What is the probability of obtaining a total of 6 for the two cards?
up vote
1
down vote
favorite
Consider the experiment of drawing two cards from a deck in which all picture cards have been removed and adding their values (with ace = $1$).
What is the sample space?
What is the probability of obtaining a total of $5$ for the two cards?
Let A be the event “total card value is 5 or less.” Find $P ( A )$
Let A be the event “total card value is 5 or less.” Find $P (A^{complement} )$
probability card-games
edited Nov 16 at 17:57
drhab
94.3k543125
asked Nov 16 at 17:34
user1222339
112
New contributor
user1222339 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
Consider the experiment of drawing two cards from a deck in which all picture cards have been removed and adding their values (with ace = $1$).
What is the sample space?
What is the probability of obtaining a total of $5$ for the two cards?
Let A be the event “total card value is 5 or less.” Find $P ( A )$
Let A be the event “total card value is 5 or less.” Find $P (A^{complement} )$
probability card-games
edited Nov 16 at 17:57
drhab
94.3k543125
asked Nov 16 at 17:34
user1222339
112
New contributor
user1222339 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
and adding their values (with ace = 1) means? are only face cards are removed or Ace as well?
– idea
Nov 16 at 17:39
The question in the header doesn't appear to match the question(s) in the body.
– lulu
Nov 16 at 17:44
Welcome to Math.SE. Take a look at both links How to ask a good question at Math.SE and for formatting MathJax. To avoid downvotes and closing you should add your own efforts to the question, and tell us where you got stuck.
– drhab
Nov 16 at 17:58
header should have said 5 and not 6 and probably not the best title. I apologize for confusion. The Ace = 1 and jack, queen, and king should be 11, 12, 13 respectfully.
– user1222339
Nov 16 at 18:12
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the experiment of drawing two cards from a deck in which all picture cards have been removed and adding their values (with ace = $1$).
What is the sample space?
What is the probability of obtaining a total of $5$ for the two cards?
Let A be the event “total card value is 5 or less.” Find $P ( A )$
Let A be the event “total card value is 5 or less.” Find $P (A^{complement} )$
probability card-games
edited Nov 16 at 17:57
drhab
94.3k543125
asked Nov 16 at 17:34
user1222339
112
New contributor
user1222339 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Consider the experiment of drawing two cards from a deck in which all picture cards have been removed and adding their values (with ace = $1$).
What is the sample space?
What is the probability of obtaining a total of $5$ for the two cards?
Let A be the event “total card value is 5 or less.” Find $P ( A )$
Let A be the event “total card value is 5 or less.” Find $P (A^{complement} )$
probability card-games
probability card-games
edited Nov 16 at 17:57
drhab
94.3k543125
asked Nov 16 at 17:34
user1222339
112
New contributor
user1222339 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 16 at 17:57
drhab
94.3k543125
asked Nov 16 at 17:34
user1222339
112
New contributor
user1222339 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 16 at 17:57
drhab
94.3k543125
edited Nov 16 at 17:57
drhab
94.3k543125
edited Nov 16 at 17:57
drhab
94.3k543125
94.3k543125
asked Nov 16 at 17:34
user1222339
112
New contributor
user1222339 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 17:34
user1222339
112
asked Nov 16 at 17:34
user1222339
112
112
New contributor
user1222339 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user1222339 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user1222339 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
and adding their values (with ace = 1) means? are only face cards are removed or Ace as well?
– idea
Nov 16 at 17:39
The question in the header doesn't appear to match the question(s) in the body.
– lulu
Nov 16 at 17:44
Welcome to Math.SE. Take a look at both links How to ask a good question at Math.SE and for formatting MathJax. To avoid downvotes and closing you should add your own efforts to the question, and tell us where you got stuck.
– drhab
Nov 16 at 17:58
header should have said 5 and not 6 and probably not the best title. I apologize for confusion. The Ace = 1 and jack, queen, and king should be 11, 12, 13 respectfully.
– user1222339
Nov 16 at 18:12
add a comment |
and adding their values (with ace = 1) means? are only face cards are removed or Ace as well?
– idea
Nov 16 at 17:39
The question in the header doesn't appear to match the question(s) in the body.
– lulu
Nov 16 at 17:44
Welcome to Math.SE. Take a look at both links How to ask a good question at Math.SE and for formatting MathJax. To avoid downvotes and closing you should add your own efforts to the question, and tell us where you got stuck.
– drhab
Nov 16 at 17:58
header should have said 5 and not 6 and probably not the best title. I apologize for confusion. The Ace = 1 and jack, queen, and king should be 11, 12, 13 respectfully.
– user1222339
Nov 16 at 18:12
and adding their values (with ace = 1) means? are only face cards are removed or Ace as well?
– idea
Nov 16 at 17:39
and adding their values (with ace = 1) means? are only face cards are removed or Ace as well?
– idea
Nov 16 at 17:39
The question in the header doesn't appear to match the question(s) in the body.
– lulu
Nov 16 at 17:44
The question in the header doesn't appear to match the question(s) in the body.
– lulu
Nov 16 at 17:44
Welcome to Math.SE. Take a look at both links How to ask a good question at Math.SE and for formatting MathJax. To avoid downvotes and closing you should add your own efforts to the question, and tell us where you got stuck.
– drhab
Nov 16 at 17:58
Welcome to Math.SE. Take a look at both links How to ask a good question at Math.SE and for formatting MathJax. To avoid downvotes and closing you should add your own efforts to the question, and tell us where you got stuck.
– drhab
Nov 16 at 17:58
header should have said 5 and not 6 and probably not the best title. I apologize for confusion. The Ace = 1 and jack, queen, and king should be 11, 12, 13 respectfully.
– user1222339
Nov 16 at 18:12
header should have said 5 and not 6 and probably not the best title. I apologize for confusion. The Ace = 1 and jack, queen, and king should be 11, 12, 13 respectfully.
– user1222339
Nov 16 at 18:12
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Assuming that all face cards are removed, you are left with $52-12=40$ cards.
So, your Sample Space is $40$ cards consisting of all numbered cards (from Ace=$1$ to $10$) in all $4$ suits.
When you draw $2$ cards to obtain a total of $5$, there are following possibilities:
$$(4,1),(3,2)$$
and there are $4$ cards of each kind.
So select $1$ of them in $^4C_1$ way, and $2$ cards can be selected from $40$ in $^{40}C_2$ ways.
$$P(text{sum=5})=frac{^4C_1cdot ^4C_1+^4C_1cdot ^4C_1}{^{40}C_2}=frac{32}{^{40}C_2}$$
For $sumleq5$, possible cases are: $$(1,1),(1,2),(1,3),(2,2),(1,4),(2,3)$$
There are $^4C_2$ ways to draw $2$ cards with same number, from $4$.
$$P(A)=frac{(^4C_2)+(^4C_1cdot ^4C_1)+(^4C_1cdot ^4C_1)+(^4C_2)+(^4C_1cdot ^4C_1)+(^4C_1cdot ^4C_1)}{^{40}C_2}=frac{76}{^{40}C_2}$$
$$P(A^C)=1-P(A)=1-frac{76}{^{40}C_2}$$
answered Nov 16 at 18:15
idea
2,03721023
The answer seems correct to me, but can you help me understand what the C stands for? what is the value? Also shouldn't it be 5 and not 4 to the left of the C? since it is the sum less than or equal to 5?
– user1222339
Nov 16 at 19:49
C stands for Combinations. $^nC_r$ means number of ways to select r objects from n. $$^nC_r=frac{n!}{r! cdot (n-r)!}$$ There, 4 represents the number of available cards having same number, eg. You have 4 1's out of which you select 1 or more. Just take a lecture on Combinations and you will understand the solution fully.
– idea
Nov 17 at 6:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Assuming that all face cards are removed, you are left with $52-12=40$ cards.
So, your Sample Space is $40$ cards consisting of all numbered cards (from Ace=$1$ to $10$) in all $4$ suits.
When you draw $2$ cards to obtain a total of $5$, there are following possibilities:
$$(4,1),(3,2)$$
and there are $4$ cards of each kind.
So select $1$ of them in $^4C_1$ way, and $2$ cards can be selected from $40$ in $^{40}C_2$ ways.
$$P(text{sum=5})=frac{^4C_1cdot ^4C_1+^4C_1cdot ^4C_1}{^{40}C_2}=frac{32}{^{40}C_2}$$
For $sumleq5$, possible cases are: $$(1,1),(1,2),(1,3),(2,2),(1,4),(2,3)$$
There are $^4C_2$ ways to draw $2$ cards with same number, from $4$.
$$P(A)=frac{(^4C_2)+(^4C_1cdot ^4C_1)+(^4C_1cdot ^4C_1)+(^4C_2)+(^4C_1cdot ^4C_1)+(^4C_1cdot ^4C_1)}{^{40}C_2}=frac{76}{^{40}C_2}$$
$$P(A^C)=1-P(A)=1-frac{76}{^{40}C_2}$$
answered Nov 16 at 18:15
idea
2,03721023
The answer seems correct to me, but can you help me understand what the C stands for? what is the value? Also shouldn't it be 5 and not 4 to the left of the C? since it is the sum less than or equal to 5?
– user1222339
Nov 16 at 19:49
C stands for Combinations. $^nC_r$ means number of ways to select r objects from n. $$^nC_r=frac{n!}{r! cdot (n-r)!}$$ There, 4 represents the number of available cards having same number, eg. You have 4 1's out of which you select 1 or more. Just take a lecture on Combinations and you will understand the solution fully.
– idea
Nov 17 at 6:30
add a comment |
up vote
1
down vote
accepted
Assuming that all face cards are removed, you are left with $52-12=40$ cards.
So, your Sample Space is $40$ cards consisting of all numbered cards (from Ace=$1$ to $10$) in all $4$ suits.
When you draw $2$ cards to obtain a total of $5$, there are following possibilities:
$$(4,1),(3,2)$$
and there are $4$ cards of each kind.
So select $1$ of them in $^4C_1$ way, and $2$ cards can be selected from $40$ in $^{40}C_2$ ways.
$$P(text{sum=5})=frac{^4C_1cdot ^4C_1+^4C_1cdot ^4C_1}{^{40}C_2}=frac{32}{^{40}C_2}$$
For $sumleq5$, possible cases are: $$(1,1),(1,2),(1,3),(2,2),(1,4),(2,3)$$
There are $^4C_2$ ways to draw $2$ cards with same number, from $4$.
$$P(A)=frac{(^4C_2)+(^4C_1cdot ^4C_1)+(^4C_1cdot ^4C_1)+(^4C_2)+(^4C_1cdot ^4C_1)+(^4C_1cdot ^4C_1)}{^{40}C_2}=frac{76}{^{40}C_2}$$
$$P(A^C)=1-P(A)=1-frac{76}{^{40}C_2}$$
answered Nov 16 at 18:15
idea
2,03721023
The answer seems correct to me, but can you help me understand what the C stands for? what is the value? Also shouldn't it be 5 and not 4 to the left of the C? since it is the sum less than or equal to 5?
– user1222339
Nov 16 at 19:49
C stands for Combinations. $^nC_r$ means number of ways to select r objects from n. $$^nC_r=frac{n!}{r! cdot (n-r)!}$$ There, 4 represents the number of available cards having same number, eg. You have 4 1's out of which you select 1 or more. Just take a lecture on Combinations and you will understand the solution fully.
– idea
Nov 17 at 6:30
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Assuming that all face cards are removed, you are left with $52-12=40$ cards.
So, your Sample Space is $40$ cards consisting of all numbered cards (from Ace=$1$ to $10$) in all $4$ suits.
When you draw $2$ cards to obtain a total of $5$, there are following possibilities:
$$(4,1),(3,2)$$
and there are $4$ cards of each kind.
So select $1$ of them in $^4C_1$ way, and $2$ cards can be selected from $40$ in $^{40}C_2$ ways.
$$P(text{sum=5})=frac{^4C_1cdot ^4C_1+^4C_1cdot ^4C_1}{^{40}C_2}=frac{32}{^{40}C_2}$$
For $sumleq5$, possible cases are: $$(1,1),(1,2),(1,3),(2,2),(1,4),(2,3)$$
There are $^4C_2$ ways to draw $2$ cards with same number, from $4$.
$$P(A)=frac{(^4C_2)+(^4C_1cdot ^4C_1)+(^4C_1cdot ^4C_1)+(^4C_2)+(^4C_1cdot ^4C_1)+(^4C_1cdot ^4C_1)}{^{40}C_2}=frac{76}{^{40}C_2}$$
$$P(A^C)=1-P(A)=1-frac{76}{^{40}C_2}$$
answered Nov 16 at 18:15
idea
2,03721023
Assuming that all face cards are removed, you are left with $52-12=40$ cards.
So, your Sample Space is $40$ cards consisting of all numbered cards (from Ace=$1$ to $10$) in all $4$ suits.
When you draw $2$ cards to obtain a total of $5$, there are following possibilities:
$$(4,1),(3,2)$$
and there are $4$ cards of each kind.
So select $1$ of them in $^4C_1$ way, and $2$ cards can be selected from $40$ in $^{40}C_2$ ways.
$$P(text{sum=5})=frac{^4C_1cdot ^4C_1+^4C_1cdot ^4C_1}{^{40}C_2}=frac{32}{^{40}C_2}$$
For $sumleq5$, possible cases are: $$(1,1),(1,2),(1,3),(2,2),(1,4),(2,3)$$
There are $^4C_2$ ways to draw $2$ cards with same number, from $4$.
$$P(A)=frac{(^4C_2)+(^4C_1cdot ^4C_1)+(^4C_1cdot ^4C_1)+(^4C_2)+(^4C_1cdot ^4C_1)+(^4C_1cdot ^4C_1)}{^{40}C_2}=frac{76}{^{40}C_2}$$
$$P(A^C)=1-P(A)=1-frac{76}{^{40}C_2}$$
answered Nov 16 at 18:15
idea
2,03721023
answered Nov 16 at 18:15
idea
2,03721023
answered Nov 16 at 18:15
idea
2,03721023
answered Nov 16 at 18:15
idea
2,03721023
2,03721023
The answer seems correct to me, but can you help me understand what the C stands for? what is the value? Also shouldn't it be 5 and not 4 to the left of the C? since it is the sum less than or equal to 5?
– user1222339
Nov 16 at 19:49
C stands for Combinations. $^nC_r$ means number of ways to select r objects from n. $$^nC_r=frac{n!}{r! cdot (n-r)!}$$ There, 4 represents the number of available cards having same number, eg. You have 4 1's out of which you select 1 or more. Just take a lecture on Combinations and you will understand the solution fully.
– idea
Nov 17 at 6:30
add a comment |
The answer seems correct to me, but can you help me understand what the C stands for? what is the value? Also shouldn't it be 5 and not 4 to the left of the C? since it is the sum less than or equal to 5?
– user1222339
Nov 16 at 19:49
C stands for Combinations. $^nC_r$ means number of ways to select r objects from n. $$^nC_r=frac{n!}{r! cdot (n-r)!}$$ There, 4 represents the number of available cards having same number, eg. You have 4 1's out of which you select 1 or more. Just take a lecture on Combinations and you will understand the solution fully.
– idea
Nov 17 at 6:30
The answer seems correct to me, but can you help me understand what the C stands for? what is the value? Also shouldn't it be 5 and not 4 to the left of the C? since it is the sum less than or equal to 5?
– user1222339
Nov 16 at 19:49
The answer seems correct to me, but can you help me understand what the C stands for? what is the value? Also shouldn't it be 5 and not 4 to the left of the C? since it is the sum less than or equal to 5?
– user1222339
Nov 16 at 19:49
C stands for Combinations. $^nC_r$ means number of ways to select r objects from n. $$^nC_r=frac{n!}{r! cdot (n-r)!}$$ There, 4 represents the number of available cards having same number, eg. You have 4 1's out of which you select 1 or more. Just take a lecture on Combinations and you will understand the solution fully.
– idea
Nov 17 at 6:30
C stands for Combinations. $^nC_r$ means number of ways to select r objects from n. $$^nC_r=frac{n!}{r! cdot (n-r)!}$$ There, 4 represents the number of available cards having same number, eg. You have 4 1's out of which you select 1 or more. Just take a lecture on Combinations and you will understand the solution fully.
– idea
Nov 17 at 6:30
add a comment |
user1222339 is a new contributor. Be nice, and check out our Code of Conduct.
user1222339 is a new contributor. Be nice, and check out our Code of Conduct.
user1222339 is a new contributor. Be nice, and check out our Code of Conduct.
user1222339 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001418%2fwhat-is-the-probability-of-obtaining-a-total-of-6-for-the-two-cards%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
and adding their values (with ace = 1) means? are only face cards are removed or Ace as well?
– idea
Nov 16 at 17:39
The question in the header doesn't appear to match the question(s) in the body.
– lulu
Nov 16 at 17:44
Welcome to Math.SE. Take a look at both links How to ask a good question at Math.SE and for formatting MathJax. To avoid downvotes and closing you should add your own efforts to the question, and tell us where you got stuck.
– drhab
Nov 16 at 17:58
header should have said 5 and not 6 and probably not the best title. I apologize for confusion. The Ace = 1 and jack, queen, and king should be 11, 12, 13 respectfully.
– user1222339
Nov 16 at 18:12