Mixing
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A bag contains 4 Black and 3 Red balls, 2 balls are drawn one by one without replacement, what is the probability of both being red?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I tried searching the internet regarding the third approach i did, but did not find any explanation for the same, here i have tried approaching this problem in 3 different ways.
approach 1:
since two balls are drawn without replacement, this can be treated as
$P (textselecting $2$ out of $7$ balls such that both are red)$
$=frac^3C_2^7C_2$
approach 2:
$P (textselect 1 R out of 3R )times$ $P (textselect 1 out of 2 remaining R )$
$=frac^3C_17C_1times$$frac^2C_16C_1$
approach 3:
$P (textselecting 1 out of 3 R and then 1 out of 2 R balls in the second draw )$
$=fracN(RR)N(RR)+N(RB)+N(BR)+N(BB)$
$=frac^3C_1times^2C_1^3C_1times^2C_1+^3C_1times^4C_1+^4C_1times^3C_1+^4C_1times^3C_1$
all the above approaches give correct answer, but i am not sure how approach 1 and approach 2 are equivalent,
clearly this is a case of conditional probability without replacement, in the first case, i take 2 balls simultaneously out of 7 and compute probability of both being red, that is same as drawing 2 balls without replacement!
is this the correct way to approach a problem?
in the approach 3 , i tried calculating the sample space where 2 balls are drawn without replacement, where there are four cases:
- Red in first and Red in second draw
- Red in first and Black in second draw
- Black in first and Red in second draw
- Black in first and Black in second draw
I am quite unsure about the validity of approach 3!
I am not sure if this is the right way to approach a problem.
probability conditional-probability
asked 2 hours ago
sonorous
205
New contributor
sonorous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
favorite
I tried searching the internet regarding the third approach i did, but did not find any explanation for the same, here i have tried approaching this problem in 3 different ways.
approach 1:
since two balls are drawn without replacement, this can be treated as
$P (textselecting $2$ out of $7$ balls such that both are red)$
$=frac^3C_2^7C_2$
approach 2:
$P (textselect 1 R out of 3R )times$ $P (textselect 1 out of 2 remaining R )$
$=frac^3C_17C_1times$$frac^2C_16C_1$
approach 3:
$P (textselecting 1 out of 3 R and then 1 out of 2 R balls in the second draw )$
$=fracN(RR)N(RR)+N(RB)+N(BR)+N(BB)$
$=frac^3C_1times^2C_1^3C_1times^2C_1+^3C_1times^4C_1+^4C_1times^3C_1+^4C_1times^3C_1$
all the above approaches give correct answer, but i am not sure how approach 1 and approach 2 are equivalent,
clearly this is a case of conditional probability without replacement, in the first case, i take 2 balls simultaneously out of 7 and compute probability of both being red, that is same as drawing 2 balls without replacement!
is this the correct way to approach a problem?
in the approach 3 , i tried calculating the sample space where 2 balls are drawn without replacement, where there are four cases:
- Red in first and Red in second draw
- Red in first and Black in second draw
- Black in first and Red in second draw
- Black in first and Black in second draw
I am quite unsure about the validity of approach 3!
I am not sure if this is the right way to approach a problem.
probability conditional-probability
asked 2 hours ago
sonorous
205
New contributor
sonorous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Logic is always the right way. All 3 approaches are logical and conform the laws of probability.
– Kristjan Kica
2 hours ago
This is a nice example of how multiple approaches to the same problem, if all are correctly thought through, will come to the same result.
– David K
18 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I tried searching the internet regarding the third approach i did, but did not find any explanation for the same, here i have tried approaching this problem in 3 different ways.
approach 1:
since two balls are drawn without replacement, this can be treated as
$P (textselecting $2$ out of $7$ balls such that both are red)$
$=frac^3C_2^7C_2$
approach 2:
$P (textselect 1 R out of 3R )times$ $P (textselect 1 out of 2 remaining R )$
$=frac^3C_17C_1times$$frac^2C_16C_1$
approach 3:
$P (textselecting 1 out of 3 R and then 1 out of 2 R balls in the second draw )$
$=fracN(RR)N(RR)+N(RB)+N(BR)+N(BB)$
$=frac^3C_1times^2C_1^3C_1times^2C_1+^3C_1times^4C_1+^4C_1times^3C_1+^4C_1times^3C_1$
all the above approaches give correct answer, but i am not sure how approach 1 and approach 2 are equivalent,
clearly this is a case of conditional probability without replacement, in the first case, i take 2 balls simultaneously out of 7 and compute probability of both being red, that is same as drawing 2 balls without replacement!
is this the correct way to approach a problem?
in the approach 3 , i tried calculating the sample space where 2 balls are drawn without replacement, where there are four cases:
- Red in first and Red in second draw
- Red in first and Black in second draw
- Black in first and Red in second draw
- Black in first and Black in second draw
I am quite unsure about the validity of approach 3!
I am not sure if this is the right way to approach a problem.
probability conditional-probability
asked 2 hours ago
sonorous
205
New contributor
sonorous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I tried searching the internet regarding the third approach i did, but did not find any explanation for the same, here i have tried approaching this problem in 3 different ways.
approach 1:
since two balls are drawn without replacement, this can be treated as
$P (textselecting $2$ out of $7$ balls such that both are red)$
$=frac^3C_2^7C_2$
approach 2:
$P (textselect 1 R out of 3R )times$ $P (textselect 1 out of 2 remaining R )$
$=frac^3C_17C_1times$$frac^2C_16C_1$
approach 3:
$P (textselecting 1 out of 3 R and then 1 out of 2 R balls in the second draw )$
$=fracN(RR)N(RR)+N(RB)+N(BR)+N(BB)$
$=frac^3C_1times^2C_1^3C_1times^2C_1+^3C_1times^4C_1+^4C_1times^3C_1+^4C_1times^3C_1$
all the above approaches give correct answer, but i am not sure how approach 1 and approach 2 are equivalent,
clearly this is a case of conditional probability without replacement, in the first case, i take 2 balls simultaneously out of 7 and compute probability of both being red, that is same as drawing 2 balls without replacement!
is this the correct way to approach a problem?
in the approach 3 , i tried calculating the sample space where 2 balls are drawn without replacement, where there are four cases:
- Red in first and Red in second draw
- Red in first and Black in second draw
- Black in first and Red in second draw
- Black in first and Black in second draw
I am quite unsure about the validity of approach 3!
I am not sure if this is the right way to approach a problem.
probability conditional-probability
probability conditional-probability
asked 2 hours ago
sonorous
205
New contributor
sonorous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
sonorous
205
New contributor
sonorous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
asked 2 hours ago
sonorous
205
New contributor
sonorous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
sonorous
205
asked 2 hours ago
sonorous
205
205
New contributor
sonorous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
sonorous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
sonorous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Logic is always the right way. All 3 approaches are logical and conform the laws of probability.
– Kristjan Kica
2 hours ago
This is a nice example of how multiple approaches to the same problem, if all are correctly thought through, will come to the same result.
– David K
18 mins ago
add a comment |Â
Logic is always the right way. All 3 approaches are logical and conform the laws of probability.
– Kristjan Kica
2 hours ago
This is a nice example of how multiple approaches to the same problem, if all are correctly thought through, will come to the same result.
– David K
18 mins ago
Logic is always the right way. All 3 approaches are logical and conform the laws of probability.
– Kristjan Kica
2 hours ago
Logic is always the right way. All 3 approaches are logical and conform the laws of probability.
– Kristjan Kica
2 hours ago
This is a nice example of how multiple approaches to the same problem, if all are correctly thought through, will come to the same result.
– David K
18 mins ago
This is a nice example of how multiple approaches to the same problem, if all are correctly thought through, will come to the same result.
– David K
18 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
All $3$ approaches are okay.
Number the red balls by $1,2,3$ and number the black balls by $4,5,6,7$
Then you could say that approach1 is linked with a probability space where: $$Omega=i,jmid i,jin1,2,3,4,5,6,7text and ineq j$$
Here $|Omega|=binom72=21$ agreeing with the denominator in approach1.
Also you could say that approach3 is linked with a probability space where: $$Omega=langle i,jranglemid i,jin1,2,3,4,5,6,7text and ineq j}$$
Here $|Omega|=7times6=42$ agreeing with the denominator in approach3.
Concerning approach2 (my favorite) both spaces can be used but actually we do not meet any impact of a choice for a probability space.
If $E_i$ denotes the event that the $i$-th ball is red for $i=1,2$ then we apply:$$P(textboth balls are red)=P(E_1cap E_2)=P(E_1)P(E_2mid E_1)=frac37frac26$$
Calculating such probability we are not even aware of any underlying probability space.
answered 2 hours ago
drhab
91.4k542124
in case of approach 2 and approach 3 , denominator is same, is it correct to infer that the denominator in approach 2 is the total number of sample points in the probability space when we approach this using conditional probability. However, it is exciting to see the variation in sample points when approached using approach 1, is it valid to say that selecting 2 red balls one after the other leads to increase in the probability space, which we see in approach 2 and 3 ?
– sonorous
1 hour ago
Actually the approach that is used does not really determine the probability space. It only points in the direction of some underlying probability space in the sense that the calculation of the probability can be (more) easily understood if that space is applied.
– drhab
1 hour ago
You are correct in stating concerning approach2 that the probability space used for approach3 "fits better" than the probability space used in approach1. If $X:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto i$ and $Y:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto j$ then $E_1=Xin1,2,3$ and $E_2=Yin1,2,3$.
– drhab
1 hour ago
add a comment |Â
up vote
0
down vote
Let $R$ represent red balls and $B$ black balls
$P(R)=3/7$
$P(B)=4/7$
Therefore $P(RR) = frac37 cdot frac26$
Ans = $6/42$
edited 10 mins ago
Brahadeesh
4,88931953
answered 25 mins ago
stanley sayanka
1
New contributor
stanley sayanka 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 |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
All $3$ approaches are okay.
Number the red balls by $1,2,3$ and number the black balls by $4,5,6,7$
Then you could say that approach1 is linked with a probability space where: $$Omega=i,jmid i,jin1,2,3,4,5,6,7text and ineq j$$
Here $|Omega|=binom72=21$ agreeing with the denominator in approach1.
Also you could say that approach3 is linked with a probability space where: $$Omega=langle i,jranglemid i,jin1,2,3,4,5,6,7text and ineq j}$$
Here $|Omega|=7times6=42$ agreeing with the denominator in approach3.
Concerning approach2 (my favorite) both spaces can be used but actually we do not meet any impact of a choice for a probability space.
If $E_i$ denotes the event that the $i$-th ball is red for $i=1,2$ then we apply:$$P(textboth balls are red)=P(E_1cap E_2)=P(E_1)P(E_2mid E_1)=frac37frac26$$
Calculating such probability we are not even aware of any underlying probability space.
answered 2 hours ago
drhab
91.4k542124
in case of approach 2 and approach 3 , denominator is same, is it correct to infer that the denominator in approach 2 is the total number of sample points in the probability space when we approach this using conditional probability. However, it is exciting to see the variation in sample points when approached using approach 1, is it valid to say that selecting 2 red balls one after the other leads to increase in the probability space, which we see in approach 2 and 3 ?
– sonorous
1 hour ago
Actually the approach that is used does not really determine the probability space. It only points in the direction of some underlying probability space in the sense that the calculation of the probability can be (more) easily understood if that space is applied.
– drhab
1 hour ago
You are correct in stating concerning approach2 that the probability space used for approach3 "fits better" than the probability space used in approach1. If $X:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto i$ and $Y:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto j$ then $E_1=Xin1,2,3$ and $E_2=Yin1,2,3$.
– drhab
1 hour ago
add a comment |Â
up vote
4
down vote
accepted
All $3$ approaches are okay.
Number the red balls by $1,2,3$ and number the black balls by $4,5,6,7$
Then you could say that approach1 is linked with a probability space where: $$Omega=i,jmid i,jin1,2,3,4,5,6,7text and ineq j$$
Here $|Omega|=binom72=21$ agreeing with the denominator in approach1.
Also you could say that approach3 is linked with a probability space where: $$Omega=langle i,jranglemid i,jin1,2,3,4,5,6,7text and ineq j}$$
Here $|Omega|=7times6=42$ agreeing with the denominator in approach3.
Concerning approach2 (my favorite) both spaces can be used but actually we do not meet any impact of a choice for a probability space.
If $E_i$ denotes the event that the $i$-th ball is red for $i=1,2$ then we apply:$$P(textboth balls are red)=P(E_1cap E_2)=P(E_1)P(E_2mid E_1)=frac37frac26$$
Calculating such probability we are not even aware of any underlying probability space.
answered 2 hours ago
drhab
91.4k542124
in case of approach 2 and approach 3 , denominator is same, is it correct to infer that the denominator in approach 2 is the total number of sample points in the probability space when we approach this using conditional probability. However, it is exciting to see the variation in sample points when approached using approach 1, is it valid to say that selecting 2 red balls one after the other leads to increase in the probability space, which we see in approach 2 and 3 ?
– sonorous
1 hour ago
Actually the approach that is used does not really determine the probability space. It only points in the direction of some underlying probability space in the sense that the calculation of the probability can be (more) easily understood if that space is applied.
– drhab
1 hour ago
You are correct in stating concerning approach2 that the probability space used for approach3 "fits better" than the probability space used in approach1. If $X:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto i$ and $Y:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto j$ then $E_1=Xin1,2,3$ and $E_2=Yin1,2,3$.
– drhab
1 hour ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
All $3$ approaches are okay.
Number the red balls by $1,2,3$ and number the black balls by $4,5,6,7$
Then you could say that approach1 is linked with a probability space where: $$Omega=i,jmid i,jin1,2,3,4,5,6,7text and ineq j$$
Here $|Omega|=binom72=21$ agreeing with the denominator in approach1.
Also you could say that approach3 is linked with a probability space where: $$Omega=langle i,jranglemid i,jin1,2,3,4,5,6,7text and ineq j}$$
Here $|Omega|=7times6=42$ agreeing with the denominator in approach3.
Concerning approach2 (my favorite) both spaces can be used but actually we do not meet any impact of a choice for a probability space.
If $E_i$ denotes the event that the $i$-th ball is red for $i=1,2$ then we apply:$$P(textboth balls are red)=P(E_1cap E_2)=P(E_1)P(E_2mid E_1)=frac37frac26$$
Calculating such probability we are not even aware of any underlying probability space.
answered 2 hours ago
drhab
91.4k542124
All $3$ approaches are okay.
Number the red balls by $1,2,3$ and number the black balls by $4,5,6,7$
Then you could say that approach1 is linked with a probability space where: $$Omega=i,jmid i,jin1,2,3,4,5,6,7text and ineq j$$
Here $|Omega|=binom72=21$ agreeing with the denominator in approach1.
Also you could say that approach3 is linked with a probability space where: $$Omega=langle i,jranglemid i,jin1,2,3,4,5,6,7text and ineq j}$$
Here $|Omega|=7times6=42$ agreeing with the denominator in approach3.
Concerning approach2 (my favorite) both spaces can be used but actually we do not meet any impact of a choice for a probability space.
If $E_i$ denotes the event that the $i$-th ball is red for $i=1,2$ then we apply:$$P(textboth balls are red)=P(E_1cap E_2)=P(E_1)P(E_2mid E_1)=frac37frac26$$
Calculating such probability we are not even aware of any underlying probability space.
answered 2 hours ago
drhab
91.4k542124
edited 1 hour ago
answered 2 hours ago
drhab
91.4k542124
answered 2 hours ago
drhab
91.4k542124
answered 2 hours ago
drhab
91.4k542124
91.4k542124
in case of approach 2 and approach 3 , denominator is same, is it correct to infer that the denominator in approach 2 is the total number of sample points in the probability space when we approach this using conditional probability. However, it is exciting to see the variation in sample points when approached using approach 1, is it valid to say that selecting 2 red balls one after the other leads to increase in the probability space, which we see in approach 2 and 3 ?
– sonorous
1 hour ago
Actually the approach that is used does not really determine the probability space. It only points in the direction of some underlying probability space in the sense that the calculation of the probability can be (more) easily understood if that space is applied.
– drhab
1 hour ago
You are correct in stating concerning approach2 that the probability space used for approach3 "fits better" than the probability space used in approach1. If $X:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto i$ and $Y:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto j$ then $E_1=Xin1,2,3$ and $E_2=Yin1,2,3$.
– drhab
1 hour ago
add a comment |Â
in case of approach 2 and approach 3 , denominator is same, is it correct to infer that the denominator in approach 2 is the total number of sample points in the probability space when we approach this using conditional probability. However, it is exciting to see the variation in sample points when approached using approach 1, is it valid to say that selecting 2 red balls one after the other leads to increase in the probability space, which we see in approach 2 and 3 ?
– sonorous
1 hour ago
Actually the approach that is used does not really determine the probability space. It only points in the direction of some underlying probability space in the sense that the calculation of the probability can be (more) easily understood if that space is applied.
– drhab
1 hour ago
You are correct in stating concerning approach2 that the probability space used for approach3 "fits better" than the probability space used in approach1. If $X:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto i$ and $Y:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto j$ then $E_1=Xin1,2,3$ and $E_2=Yin1,2,3$.
– drhab
1 hour ago
in case of approach 2 and approach 3 , denominator is same, is it correct to infer that the denominator in approach 2 is the total number of sample points in the probability space when we approach this using conditional probability. However, it is exciting to see the variation in sample points when approached using approach 1, is it valid to say that selecting 2 red balls one after the other leads to increase in the probability space, which we see in approach 2 and 3 ?
– sonorous
1 hour ago
in case of approach 2 and approach 3 , denominator is same, is it correct to infer that the denominator in approach 2 is the total number of sample points in the probability space when we approach this using conditional probability. However, it is exciting to see the variation in sample points when approached using approach 1, is it valid to say that selecting 2 red balls one after the other leads to increase in the probability space, which we see in approach 2 and 3 ?
– sonorous
1 hour ago
Actually the approach that is used does not really determine the probability space. It only points in the direction of some underlying probability space in the sense that the calculation of the probability can be (more) easily understood if that space is applied.
– drhab
1 hour ago
Actually the approach that is used does not really determine the probability space. It only points in the direction of some underlying probability space in the sense that the calculation of the probability can be (more) easily understood if that space is applied.
– drhab
1 hour ago
You are correct in stating concerning approach2 that the probability space used for approach3 "fits better" than the probability space used in approach1. If $X:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto i$ and $Y:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto j$ then $E_1=Xin1,2,3$ and $E_2=Yin1,2,3$.
– drhab
1 hour ago
You are correct in stating concerning approach2 that the probability space used for approach3 "fits better" than the probability space used in approach1. If $X:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto i$ and $Y:Omegatomathbb R$ is prescribed by $langle i,jranglemapsto j$ then $E_1=Xin1,2,3$ and $E_2=Yin1,2,3$.
– drhab
1 hour ago
add a comment |Â
up vote
0
down vote
Let $R$ represent red balls and $B$ black balls
$P(R)=3/7$
$P(B)=4/7$
Therefore $P(RR) = frac37 cdot frac26$
Ans = $6/42$
edited 10 mins ago
Brahadeesh
4,88931953
answered 25 mins ago
stanley sayanka
1
New contributor
stanley sayanka 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 |Â
up vote
0
down vote
Let $R$ represent red balls and $B$ black balls
$P(R)=3/7$
$P(B)=4/7$
Therefore $P(RR) = frac37 cdot frac26$
Ans = $6/42$
edited 10 mins ago
Brahadeesh
4,88931953
answered 25 mins ago
stanley sayanka
1
New contributor
stanley sayanka 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 |Â
up vote
0
down vote
up vote
0
down vote
Let $R$ represent red balls and $B$ black balls
$P(R)=3/7$
$P(B)=4/7$
Therefore $P(RR) = frac37 cdot frac26$
Ans = $6/42$
edited 10 mins ago
Brahadeesh
4,88931953
answered 25 mins ago
stanley sayanka
1
New contributor
stanley sayanka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $R$ represent red balls and $B$ black balls
$P(R)=3/7$
$P(B)=4/7$
Therefore $P(RR) = frac37 cdot frac26$
Ans = $6/42$
edited 10 mins ago
Brahadeesh
4,88931953
answered 25 mins ago
stanley sayanka
1
New contributor
stanley sayanka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 10 mins ago
Brahadeesh
4,88931953
edited 10 mins ago
Brahadeesh
4,88931953
edited 10 mins ago
Brahadeesh
4,88931953
4,88931953
answered 25 mins ago
stanley sayanka
1
New contributor
stanley sayanka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 25 mins ago
stanley sayanka
1
answered 25 mins ago
stanley sayanka
1
1
New contributor
stanley sayanka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
stanley sayanka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
stanley sayanka 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 |Â
add a comment |Â
sonorous is a new contributor. Be nice, and check out our Code of Conduct.
sonorous is a new contributor. Be nice, and check out our Code of Conduct.
sonorous is a new contributor. Be nice, and check out our Code of Conduct.
sonorous is a new contributor. Be nice, and check out our Code of Conduct.
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Logic is always the right way. All 3 approaches are logical and conform the laws of probability.
– Kristjan Kica
2 hours ago
This is a nice example of how multiple approaches to the same problem, if all are correctly thought through, will come to the same result.
– David K
18 mins ago