Mixing
![How to simplify $intsqrt[4]1-8x^2+8x^4-4xsqrtx^2-1+8x^3sqrtx^2-1dx$?](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least one is yellow OR the first two draws are the same?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least one is yellow OR the first two draws are the same?
I'm not sure my intuition for the solution of the problem is correct, in particular, I have some issue with the addition rule part.
I started out this problem by thinking about the two scenarios:
1) the first two are yellow
2) the first two are blue
I'm a bit confused about the first scenario: P(at least 1 yellow) OR P(first two are the same) if the first two are yellow, that also satisfy the P(at least 1 yellow) so in this case does the probability equal to 1?
Is this similar to flipping a coin P(head) or P(tail)? Or perhaps it's only a "double counted intersection" that we have to subtract out?
I also thought about using compliment to solve this problem but I'm not quite sure if it works this way:
1- [P(no yellow) or P(first two are different)]
thank you!
probability
asked Jul 26 at 21:26
pino231
3268
add a comment |Â
up vote
1
down vote
favorite
20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least one is yellow OR the first two draws are the same?
I'm not sure my intuition for the solution of the problem is correct, in particular, I have some issue with the addition rule part.
I started out this problem by thinking about the two scenarios:
1) the first two are yellow
2) the first two are blue
I'm a bit confused about the first scenario: P(at least 1 yellow) OR P(first two are the same) if the first two are yellow, that also satisfy the P(at least 1 yellow) so in this case does the probability equal to 1?
Is this similar to flipping a coin P(head) or P(tail)? Or perhaps it's only a "double counted intersection" that we have to subtract out?
I also thought about using compliment to solve this problem but I'm not quite sure if it works this way:
1- [P(no yellow) or P(first two are different)]
thank you!
probability
asked Jul 26 at 21:26
pino231
3268
2
Is it a success if both conditions hold? If so, then the answer is $1$ (if there are no yellows drawn then every draw is blue, so, in particular, the first two draws are blue).
– lulu
Jul 26 at 21:31
2
If the first two draws are not the same then there will be at least one yellow
– Henry
Jul 26 at 23:08
OHHHHHH, I get it now, because we only have 2 types of balls to draw from. No matter what we draw we always satisfy at least 1 of the 2 conditions thus the probability is 1!
– pino231
Jul 27 at 2:16
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least one is yellow OR the first two draws are the same?
I'm not sure my intuition for the solution of the problem is correct, in particular, I have some issue with the addition rule part.
I started out this problem by thinking about the two scenarios:
1) the first two are yellow
2) the first two are blue
I'm a bit confused about the first scenario: P(at least 1 yellow) OR P(first two are the same) if the first two are yellow, that also satisfy the P(at least 1 yellow) so in this case does the probability equal to 1?
Is this similar to flipping a coin P(head) or P(tail)? Or perhaps it's only a "double counted intersection" that we have to subtract out?
I also thought about using compliment to solve this problem but I'm not quite sure if it works this way:
1- [P(no yellow) or P(first two are different)]
thank you!
probability
asked Jul 26 at 21:26
pino231
3268
20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least one is yellow OR the first two draws are the same?
I'm not sure my intuition for the solution of the problem is correct, in particular, I have some issue with the addition rule part.
I started out this problem by thinking about the two scenarios:
1) the first two are yellow
2) the first two are blue
I'm a bit confused about the first scenario: P(at least 1 yellow) OR P(first two are the same) if the first two are yellow, that also satisfy the P(at least 1 yellow) so in this case does the probability equal to 1?
Is this similar to flipping a coin P(head) or P(tail)? Or perhaps it's only a "double counted intersection" that we have to subtract out?
I also thought about using compliment to solve this problem but I'm not quite sure if it works this way:
1- [P(no yellow) or P(first two are different)]
thank you!
probability
asked Jul 26 at 21:26
pino231
3268
edited Jul 26 at 23:36
asked Jul 26 at 21:26
pino231
3268
asked Jul 26 at 21:26
pino231
3268
asked Jul 26 at 21:26
pino231
3268
3268
2
Is it a success if both conditions hold? If so, then the answer is $1$ (if there are no yellows drawn then every draw is blue, so, in particular, the first two draws are blue).
– lulu
Jul 26 at 21:31
2
If the first two draws are not the same then there will be at least one yellow
– Henry
Jul 26 at 23:08
OHHHHHH, I get it now, because we only have 2 types of balls to draw from. No matter what we draw we always satisfy at least 1 of the 2 conditions thus the probability is 1!
– pino231
Jul 27 at 2:16
add a comment |Â
2
Is it a success if both conditions hold? If so, then the answer is $1$ (if there are no yellows drawn then every draw is blue, so, in particular, the first two draws are blue).
– lulu
Jul 26 at 21:31
2
If the first two draws are not the same then there will be at least one yellow
– Henry
Jul 26 at 23:08
OHHHHHH, I get it now, because we only have 2 types of balls to draw from. No matter what we draw we always satisfy at least 1 of the 2 conditions thus the probability is 1!
– pino231
Jul 27 at 2:16
2
2
Is it a success if both conditions hold? If so, then the answer is $1$ (if there are no yellows drawn then every draw is blue, so, in particular, the first two draws are blue).
– lulu
Jul 26 at 21:31
Is it a success if both conditions hold? If so, then the answer is $1$ (if there are no yellows drawn then every draw is blue, so, in particular, the first two draws are blue).
– lulu
Jul 26 at 21:31
2
2
If the first two draws are not the same then there will be at least one yellow
– Henry
Jul 26 at 23:08
If the first two draws are not the same then there will be at least one yellow
– Henry
Jul 26 at 23:08
OHHHHHH, I get it now, because we only have 2 types of balls to draw from. No matter what we draw we always satisfy at least 1 of the 2 conditions thus the probability is 1!
– pino231
Jul 27 at 2:16
OHHHHHH, I get it now, because we only have 2 types of balls to draw from. No matter what we draw we always satisfy at least 1 of the 2 conditions thus the probability is 1!
– pino231
Jul 27 at 2:16
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
OR in mathematics is the inclusive or, so if either condition or both is satisfied the sentence is satisfied. OR in English is ambiguous, it can be inclusive or exclusive. Given the problem, I would take it as inclusive here. As you say, that makes the probability $1$. Either the first two balls are blue and hence the same color or at least one of them is yellow. We win either way.
answered Jul 27 at 0:22
Ross Millikan
275k21186351
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
OR in mathematics is the inclusive or, so if either condition or both is satisfied the sentence is satisfied. OR in English is ambiguous, it can be inclusive or exclusive. Given the problem, I would take it as inclusive here. As you say, that makes the probability $1$. Either the first two balls are blue and hence the same color or at least one of them is yellow. We win either way.
answered Jul 27 at 0:22
Ross Millikan
275k21186351
add a comment |Â
up vote
4
down vote
accepted
OR in mathematics is the inclusive or, so if either condition or both is satisfied the sentence is satisfied. OR in English is ambiguous, it can be inclusive or exclusive. Given the problem, I would take it as inclusive here. As you say, that makes the probability $1$. Either the first two balls are blue and hence the same color or at least one of them is yellow. We win either way.
answered Jul 27 at 0:22
Ross Millikan
275k21186351
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
OR in mathematics is the inclusive or, so if either condition or both is satisfied the sentence is satisfied. OR in English is ambiguous, it can be inclusive or exclusive. Given the problem, I would take it as inclusive here. As you say, that makes the probability $1$. Either the first two balls are blue and hence the same color or at least one of them is yellow. We win either way.
answered Jul 27 at 0:22
Ross Millikan
275k21186351
OR in mathematics is the inclusive or, so if either condition or both is satisfied the sentence is satisfied. OR in English is ambiguous, it can be inclusive or exclusive. Given the problem, I would take it as inclusive here. As you say, that makes the probability $1$. Either the first two balls are blue and hence the same color or at least one of them is yellow. We win either way.
answered Jul 27 at 0:22
Ross Millikan
275k21186351
answered Jul 27 at 0:22
Ross Millikan
275k21186351
answered Jul 27 at 0:22
Ross Millikan
275k21186351
answered Jul 27 at 0:22
Ross Millikan
275k21186351
275k21186351
add a comment |Â
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%2f2863837%2f20-blue-balls-and-11-yellow-drawing-6-times-with-no-replacement-what-is-the-ch%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
2
Is it a success if both conditions hold? If so, then the answer is $1$ (if there are no yellows drawn then every draw is blue, so, in particular, the first two draws are blue).
– lulu
Jul 26 at 21:31
2
If the first two draws are not the same then there will be at least one yellow
– Henry
Jul 26 at 23:08
OHHHHHH, I get it now, because we only have 2 types of balls to draw from. No matter what we draw we always satisfy at least 1 of the 2 conditions thus the probability is 1!
– pino231
Jul 27 at 2:16