A bag contains 3 red, 4 blue, and 5 green balls. [closed]

Clash Royale CLAN TAG#URR8PPP

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Peter draws a ball from the bag, and then Angelina draws a ball. What is the probability that Angelina got a green ball?

So far I have this:

Scenario A: 1st ball is not green, 2nd green: 7/12 * 5/11 = 35/132

Scenario B: 1st ball is green, 2nd green: 5/12 * 4/11 = 20/132

--> $$frac55132 = frac512$$

probability combinatorics

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edited Jul 25 at 18:52

asked Jul 25 at 18:37

Wolf

1466

closed as off-topic by asdf, gt6989b, Parcly Taxel, Xander Henderson, Asaf Karagila Jul 26 at 9:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – asdf, gt6989b, Parcly Taxel, Xander Henderson, Asaf Karagila

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  So the first ball is green or it is not green...
  – Robert Z
  Jul 25 at 18:46
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Yes, you are correct!
  – Robert Z
  Jul 25 at 18:47
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  correction Wolf: : You have $$frac20 + 35132 = frac55132= frac 512$$
  – amWhy
  Jul 25 at 18:48
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $$P=P_1+P_2=frac 511frac 712+frac 411frac 512=frac512 times 11(4+7)=frac512 times 11(11)=frac 512$$
  – Isham
  Jul 25 at 18:50
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Possible duplicate of Second marble is of same color
  – Rahul Goswami
  Jul 25 at 19:11
  

  
  
  
  

 | 
show 9 more comments

up vote
8
down vote

favorite

1

Peter draws a ball from the bag, and then Angelina draws a ball. What is the probability that Angelina got a green ball?

So far I have this:

Scenario A: 1st ball is not green, 2nd green: 7/12 * 5/11 = 35/132

Scenario B: 1st ball is green, 2nd green: 5/12 * 4/11 = 20/132

--> $$frac55132 = frac512$$

probability combinatorics

share|cite|improve this question

edited Jul 25 at 18:52

asked Jul 25 at 18:37

Wolf

1466

closed as off-topic by asdf, gt6989b, Parcly Taxel, Xander Henderson, Asaf Karagila Jul 26 at 9:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – asdf, gt6989b, Parcly Taxel, Xander Henderson, Asaf Karagila

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  So the first ball is green or it is not green...
  – Robert Z
  Jul 25 at 18:46
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Yes, you are correct!
  – Robert Z
  Jul 25 at 18:47
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  correction Wolf: : You have $$frac20 + 35132 = frac55132= frac 512$$
  – amWhy
  Jul 25 at 18:48
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $$P=P_1+P_2=frac 511frac 712+frac 411frac 512=frac512 times 11(4+7)=frac512 times 11(11)=frac 512$$
  – Isham
  Jul 25 at 18:50
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Possible duplicate of Second marble is of same color
  – Rahul Goswami
  Jul 25 at 19:11
  

  
  
  
  

 | 
show 9 more comments

up vote
8
down vote

favorite

1

up vote
8
down vote

favorite

1

1

Peter draws a ball from the bag, and then Angelina draws a ball. What is the probability that Angelina got a green ball?

So far I have this:

Scenario A: 1st ball is not green, 2nd green: 7/12 * 5/11 = 35/132

Scenario B: 1st ball is green, 2nd green: 5/12 * 4/11 = 20/132

--> $$frac55132 = frac512$$

probability combinatorics

share|cite|improve this question

edited Jul 25 at 18:52

asked Jul 25 at 18:37

Wolf

1466

Peter draws a ball from the bag, and then Angelina draws a ball. What is the probability that Angelina got a green ball?

So far I have this:

Scenario A: 1st ball is not green, 2nd green: 7/12 * 5/11 = 35/132

Scenario B: 1st ball is green, 2nd green: 5/12 * 4/11 = 20/132

--> $$frac55132 = frac512$$

probability combinatorics

share|cite|improve this question

edited Jul 25 at 18:52

asked Jul 25 at 18:37

Wolf

1466

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 18:52

edited Jul 25 at 18:52

edited Jul 25 at 18:52

asked Jul 25 at 18:37

Wolf

1466

asked Jul 25 at 18:37

Wolf

1466

asked Jul 25 at 18:37

Wolf

1466

1466

closed as off-topic by asdf, gt6989b, Parcly Taxel, Xander Henderson, Asaf Karagila Jul 26 at 9:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – asdf, gt6989b, Parcly Taxel, Xander Henderson, Asaf Karagila

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by asdf, gt6989b, Parcly Taxel, Xander Henderson, Asaf Karagila Jul 26 at 9:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – asdf, gt6989b, Parcly Taxel, Xander Henderson, Asaf Karagila

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  So the first ball is green or it is not green...
  – Robert Z
  Jul 25 at 18:46
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Yes, you are correct!
  – Robert Z
  Jul 25 at 18:47
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  correction Wolf: : You have $$frac20 + 35132 = frac55132= frac 512$$
  – amWhy
  Jul 25 at 18:48
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $$P=P_1+P_2=frac 511frac 712+frac 411frac 512=frac512 times 11(4+7)=frac512 times 11(11)=frac 512$$
  – Isham
  Jul 25 at 18:50
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Possible duplicate of Second marble is of same color
  – Rahul Goswami
  Jul 25 at 19:11
  

  
  
  
  

 | 
show 9 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  So the first ball is green or it is not green...
  – Robert Z
  Jul 25 at 18:46
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Yes, you are correct!
  – Robert Z
  Jul 25 at 18:47
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  correction Wolf: : You have $$frac20 + 35132 = frac55132= frac 512$$
  – amWhy
  Jul 25 at 18:48
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $$P=P_1+P_2=frac 511frac 712+frac 411frac 512=frac512 times 11(4+7)=frac512 times 11(11)=frac 512$$
  – Isham
  Jul 25 at 18:50
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Possible duplicate of Second marble is of same color
  – Rahul Goswami
  Jul 25 at 19:11
  

  
  
  
  

1

1

So the first ball is green or it is not green...
– Robert Z
Jul 25 at 18:46

So the first ball is green or it is not green...
– Robert Z
Jul 25 at 18:46

3

3

Yes, you are correct!
– Robert Z
Jul 25 at 18:47

Yes, you are correct!
– Robert Z
Jul 25 at 18:47

2

2

correction Wolf: : You have $$frac20 + 35132 = frac55132= frac 512$$
– amWhy
Jul 25 at 18:48

correction Wolf: : You have $$frac20 + 35132 = frac55132= frac 512$$
– amWhy
Jul 25 at 18:48

2

2

$$P=P_1+P_2=frac 511frac 712+frac 411frac 512=frac512 times 11(4+7)=frac512 times 11(11)=frac 512$$
– Isham
Jul 25 at 18:50

$$P=P_1+P_2=frac 511frac 712+frac 411frac 512=frac512 times 11(4+7)=frac512 times 11(11)=frac 512$$
– Isham
Jul 25 at 18:50

4

4

Possible duplicate of Second marble is of same color
– Rahul Goswami
Jul 25 at 19:11

Possible duplicate of Second marble is of same color
– Rahul Goswami
Jul 25 at 19:11

 | 
show 9 more comments

3 Answers
3

active

oldest

votes

up vote
20
down vote



accepted

Your approach is fine, but it's worth realizing that unless you know something about the ball Peter has chosen, he may as well not have chosen at all, in which case Angelina simply has a $5/12$ chance of choosing a green ball.

Where this point really becomes worth understanding is if, for example, not only Peter, but also Olivia, Ned, Melissa, and Larry choose balls before Angelina. The tree of possibilities is tedious and messy, but the answer is still $5/12$.

share|cite|improve this answer

edited Jul 25 at 18:59

answered Jul 25 at 18:51

Barry Cipra

56.4k652118

add a comment | 

up vote
8
down vote

Angela's draw is actually independent from the order she draws the ball whether she is first, second, ..., or twelfth (surprisingly enough). So, the probability of her drawing a green ball is actually $dfrac512$. To give you a better explanation, suppose there are $k$ green balls and $n$ total balls. Then, the probability that Angela draws a green ball is:

$$requirecancel dfrackndfrack-1n-1+dfracn-kndfrackn-1 = dfrack(k-1)+k(n-k)n(n-1) = dfrack cancel(n-1)n cancel(n-1) = dfrackn$$

Which is the same probability that Peter has for drawing a green ball.

share|cite|improve this answer

edited Jul 25 at 22:50

answered Jul 25 at 18:51

InterstellarProbe

2,680516

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Angela's draw is actually independent from Peter's draw" - it's not. You're probably misusing the word "independent". The probability distribution for Angelina's draw is the same as it would be if Peter did not draw, but that's not what "independent" means. If they were independent, we would have equalities like P(Angelina draws red AND Peter draws red) = P(Angelina draws red)*P(Peter draws red).
  – user2357112
  Jul 25 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user2357112 I corrected my wording. Thank you for pointing it out.
  – InterstellarProbe
  Jul 25 at 22:51
  

  
  
  
  

add a comment | 

up vote
2
down vote

HINT

Brute force: condition on the color of the first drawn ball, there are 3 cases. Find the probability of each case happening, and what is the resulting probability in each case.

Then group them together in one final expression.

share|cite|improve this answer

answered Jul 25 at 18:40

gt6989b

30.3k22148

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
20
down vote



accepted

Your approach is fine, but it's worth realizing that unless you know something about the ball Peter has chosen, he may as well not have chosen at all, in which case Angelina simply has a $5/12$ chance of choosing a green ball.

Where this point really becomes worth understanding is if, for example, not only Peter, but also Olivia, Ned, Melissa, and Larry choose balls before Angelina. The tree of possibilities is tedious and messy, but the answer is still $5/12$.

share|cite|improve this answer

edited Jul 25 at 18:59

answered Jul 25 at 18:51

Barry Cipra

56.4k652118

add a comment | 

up vote
20
down vote



accepted

Your approach is fine, but it's worth realizing that unless you know something about the ball Peter has chosen, he may as well not have chosen at all, in which case Angelina simply has a $5/12$ chance of choosing a green ball.

Where this point really becomes worth understanding is if, for example, not only Peter, but also Olivia, Ned, Melissa, and Larry choose balls before Angelina. The tree of possibilities is tedious and messy, but the answer is still $5/12$.

share|cite|improve this answer

edited Jul 25 at 18:59

answered Jul 25 at 18:51

Barry Cipra

56.4k652118

add a comment | 

up vote
20
down vote



accepted

up vote
20
down vote



accepted

Your approach is fine, but it's worth realizing that unless you know something about the ball Peter has chosen, he may as well not have chosen at all, in which case Angelina simply has a $5/12$ chance of choosing a green ball.

Where this point really becomes worth understanding is if, for example, not only Peter, but also Olivia, Ned, Melissa, and Larry choose balls before Angelina. The tree of possibilities is tedious and messy, but the answer is still $5/12$.

share|cite|improve this answer

edited Jul 25 at 18:59

answered Jul 25 at 18:51

Barry Cipra

56.4k652118

Your approach is fine, but it's worth realizing that unless you know something about the ball Peter has chosen, he may as well not have chosen at all, in which case Angelina simply has a $5/12$ chance of choosing a green ball.

Where this point really becomes worth understanding is if, for example, not only Peter, but also Olivia, Ned, Melissa, and Larry choose balls before Angelina. The tree of possibilities is tedious and messy, but the answer is still $5/12$.

share|cite|improve this answer

edited Jul 25 at 18:59

answered Jul 25 at 18:51

Barry Cipra

56.4k652118

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 18:59

edited Jul 25 at 18:59

edited Jul 25 at 18:59

answered Jul 25 at 18:51

Barry Cipra

56.4k652118

answered Jul 25 at 18:51

Barry Cipra

56.4k652118

answered Jul 25 at 18:51

Barry Cipra

56.4k652118

56.4k652118

add a comment | 

add a comment | 

up vote
8
down vote

Angela's draw is actually independent from the order she draws the ball whether she is first, second, ..., or twelfth (surprisingly enough). So, the probability of her drawing a green ball is actually $dfrac512$. To give you a better explanation, suppose there are $k$ green balls and $n$ total balls. Then, the probability that Angela draws a green ball is:

$$requirecancel dfrackndfrack-1n-1+dfracn-kndfrackn-1 = dfrack(k-1)+k(n-k)n(n-1) = dfrack cancel(n-1)n cancel(n-1) = dfrackn$$

Which is the same probability that Peter has for drawing a green ball.

share|cite|improve this answer

edited Jul 25 at 22:50

answered Jul 25 at 18:51

InterstellarProbe

2,680516

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Angela's draw is actually independent from Peter's draw" - it's not. You're probably misusing the word "independent". The probability distribution for Angelina's draw is the same as it would be if Peter did not draw, but that's not what "independent" means. If they were independent, we would have equalities like P(Angelina draws red AND Peter draws red) = P(Angelina draws red)*P(Peter draws red).
  – user2357112
  Jul 25 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user2357112 I corrected my wording. Thank you for pointing it out.
  – InterstellarProbe
  Jul 25 at 22:51
  

  
  
  
  

add a comment | 

up vote
8
down vote

Angela's draw is actually independent from the order she draws the ball whether she is first, second, ..., or twelfth (surprisingly enough). So, the probability of her drawing a green ball is actually $dfrac512$. To give you a better explanation, suppose there are $k$ green balls and $n$ total balls. Then, the probability that Angela draws a green ball is:

$$requirecancel dfrackndfrack-1n-1+dfracn-kndfrackn-1 = dfrack(k-1)+k(n-k)n(n-1) = dfrack cancel(n-1)n cancel(n-1) = dfrackn$$

Which is the same probability that Peter has for drawing a green ball.

share|cite|improve this answer

edited Jul 25 at 22:50

answered Jul 25 at 18:51

InterstellarProbe

2,680516

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Angela's draw is actually independent from Peter's draw" - it's not. You're probably misusing the word "independent". The probability distribution for Angelina's draw is the same as it would be if Peter did not draw, but that's not what "independent" means. If they were independent, we would have equalities like P(Angelina draws red AND Peter draws red) = P(Angelina draws red)*P(Peter draws red).
  – user2357112
  Jul 25 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user2357112 I corrected my wording. Thank you for pointing it out.
  – InterstellarProbe
  Jul 25 at 22:51
  

  
  
  
  

add a comment | 

up vote
8
down vote

up vote
8
down vote

Angela's draw is actually independent from the order she draws the ball whether she is first, second, ..., or twelfth (surprisingly enough). So, the probability of her drawing a green ball is actually $dfrac512$. To give you a better explanation, suppose there are $k$ green balls and $n$ total balls. Then, the probability that Angela draws a green ball is:

$$requirecancel dfrackndfrack-1n-1+dfracn-kndfrackn-1 = dfrack(k-1)+k(n-k)n(n-1) = dfrack cancel(n-1)n cancel(n-1) = dfrackn$$

Which is the same probability that Peter has for drawing a green ball.

share|cite|improve this answer

edited Jul 25 at 22:50

answered Jul 25 at 18:51

InterstellarProbe

2,680516

Angela's draw is actually independent from the order she draws the ball whether she is first, second, ..., or twelfth (surprisingly enough). So, the probability of her drawing a green ball is actually $dfrac512$. To give you a better explanation, suppose there are $k$ green balls and $n$ total balls. Then, the probability that Angela draws a green ball is:

$$requirecancel dfrackndfrack-1n-1+dfracn-kndfrackn-1 = dfrack(k-1)+k(n-k)n(n-1) = dfrack cancel(n-1)n cancel(n-1) = dfrackn$$

Which is the same probability that Peter has for drawing a green ball.

share|cite|improve this answer

edited Jul 25 at 22:50

answered Jul 25 at 18:51

InterstellarProbe

2,680516

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 22:50

edited Jul 25 at 22:50

edited Jul 25 at 22:50

answered Jul 25 at 18:51

InterstellarProbe

2,680516

answered Jul 25 at 18:51

InterstellarProbe

2,680516

answered Jul 25 at 18:51

InterstellarProbe

2,680516

2,680516

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Angela's draw is actually independent from Peter's draw" - it's not. You're probably misusing the word "independent". The probability distribution for Angelina's draw is the same as it would be if Peter did not draw, but that's not what "independent" means. If they were independent, we would have equalities like P(Angelina draws red AND Peter draws red) = P(Angelina draws red)*P(Peter draws red).
  – user2357112
  Jul 25 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user2357112 I corrected my wording. Thank you for pointing it out.
  – InterstellarProbe
  Jul 25 at 22:51
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Angela's draw is actually independent from Peter's draw" - it's not. You're probably misusing the word "independent". The probability distribution for Angelina's draw is the same as it would be if Peter did not draw, but that's not what "independent" means. If they were independent, we would have equalities like P(Angelina draws red AND Peter draws red) = P(Angelina draws red)*P(Peter draws red).
  – user2357112
  Jul 25 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user2357112 I corrected my wording. Thank you for pointing it out.
  – InterstellarProbe
  Jul 25 at 22:51
  

  
  
  
  

1

1

"Angela's draw is actually independent from Peter's draw" - it's not. You're probably misusing the word "independent". The probability distribution for Angelina's draw is the same as it would be if Peter did not draw, but that's not what "independent" means. If they were independent, we would have equalities like P(Angelina draws red AND Peter draws red) = P(Angelina draws red)*P(Peter draws red).
– user2357112
Jul 25 at 22:04

"Angela's draw is actually independent from Peter's draw" - it's not. You're probably misusing the word "independent". The probability distribution for Angelina's draw is the same as it would be if Peter did not draw, but that's not what "independent" means. If they were independent, we would have equalities like P(Angelina draws red AND Peter draws red) = P(Angelina draws red)*P(Peter draws red).
– user2357112
Jul 25 at 22:04

@user2357112 I corrected my wording. Thank you for pointing it out.
– InterstellarProbe
Jul 25 at 22:51

@user2357112 I corrected my wording. Thank you for pointing it out.
– InterstellarProbe
Jul 25 at 22:51

add a comment | 

up vote
2
down vote

HINT

Brute force: condition on the color of the first drawn ball, there are 3 cases. Find the probability of each case happening, and what is the resulting probability in each case.

Then group them together in one final expression.

share|cite|improve this answer

answered Jul 25 at 18:40

gt6989b

30.3k22148

add a comment | 

up vote
2
down vote

HINT

Brute force: condition on the color of the first drawn ball, there are 3 cases. Find the probability of each case happening, and what is the resulting probability in each case.

Then group them together in one final expression.

share|cite|improve this answer

answered Jul 25 at 18:40

gt6989b

30.3k22148

add a comment | 

up vote
2
down vote

up vote
2
down vote

HINT

Brute force: condition on the color of the first drawn ball, there are 3 cases. Find the probability of each case happening, and what is the resulting probability in each case.

Then group them together in one final expression.

share|cite|improve this answer

answered Jul 25 at 18:40

gt6989b

30.3k22148

HINT

Brute force: condition on the color of the first drawn ball, there are 3 cases. Find the probability of each case happening, and what is the resulting probability in each case.

Then group them together in one final expression.

share|cite|improve this answer

answered Jul 25 at 18:40

gt6989b

30.3k22148

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 18:40

gt6989b

30.3k22148

answered Jul 25 at 18:40

gt6989b

30.3k22148

answered Jul 25 at 18:40

gt6989b

30.3k22148

30.3k22148

add a comment | 

add a comment |