Mixing
There are 10 marbles in a bag. $6$ are red and $4$ are blue. You must chose at least 1 red marble. In how many ways can you chose three total marbles.
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
I thought the answer is $^9C_2$ since the first (red) marble didn't count. You have to pick a red marble which reduces the total count from 10 to 9. The answer is 116 possible ways.
probability permutations combinations
edited Aug 10 at 0:50
Key Flex
1
asked Aug 9 at 22:35
user9995331
1114
add a comment |Â
up vote
4
down vote
favorite
I thought the answer is $^9C_2$ since the first (red) marble didn't count. You have to pick a red marble which reduces the total count from 10 to 9. The answer is 116 possible ways.
probability permutations combinations
edited Aug 10 at 0:50
Key Flex
1
asked Aug 9 at 22:35
user9995331
1114
Why someone wants to close this interesting topic? Explain please your step.
– Michael Rozenberg
Aug 10 at 9:37
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I thought the answer is $^9C_2$ since the first (red) marble didn't count. You have to pick a red marble which reduces the total count from 10 to 9. The answer is 116 possible ways.
probability permutations combinations
edited Aug 10 at 0:50
Key Flex
1
asked Aug 9 at 22:35
user9995331
1114
I thought the answer is $^9C_2$ since the first (red) marble didn't count. You have to pick a red marble which reduces the total count from 10 to 9. The answer is 116 possible ways.
probability permutations combinations
edited Aug 10 at 0:50
Key Flex
1
asked Aug 9 at 22:35
user9995331
1114
edited Aug 10 at 0:50
Key Flex
1
edited Aug 10 at 0:50
Key Flex
1
edited Aug 10 at 0:50
Key Flex
1
1
asked Aug 9 at 22:35
user9995331
1114
asked Aug 9 at 22:35
user9995331
1114
asked Aug 9 at 22:35
user9995331
1114
1114
Why someone wants to close this interesting topic? Explain please your step.
– Michael Rozenberg
Aug 10 at 9:37
add a comment |Â
Why someone wants to close this interesting topic? Explain please your step.
– Michael Rozenberg
Aug 10 at 9:37
Why someone wants to close this interesting topic? Explain please your step.
– Michael Rozenberg
Aug 10 at 9:37
Why someone wants to close this interesting topic? Explain please your step.
– Michael Rozenberg
Aug 10 at 9:37
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
5
down vote
accepted
There are total of $10$ balls in which $6$ are red and $4$ are blue.
First you can choose $1$ Red and $2$ blue in $dbinom61dbinom42=36$ ways
Second you can choose $2$ Red and $1$ blue in $dbinom62dbinom41=60$ ways
Third you can choose all $3$ red in $dbinom63=20$ ways
In total you can choose in $dbinom61dbinom42+dbinom62dbinom41+dbinom63=36+60+20=116$ ways
answered Aug 9 at 22:45
Key Flex
1
add a comment |Â
up vote
5
down vote
The overall numbers of ways to choose 3 marbles are
$$N_1=binom103=120$$
The overall numbers of ways to choose 3 blue marbles are
$$N_2=binom43=binom41=4$$
therefore $N=N_1-N_2=116$.
answered Aug 9 at 22:43
gimusi
69.4k73685
add a comment |Â
up vote
4
down vote
Hint: In how many ways can you choose three marbles, none of which are red?
answered Aug 9 at 22:43
rogerl
16.6k22745
2
4 C 3 would be picking all blue. Total number is 10 C 3. Take the difference?
– user9995331
Aug 9 at 22:53
add a comment |Â
up vote
4
down vote
you want to choose three marbles and must choose at least a red marble.
Here are the ways in you can do that
1Red 2Blue
2Red 1Blue
3Red
$$total=binom61binom42+binom62binom41+binom63=116 ways$$
answered Aug 9 at 22:46
Deepesh Meena
2,678720
Why do you need to account for the 'other' possibilities the chose function is implied? The ball is either red or blue. I am confused why you multiply them in this instance.
– user9995331
Aug 9 at 23:17
you need to choose three balls but you also want to choose at least 1 read thus every time you need to choose a red with blue balls possibilities are 1red and 2 blue balls , 2 red and 1 blue balls , 3 red balls
– Deepesh Meena
Aug 9 at 23:18
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
There are total of $10$ balls in which $6$ are red and $4$ are blue.
First you can choose $1$ Red and $2$ blue in $dbinom61dbinom42=36$ ways
Second you can choose $2$ Red and $1$ blue in $dbinom62dbinom41=60$ ways
Third you can choose all $3$ red in $dbinom63=20$ ways
In total you can choose in $dbinom61dbinom42+dbinom62dbinom41+dbinom63=36+60+20=116$ ways
answered Aug 9 at 22:45
Key Flex
1
add a comment |Â
up vote
5
down vote
accepted
There are total of $10$ balls in which $6$ are red and $4$ are blue.
First you can choose $1$ Red and $2$ blue in $dbinom61dbinom42=36$ ways
Second you can choose $2$ Red and $1$ blue in $dbinom62dbinom41=60$ ways
Third you can choose all $3$ red in $dbinom63=20$ ways
In total you can choose in $dbinom61dbinom42+dbinom62dbinom41+dbinom63=36+60+20=116$ ways
answered Aug 9 at 22:45
Key Flex
1
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
There are total of $10$ balls in which $6$ are red and $4$ are blue.
First you can choose $1$ Red and $2$ blue in $dbinom61dbinom42=36$ ways
Second you can choose $2$ Red and $1$ blue in $dbinom62dbinom41=60$ ways
Third you can choose all $3$ red in $dbinom63=20$ ways
In total you can choose in $dbinom61dbinom42+dbinom62dbinom41+dbinom63=36+60+20=116$ ways
answered Aug 9 at 22:45
Key Flex
1
There are total of $10$ balls in which $6$ are red and $4$ are blue.
First you can choose $1$ Red and $2$ blue in $dbinom61dbinom42=36$ ways
Second you can choose $2$ Red and $1$ blue in $dbinom62dbinom41=60$ ways
Third you can choose all $3$ red in $dbinom63=20$ ways
In total you can choose in $dbinom61dbinom42+dbinom62dbinom41+dbinom63=36+60+20=116$ ways
answered Aug 9 at 22:45
Key Flex
1
edited Aug 9 at 22:52
answered Aug 9 at 22:45
Key Flex
1
answered Aug 9 at 22:45
Key Flex
1
answered Aug 9 at 22:45
Key Flex
1
1
add a comment |Â
add a comment |Â
up vote
5
down vote
The overall numbers of ways to choose 3 marbles are
$$N_1=binom103=120$$
The overall numbers of ways to choose 3 blue marbles are
$$N_2=binom43=binom41=4$$
therefore $N=N_1-N_2=116$.
answered Aug 9 at 22:43
gimusi
69.4k73685
add a comment |Â
up vote
5
down vote
The overall numbers of ways to choose 3 marbles are
$$N_1=binom103=120$$
The overall numbers of ways to choose 3 blue marbles are
$$N_2=binom43=binom41=4$$
therefore $N=N_1-N_2=116$.
answered Aug 9 at 22:43
gimusi
69.4k73685
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The overall numbers of ways to choose 3 marbles are
$$N_1=binom103=120$$
The overall numbers of ways to choose 3 blue marbles are
$$N_2=binom43=binom41=4$$
therefore $N=N_1-N_2=116$.
answered Aug 9 at 22:43
gimusi
69.4k73685
The overall numbers of ways to choose 3 marbles are
$$N_1=binom103=120$$
The overall numbers of ways to choose 3 blue marbles are
$$N_2=binom43=binom41=4$$
therefore $N=N_1-N_2=116$.
answered Aug 9 at 22:43
gimusi
69.4k73685
edited Aug 9 at 22:48
answered Aug 9 at 22:43
gimusi
69.4k73685
answered Aug 9 at 22:43
gimusi
69.4k73685
answered Aug 9 at 22:43
gimusi
69.4k73685
69.4k73685
add a comment |Â
add a comment |Â
up vote
4
down vote
Hint: In how many ways can you choose three marbles, none of which are red?
answered Aug 9 at 22:43
rogerl
16.6k22745
2
4 C 3 would be picking all blue. Total number is 10 C 3. Take the difference?
– user9995331
Aug 9 at 22:53
add a comment |Â
up vote
4
down vote
Hint: In how many ways can you choose three marbles, none of which are red?
answered Aug 9 at 22:43
rogerl
16.6k22745
2
4 C 3 would be picking all blue. Total number is 10 C 3. Take the difference?
– user9995331
Aug 9 at 22:53
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Hint: In how many ways can you choose three marbles, none of which are red?
answered Aug 9 at 22:43
rogerl
16.6k22745
Hint: In how many ways can you choose three marbles, none of which are red?
answered Aug 9 at 22:43
rogerl
16.6k22745
answered Aug 9 at 22:43
rogerl
16.6k22745
answered Aug 9 at 22:43
rogerl
16.6k22745
answered Aug 9 at 22:43
rogerl
16.6k22745
16.6k22745
2
4 C 3 would be picking all blue. Total number is 10 C 3. Take the difference?
– user9995331
Aug 9 at 22:53
add a comment |Â
2
4 C 3 would be picking all blue. Total number is 10 C 3. Take the difference?
– user9995331
Aug 9 at 22:53
2
2
4 C 3 would be picking all blue. Total number is 10 C 3. Take the difference?
– user9995331
Aug 9 at 22:53
4 C 3 would be picking all blue. Total number is 10 C 3. Take the difference?
– user9995331
Aug 9 at 22:53
add a comment |Â
up vote
4
down vote
you want to choose three marbles and must choose at least a red marble.
Here are the ways in you can do that
1Red 2Blue
2Red 1Blue
3Red
$$total=binom61binom42+binom62binom41+binom63=116 ways$$
answered Aug 9 at 22:46
Deepesh Meena
2,678720
Why do you need to account for the 'other' possibilities the chose function is implied? The ball is either red or blue. I am confused why you multiply them in this instance.
– user9995331
Aug 9 at 23:17
you need to choose three balls but you also want to choose at least 1 read thus every time you need to choose a red with blue balls possibilities are 1red and 2 blue balls , 2 red and 1 blue balls , 3 red balls
– Deepesh Meena
Aug 9 at 23:18
add a comment |Â
up vote
4
down vote
you want to choose three marbles and must choose at least a red marble.
Here are the ways in you can do that
1Red 2Blue
2Red 1Blue
3Red
$$total=binom61binom42+binom62binom41+binom63=116 ways$$
answered Aug 9 at 22:46
Deepesh Meena
2,678720
Why do you need to account for the 'other' possibilities the chose function is implied? The ball is either red or blue. I am confused why you multiply them in this instance.
– user9995331
Aug 9 at 23:17
you need to choose three balls but you also want to choose at least 1 read thus every time you need to choose a red with blue balls possibilities are 1red and 2 blue balls , 2 red and 1 blue balls , 3 red balls
– Deepesh Meena
Aug 9 at 23:18
add a comment |Â
up vote
4
down vote
up vote
4
down vote
you want to choose three marbles and must choose at least a red marble.
Here are the ways in you can do that
1Red 2Blue
2Red 1Blue
3Red
$$total=binom61binom42+binom62binom41+binom63=116 ways$$
answered Aug 9 at 22:46
Deepesh Meena
2,678720
you want to choose three marbles and must choose at least a red marble.
Here are the ways in you can do that
1Red 2Blue
2Red 1Blue
3Red
$$total=binom61binom42+binom62binom41+binom63=116 ways$$
answered Aug 9 at 22:46
Deepesh Meena
2,678720
edited Aug 9 at 22:53
answered Aug 9 at 22:46
Deepesh Meena
2,678720
answered Aug 9 at 22:46
Deepesh Meena
2,678720
answered Aug 9 at 22:46
Deepesh Meena
2,678720
2,678720
Why do you need to account for the 'other' possibilities the chose function is implied? The ball is either red or blue. I am confused why you multiply them in this instance.
– user9995331
Aug 9 at 23:17
you need to choose three balls but you also want to choose at least 1 read thus every time you need to choose a red with blue balls possibilities are 1red and 2 blue balls , 2 red and 1 blue balls , 3 red balls
– Deepesh Meena
Aug 9 at 23:18
add a comment |Â
Why do you need to account for the 'other' possibilities the chose function is implied? The ball is either red or blue. I am confused why you multiply them in this instance.
– user9995331
Aug 9 at 23:17
you need to choose three balls but you also want to choose at least 1 read thus every time you need to choose a red with blue balls possibilities are 1red and 2 blue balls , 2 red and 1 blue balls , 3 red balls
– Deepesh Meena
Aug 9 at 23:18
Why do you need to account for the 'other' possibilities the chose function is implied? The ball is either red or blue. I am confused why you multiply them in this instance.
– user9995331
Aug 9 at 23:17
Why do you need to account for the 'other' possibilities the chose function is implied? The ball is either red or blue. I am confused why you multiply them in this instance.
– user9995331
Aug 9 at 23:17
you need to choose three balls but you also want to choose at least 1 read thus every time you need to choose a red with blue balls possibilities are 1red and 2 blue balls , 2 red and 1 blue balls , 3 red balls
– Deepesh Meena
Aug 9 at 23:18
you need to choose three balls but you also want to choose at least 1 read thus every time you need to choose a red with blue balls possibilities are 1red and 2 blue balls , 2 red and 1 blue balls , 3 red balls
– Deepesh Meena
Aug 9 at 23:18
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2877772%2fthere-are-10-marbles-in-a-bag-6-are-red-and-4-are-blue-you-must-chose-at-l%23new-answer', 'question_page');
);
Post as a guest
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
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
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
Why someone wants to close this interesting topic? Explain please your step.
– Michael Rozenberg
Aug 10 at 9:37