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Find the probability that the card number 18 is the second jack that you deal.
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You deal from a well-shuffled $52$-card deck, one card at a time. Find the probability that the card number 18 is the second jack that you deal. Include at least $4$ digits after the decimal point in your answer.
I have tried this numerous different ways, but I cant seem to get it.
I've tried
$$fracdbinom4816 cdot dfrac452dbinom5216 cdot dfrac335$$
and I don't know why this doesn't work.
probability combinatorics
edited 1 hour ago
N. F. Taussig
40.1k93253
asked 1 hour ago
Nathaniel Sexton
411
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
3
down vote
favorite
You deal from a well-shuffled $52$-card deck, one card at a time. Find the probability that the card number 18 is the second jack that you deal. Include at least $4$ digits after the decimal point in your answer.
I have tried this numerous different ways, but I cant seem to get it.
I've tried
$$fracdbinom4816 cdot dfrac452dbinom5216 cdot dfrac335$$
and I don't know why this doesn't work.
probability combinatorics
edited 1 hour ago
N. F. Taussig
40.1k93253
asked 1 hour ago
Nathaniel Sexton
411
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
1 hour ago
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up vote
3
down vote
favorite
up vote
3
down vote
favorite
You deal from a well-shuffled $52$-card deck, one card at a time. Find the probability that the card number 18 is the second jack that you deal. Include at least $4$ digits after the decimal point in your answer.
I have tried this numerous different ways, but I cant seem to get it.
I've tried
$$fracdbinom4816 cdot dfrac452dbinom5216 cdot dfrac335$$
and I don't know why this doesn't work.
probability combinatorics
edited 1 hour ago
N. F. Taussig
40.1k93253
asked 1 hour ago
Nathaniel Sexton
411
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You deal from a well-shuffled $52$-card deck, one card at a time. Find the probability that the card number 18 is the second jack that you deal. Include at least $4$ digits after the decimal point in your answer.
I have tried this numerous different ways, but I cant seem to get it.
I've tried
$$fracdbinom4816 cdot dfrac452dbinom5216 cdot dfrac335$$
and I don't know why this doesn't work.
probability combinatorics
probability combinatorics
edited 1 hour ago
N. F. Taussig
40.1k93253
asked 1 hour ago
Nathaniel Sexton
411
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
N. F. Taussig
40.1k93253
asked 1 hour ago
Nathaniel Sexton
411
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
N. F. Taussig
40.1k93253
edited 1 hour ago
N. F. Taussig
40.1k93253
edited 1 hour ago
N. F. Taussig
40.1k93253
40.1k93253
asked 1 hour ago
Nathaniel Sexton
411
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Nathaniel Sexton
411
asked 1 hour ago
Nathaniel Sexton
411
411
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
1 hour ago
add a comment |Â
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
1 hour ago
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
1 hour ago
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
1 hour ago
add a comment |Â
3 Answers
3
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oldest
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up vote
3
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If the $18$th card is the second Jack you deal, then there must be one Jack and 16 non-Jacks among the first $17$ cards, then a second Jack at the $18$th card. The probability of obtaining one of the four Jacks and $16$ of the $48$ non-Jacks in the first $17$ deals is
$$fracdbinom41dbinom4816dbinom5217$$
The probability of dealing one of the three remaining jacks from among the $52 - 17 = 35$ remaining cards is $3/35$. Hence, the probability that the second Jack you deal occurs on the $18$th deal is
$$fracdbinom41dbinom4816dbinom5217 cdot frac335$$
answered 1 hour ago
N. F. Taussig
40.1k93253
add a comment |Â
up vote
2
down vote
There are $17$ ways to choose when the first Jack is drawn.
$$17tag1$$
Given this, there are $4$ ways to choose which Jack that is.
$$4 tag2$$
There are $16$ non-Jack cards drawn before the second Jack, and there are $48$ choices for the first of these cards, $47$ for the second, and so on ...
$$48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33 tag3$$
Finally there are three choices for which Jack is drawn second:
$$3 tag4$$
The number of successful draws is the product of values $(1)$ through $(4)$ above.
The total number of possible draws of $18$ cards in order is:
$$52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$
The final probability is the number of successful possibilities, divided by the total number of possibilities:
$$displaystylefrac17 ,cdot, 4 ,cdot,left(,48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33,right), cdot , 3 52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$
answered 1 hour ago
Zubin Mukerjee
14.5k32456
add a comment |Â
up vote
0
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The first $17$ cards can be a single jack ($1$ from $4$) plus any $16$ from $48$. The $18$th card can be one of $3$ jacks from the $35$ remaining cards. So I get: $$4cdot fracbinom4816binom5217cdot frac335 = 0.03523$$
answered 1 hour ago
Phil H
2,4922311
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
If the $18$th card is the second Jack you deal, then there must be one Jack and 16 non-Jacks among the first $17$ cards, then a second Jack at the $18$th card. The probability of obtaining one of the four Jacks and $16$ of the $48$ non-Jacks in the first $17$ deals is
$$fracdbinom41dbinom4816dbinom5217$$
The probability of dealing one of the three remaining jacks from among the $52 - 17 = 35$ remaining cards is $3/35$. Hence, the probability that the second Jack you deal occurs on the $18$th deal is
$$fracdbinom41dbinom4816dbinom5217 cdot frac335$$
answered 1 hour ago
N. F. Taussig
40.1k93253
add a comment |Â
up vote
3
down vote
If the $18$th card is the second Jack you deal, then there must be one Jack and 16 non-Jacks among the first $17$ cards, then a second Jack at the $18$th card. The probability of obtaining one of the four Jacks and $16$ of the $48$ non-Jacks in the first $17$ deals is
$$fracdbinom41dbinom4816dbinom5217$$
The probability of dealing one of the three remaining jacks from among the $52 - 17 = 35$ remaining cards is $3/35$. Hence, the probability that the second Jack you deal occurs on the $18$th deal is
$$fracdbinom41dbinom4816dbinom5217 cdot frac335$$
answered 1 hour ago
N. F. Taussig
40.1k93253
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If the $18$th card is the second Jack you deal, then there must be one Jack and 16 non-Jacks among the first $17$ cards, then a second Jack at the $18$th card. The probability of obtaining one of the four Jacks and $16$ of the $48$ non-Jacks in the first $17$ deals is
$$fracdbinom41dbinom4816dbinom5217$$
The probability of dealing one of the three remaining jacks from among the $52 - 17 = 35$ remaining cards is $3/35$. Hence, the probability that the second Jack you deal occurs on the $18$th deal is
$$fracdbinom41dbinom4816dbinom5217 cdot frac335$$
answered 1 hour ago
N. F. Taussig
40.1k93253
If the $18$th card is the second Jack you deal, then there must be one Jack and 16 non-Jacks among the first $17$ cards, then a second Jack at the $18$th card. The probability of obtaining one of the four Jacks and $16$ of the $48$ non-Jacks in the first $17$ deals is
$$fracdbinom41dbinom4816dbinom5217$$
The probability of dealing one of the three remaining jacks from among the $52 - 17 = 35$ remaining cards is $3/35$. Hence, the probability that the second Jack you deal occurs on the $18$th deal is
$$fracdbinom41dbinom4816dbinom5217 cdot frac335$$
answered 1 hour ago
N. F. Taussig
40.1k93253
answered 1 hour ago
N. F. Taussig
40.1k93253
answered 1 hour ago
N. F. Taussig
40.1k93253
answered 1 hour ago
N. F. Taussig
40.1k93253
40.1k93253
add a comment |Â
add a comment |Â
up vote
2
down vote
There are $17$ ways to choose when the first Jack is drawn.
$$17tag1$$
Given this, there are $4$ ways to choose which Jack that is.
$$4 tag2$$
There are $16$ non-Jack cards drawn before the second Jack, and there are $48$ choices for the first of these cards, $47$ for the second, and so on ...
$$48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33 tag3$$
Finally there are three choices for which Jack is drawn second:
$$3 tag4$$
The number of successful draws is the product of values $(1)$ through $(4)$ above.
The total number of possible draws of $18$ cards in order is:
$$52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$
The final probability is the number of successful possibilities, divided by the total number of possibilities:
$$displaystylefrac17 ,cdot, 4 ,cdot,left(,48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33,right), cdot , 3 52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$
answered 1 hour ago
Zubin Mukerjee
14.5k32456
add a comment |Â
up vote
2
down vote
There are $17$ ways to choose when the first Jack is drawn.
$$17tag1$$
Given this, there are $4$ ways to choose which Jack that is.
$$4 tag2$$
There are $16$ non-Jack cards drawn before the second Jack, and there are $48$ choices for the first of these cards, $47$ for the second, and so on ...
$$48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33 tag3$$
Finally there are three choices for which Jack is drawn second:
$$3 tag4$$
The number of successful draws is the product of values $(1)$ through $(4)$ above.
The total number of possible draws of $18$ cards in order is:
$$52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$
The final probability is the number of successful possibilities, divided by the total number of possibilities:
$$displaystylefrac17 ,cdot, 4 ,cdot,left(,48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33,right), cdot , 3 52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$
answered 1 hour ago
Zubin Mukerjee
14.5k32456
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There are $17$ ways to choose when the first Jack is drawn.
$$17tag1$$
Given this, there are $4$ ways to choose which Jack that is.
$$4 tag2$$
There are $16$ non-Jack cards drawn before the second Jack, and there are $48$ choices for the first of these cards, $47$ for the second, and so on ...
$$48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33 tag3$$
Finally there are three choices for which Jack is drawn second:
$$3 tag4$$
The number of successful draws is the product of values $(1)$ through $(4)$ above.
The total number of possible draws of $18$ cards in order is:
$$52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$
The final probability is the number of successful possibilities, divided by the total number of possibilities:
$$displaystylefrac17 ,cdot, 4 ,cdot,left(,48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33,right), cdot , 3 52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$
answered 1 hour ago
Zubin Mukerjee
14.5k32456
There are $17$ ways to choose when the first Jack is drawn.
$$17tag1$$
Given this, there are $4$ ways to choose which Jack that is.
$$4 tag2$$
There are $16$ non-Jack cards drawn before the second Jack, and there are $48$ choices for the first of these cards, $47$ for the second, and so on ...
$$48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33 tag3$$
Finally there are three choices for which Jack is drawn second:
$$3 tag4$$
The number of successful draws is the product of values $(1)$ through $(4)$ above.
The total number of possible draws of $18$ cards in order is:
$$52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$
The final probability is the number of successful possibilities, divided by the total number of possibilities:
$$displaystylefrac17 ,cdot, 4 ,cdot,left(,48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33,right), cdot , 3 52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$
answered 1 hour ago
Zubin Mukerjee
14.5k32456
answered 1 hour ago
Zubin Mukerjee
14.5k32456
answered 1 hour ago
Zubin Mukerjee
14.5k32456
answered 1 hour ago
Zubin Mukerjee
14.5k32456
14.5k32456
add a comment |Â
add a comment |Â
up vote
0
down vote
The first $17$ cards can be a single jack ($1$ from $4$) plus any $16$ from $48$. The $18$th card can be one of $3$ jacks from the $35$ remaining cards. So I get: $$4cdot fracbinom4816binom5217cdot frac335 = 0.03523$$
answered 1 hour ago
Phil H
2,4922311
add a comment |Â
up vote
0
down vote
The first $17$ cards can be a single jack ($1$ from $4$) plus any $16$ from $48$. The $18$th card can be one of $3$ jacks from the $35$ remaining cards. So I get: $$4cdot fracbinom4816binom5217cdot frac335 = 0.03523$$
answered 1 hour ago
Phil H
2,4922311
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The first $17$ cards can be a single jack ($1$ from $4$) plus any $16$ from $48$. The $18$th card can be one of $3$ jacks from the $35$ remaining cards. So I get: $$4cdot fracbinom4816binom5217cdot frac335 = 0.03523$$
answered 1 hour ago
Phil H
2,4922311
The first $17$ cards can be a single jack ($1$ from $4$) plus any $16$ from $48$. The $18$th card can be one of $3$ jacks from the $35$ remaining cards. So I get: $$4cdot fracbinom4816binom5217cdot frac335 = 0.03523$$
answered 1 hour ago
Phil H
2,4922311
answered 1 hour ago
Phil H
2,4922311
answered 1 hour ago
Phil H
2,4922311
answered 1 hour ago
Phil H
2,4922311
2,4922311
add a comment |Â
add a comment |Â
Nathaniel Sexton is a new contributor. Be nice, and check out our Code of Conduct.
Nathaniel Sexton is a new contributor. Be nice, and check out our Code of Conduct.
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