douma
Dec 10 '18 at 17:39
$begingroup$
Edited the title to match the body. (Sorry, I accidentally typed return and missed out on the opportunity to comment on the edit.)
$endgroup$
– Brian Tung
Dec 10 '18 at 17:49
$begingroup$
Edited the title to match the body. (Sorry, I accidentally typed return and missed out on the opportunity to comment on the edit.)
$endgroup$
– Brian Tung
Dec 10 '18 at 17:49
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You have a deck of $52$ cards and remove $2$ cards which are both kings. You now take a new card out of the $50$ remaining cards and are asked the probability of it being a king. There are $2$ kings left in the pack of $50$ cards. The answer is
$$ frac{2}{50}. $$
Where are you going wrong ?
"I was thinking $4$ cards can be selected in $52$C$4$ ways"
Correct but .... Why are you thinking about $4$ cards ! The question asks you what is the probability upon taking the third card, what happens next is irrelevant to the question. You should focus on extracting the relevant information from the problem. It's sometimes a good idea to imagine yourself taking out the cards of the deck to identify the relevant information.
answered Dec 10 '18 at 17:41
DigitalisDigitalis
528216
$endgroup$
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
You almost have it.
Since you want to consider the 4 card subsets, and since you knew that your subset had 2 kings already, you want to look only at 4 card subsets that have 2 kings. If you have 2 kings you can choose the other 2 cards in :
$$50C2$$
ways.
Next you want to fix your third card as king. You can choose the fourth card in $49$ ways.
So the required probability would be
$$dfrac{49}{50C2}$$
answered Dec 10 '18 at 17:45
rsadhvikarsadhvika
1,7081228
$endgroup$
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
After taking two kings there are 50 cards left with 2 kings among them, so the probability that the next card is a king is $2/50$.
answered Dec 10 '18 at 17:31
mlerma54mlerma54
1,177148
$endgroup$
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have a deck of $52$ cards and remove $2$ cards which are both kings. You now take a new card out of the $50$ remaining cards and are asked the probability of it being a king. There are $2$ kings left in the pack of $50$ cards. The answer is
$$ frac{2}{50}. $$
Where are you going wrong ?
"I was thinking $4$ cards can be selected in $52$C$4$ ways"
Correct but .... Why are you thinking about $4$ cards ! The question asks you what is the probability upon taking the third card, what happens next is irrelevant to the question. You should focus on extracting the relevant information from the problem. It's sometimes a good idea to imagine yourself taking out the cards of the deck to identify the relevant information.
answered Dec 10 '18 at 17:41
DigitalisDigitalis
528216
$endgroup$
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
You have a deck of $52$ cards and remove $2$ cards which are both kings. You now take a new card out of the $50$ remaining cards and are asked the probability of it being a king. There are $2$ kings left in the pack of $50$ cards. The answer is
$$ frac{2}{50}. $$
Where are you going wrong ?
"I was thinking $4$ cards can be selected in $52$C$4$ ways"
Correct but .... Why are you thinking about $4$ cards ! The question asks you what is the probability upon taking the third card, what happens next is irrelevant to the question. You should focus on extracting the relevant information from the problem. It's sometimes a good idea to imagine yourself taking out the cards of the deck to identify the relevant information.
answered Dec 10 '18 at 17:41
DigitalisDigitalis
528216
$endgroup$
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
You have a deck of $52$ cards and remove $2$ cards which are both kings. You now take a new card out of the $50$ remaining cards and are asked the probability of it being a king. There are $2$ kings left in the pack of $50$ cards. The answer is
$$ frac{2}{50}. $$
Where are you going wrong ?
"I was thinking $4$ cards can be selected in $52$C$4$ ways"
Correct but .... Why are you thinking about $4$ cards ! The question asks you what is the probability upon taking the third card, what happens next is irrelevant to the question. You should focus on extracting the relevant information from the problem. It's sometimes a good idea to imagine yourself taking out the cards of the deck to identify the relevant information.
answered Dec 10 '18 at 17:41
DigitalisDigitalis
528216
$endgroup$
You have a deck of $52$ cards and remove $2$ cards which are both kings. You now take a new card out of the $50$ remaining cards and are asked the probability of it being a king. There are $2$ kings left in the pack of $50$ cards. The answer is
$$ frac{2}{50}. $$
Where are you going wrong ?
"I was thinking $4$ cards can be selected in $52$C$4$ ways"
Correct but .... Why are you thinking about $4$ cards ! The question asks you what is the probability upon taking the third card, what happens next is irrelevant to the question. You should focus on extracting the relevant information from the problem. It's sometimes a good idea to imagine yourself taking out the cards of the deck to identify the relevant information.
answered Dec 10 '18 at 17:41
DigitalisDigitalis
528216
answered Dec 10 '18 at 17:41
DigitalisDigitalis
528216
answered Dec 10 '18 at 17:41
DigitalisDigitalis
528216
answered Dec 10 '18 at 17:41
DigitalisDigitalis
528216
528216
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
You almost have it.
Since you want to consider the 4 card subsets, and since you knew that your subset had 2 kings already, you want to look only at 4 card subsets that have 2 kings. If you have 2 kings you can choose the other 2 cards in :
$$50C2$$
ways.
Next you want to fix your third card as king. You can choose the fourth card in $49$ ways.
So the required probability would be
$$dfrac{49}{50C2}$$
answered Dec 10 '18 at 17:45
rsadhvikarsadhvika
1,7081228
$endgroup$
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
You almost have it.
Since you want to consider the 4 card subsets, and since you knew that your subset had 2 kings already, you want to look only at 4 card subsets that have 2 kings. If you have 2 kings you can choose the other 2 cards in :
$$50C2$$
ways.
Next you want to fix your third card as king. You can choose the fourth card in $49$ ways.
So the required probability would be
$$dfrac{49}{50C2}$$
answered Dec 10 '18 at 17:45
rsadhvikarsadhvika
1,7081228
$endgroup$
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
You almost have it.
Since you want to consider the 4 card subsets, and since you knew that your subset had 2 kings already, you want to look only at 4 card subsets that have 2 kings. If you have 2 kings you can choose the other 2 cards in :
$$50C2$$
ways.
Next you want to fix your third card as king. You can choose the fourth card in $49$ ways.
So the required probability would be
$$dfrac{49}{50C2}$$
answered Dec 10 '18 at 17:45
rsadhvikarsadhvika
1,7081228
$endgroup$
You almost have it.
Since you want to consider the 4 card subsets, and since you knew that your subset had 2 kings already, you want to look only at 4 card subsets that have 2 kings. If you have 2 kings you can choose the other 2 cards in :
$$50C2$$
ways.
Next you want to fix your third card as king. You can choose the fourth card in $49$ ways.
So the required probability would be
$$dfrac{49}{50C2}$$
answered Dec 10 '18 at 17:45
rsadhvikarsadhvika
1,7081228
answered Dec 10 '18 at 17:45
rsadhvikarsadhvika
1,7081228
answered Dec 10 '18 at 17:45
rsadhvikarsadhvika
1,7081228
answered Dec 10 '18 at 17:45
rsadhvikarsadhvika
1,7081228
1,7081228
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
After taking two kings there are 50 cards left with 2 kings among them, so the probability that the next card is a king is $2/50$.
answered Dec 10 '18 at 17:31
mlerma54mlerma54
1,177148
$endgroup$
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
After taking two kings there are 50 cards left with 2 kings among them, so the probability that the next card is a king is $2/50$.
answered Dec 10 '18 at 17:31
mlerma54mlerma54
1,177148
$endgroup$
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
After taking two kings there are 50 cards left with 2 kings among them, so the probability that the next card is a king is $2/50$.
answered Dec 10 '18 at 17:31
mlerma54mlerma54
1,177148
$endgroup$
After taking two kings there are 50 cards left with 2 kings among them, so the probability that the next card is a king is $2/50$.
answered Dec 10 '18 at 17:31
mlerma54mlerma54
1,177148
answered Dec 10 '18 at 17:31
mlerma54mlerma54
1,177148
answered Dec 10 '18 at 17:31
mlerma54mlerma54
1,177148
answered Dec 10 '18 at 17:31
mlerma54mlerma54
1,177148
1,177148
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
$begingroup$
thank you so much.
$endgroup$
– swapnil
Dec 13 '18 at 9:41
add a comment |
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$begingroup$
4P3 doesn't look correct as you're not using the fact that there were 2 kings already.
$endgroup$
– rsadhvika
Dec 10 '18 at 17:37
2
$begingroup$
The given answer is correct because you have $2$ kings left in a deck of $50$ cards. You seem to be trying to calculate the probability that the first three cards were kings. This problem is different because you are given that the first two cards are kings.
$endgroup$
– John Douma
Dec 10 '18 at 17:39
$begingroup$
Edited the title to match the body. (Sorry, I accidentally typed return and missed out on the opportunity to comment on the edit.)
$endgroup$
– Brian Tung
Dec 10 '18 at 17:49