Mixing
Probability Inclusion-Exclusion With 3 Events
$begingroup$
The question is: An urn contains 4 balls: 1 white, 1 green, and 2 red. We draw 3 balls with replacement. Find the probability we did not see all three colors.
I need to define the events as
W= {white ball did not appear} and similarly for R and G, while specifically using inclusion-exclusion to solve the problem.
My first thought was to use the identity P(A $cup$ B) = P(A) + P(B) - P(A $cap$ B) and solve for (A $cap$ B) but I would only be able to use it for two colors at a time.
probability probability-theory problem-solving inclusion-exclusion
asked Jan 26 at 21:22
jerrythespherejerrythesphere
1
$endgroup$
add a comment |
$begingroup$
The question is: An urn contains 4 balls: 1 white, 1 green, and 2 red. We draw 3 balls with replacement. Find the probability we did not see all three colors.
I need to define the events as
W= {white ball did not appear} and similarly for R and G, while specifically using inclusion-exclusion to solve the problem.
My first thought was to use the identity P(A $cup$ B) = P(A) + P(B) - P(A $cap$ B) and solve for (A $cap$ B) but I would only be able to use it for two colors at a time.
probability probability-theory problem-solving inclusion-exclusion
asked Jan 26 at 21:22
jerrythespherejerrythesphere
1
$endgroup$
2
$begingroup$
It is much easier to use the converse probability: probability not seeing all three colors is equal to One minus the probability seeing all three colors.
$endgroup$
– callculus
Jan 26 at 21:31
$begingroup$
I have to use inclusion-exclusion; this is a multi-part problem and it specifies doing it this way
$endgroup$
– jerrythesphere
Jan 26 at 22:04
$begingroup$
You have to show some effort. What result do you get if you follow my advice?
$endgroup$
– callculus
Jan 26 at 22:07
$begingroup$
OK. Now it´s clear that you have to use the inclusion-exclusion principle. But with my advice you can double check the result which you get with the inclusion-exclusion principle. You just have to start to do something.
$endgroup$
– callculus
Jan 26 at 22:19
add a comment |
$begingroup$
The question is: An urn contains 4 balls: 1 white, 1 green, and 2 red. We draw 3 balls with replacement. Find the probability we did not see all three colors.
I need to define the events as
W= {white ball did not appear} and similarly for R and G, while specifically using inclusion-exclusion to solve the problem.
My first thought was to use the identity P(A $cup$ B) = P(A) + P(B) - P(A $cap$ B) and solve for (A $cap$ B) but I would only be able to use it for two colors at a time.
probability probability-theory problem-solving inclusion-exclusion
asked Jan 26 at 21:22
jerrythespherejerrythesphere
1
$endgroup$
The question is: An urn contains 4 balls: 1 white, 1 green, and 2 red. We draw 3 balls with replacement. Find the probability we did not see all three colors.
I need to define the events as
W= {white ball did not appear} and similarly for R and G, while specifically using inclusion-exclusion to solve the problem.
My first thought was to use the identity P(A $cup$ B) = P(A) + P(B) - P(A $cap$ B) and solve for (A $cap$ B) but I would only be able to use it for two colors at a time.
probability probability-theory problem-solving inclusion-exclusion
probability probability-theory problem-solving inclusion-exclusion
asked Jan 26 at 21:22
jerrythespherejerrythesphere
1
asked Jan 26 at 21:22
jerrythespherejerrythesphere
1
asked Jan 26 at 21:22
jerrythespherejerrythesphere
1
asked Jan 26 at 21:22
jerrythespherejerrythesphere
1
asked Jan 26 at 21:22
jerrythespherejerrythesphere
1
1
2
$begingroup$
It is much easier to use the converse probability: probability not seeing all three colors is equal to One minus the probability seeing all three colors.
$endgroup$
– callculus
Jan 26 at 21:31
$begingroup$
I have to use inclusion-exclusion; this is a multi-part problem and it specifies doing it this way
$endgroup$
– jerrythesphere
Jan 26 at 22:04
$begingroup$
You have to show some effort. What result do you get if you follow my advice?
$endgroup$
– callculus
Jan 26 at 22:07
$begingroup$
OK. Now it´s clear that you have to use the inclusion-exclusion principle. But with my advice you can double check the result which you get with the inclusion-exclusion principle. You just have to start to do something.
$endgroup$
– callculus
Jan 26 at 22:19
add a comment |
2
$begingroup$
It is much easier to use the converse probability: probability not seeing all three colors is equal to One minus the probability seeing all three colors.
$endgroup$
– callculus
Jan 26 at 21:31
$begingroup$
I have to use inclusion-exclusion; this is a multi-part problem and it specifies doing it this way
$endgroup$
– jerrythesphere
Jan 26 at 22:04
$begingroup$
You have to show some effort. What result do you get if you follow my advice?
$endgroup$
– callculus
Jan 26 at 22:07
$begingroup$
OK. Now it´s clear that you have to use the inclusion-exclusion principle. But with my advice you can double check the result which you get with the inclusion-exclusion principle. You just have to start to do something.
$endgroup$
– callculus
Jan 26 at 22:19
2
2
$begingroup$
It is much easier to use the converse probability: probability not seeing all three colors is equal to One minus the probability seeing all three colors.
$endgroup$
– callculus
Jan 26 at 21:31
$begingroup$
It is much easier to use the converse probability: probability not seeing all three colors is equal to One minus the probability seeing all three colors.
$endgroup$
– callculus
Jan 26 at 21:31
$begingroup$
I have to use inclusion-exclusion; this is a multi-part problem and it specifies doing it this way
$endgroup$
– jerrythesphere
Jan 26 at 22:04
$begingroup$
I have to use inclusion-exclusion; this is a multi-part problem and it specifies doing it this way
$endgroup$
– jerrythesphere
Jan 26 at 22:04
$begingroup$
You have to show some effort. What result do you get if you follow my advice?
$endgroup$
– callculus
Jan 26 at 22:07
$begingroup$
You have to show some effort. What result do you get if you follow my advice?
$endgroup$
– callculus
Jan 26 at 22:07
$begingroup$
OK. Now it´s clear that you have to use the inclusion-exclusion principle. But with my advice you can double check the result which you get with the inclusion-exclusion principle. You just have to start to do something.
$endgroup$
– callculus
Jan 26 at 22:19
$begingroup$
OK. Now it´s clear that you have to use the inclusion-exclusion principle. But with my advice you can double check the result which you get with the inclusion-exclusion principle. You just have to start to do something.
$endgroup$
– callculus
Jan 26 at 22:19
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here is how the principle of inclusion-exclusion looks with three events:
$begin{align*}
Pr(Wcup Rcup G) &= Pr(W)+ Pr(R)+ Pr(G)\
&quad-Pr(Wcap R)-Pr(Wcap G)-Pr(Gcap R)\
&qquad+ Pr(Wcap Rcap G)
end{align*}$
It’s up to you to compute each of the terms on the RHS.
answered Jan 26 at 22:09
Laars HeleniusLaars Helenius
6,3111423
$endgroup$
$begingroup$
I understand that P(W), P(G) and P(R) represent the probabilities that white, red and green do not show up, but what would P(W $cup$ R) represent? I assume it would be the outcomes that W and R have in common meaning only having green appear, so P(W $cup$ R) = $frac{2*2}{4*4}$
$endgroup$
– jerrythesphere
Jan 27 at 17:33
$begingroup$
@jerrythesphere For $P(Wcap R)$ you just have to multiply the probabilities that the i-th draw is green. The probablitly to pick a green marble at the i-th draw is $P(g)=frac{1}{1+1+2}=frac14$. Similar for $P(w)$ and $P(r)$. The sum of these probabilities has to be $1$.
$endgroup$
– callculus
Jan 27 at 18:12
add a comment |
$begingroup$
The principle of inclusion and exclusion works for any number of events:
$P(Acup Bcup C)=P(A)+P(B)+P(C)-P(Acap B)-P(Acap C)-P(Bcap C)+P(Acap Bcap C)$$
answered Jan 26 at 22:09
saulspatzsaulspatz
17.1k31435
$endgroup$
add a comment |
$begingroup$
Let $W,R,G$ the events, that the white, red and green marbles do not show up.
Then the number of combinations for W are:
$a) gggrightarrow 1, b) ggrrightarrow 3, c) rrgrightarrow 3, d) rrrrightarrow 1$
The corresponding probabilities are
$a) frac{1}{4}cdot frac{1}{4}cdot frac{1}{4}= frac{1}{64}$
$b) 3cdot frac{1}{4}cdot frac{1}{4}cdot frac{1}{2}= frac{3}{32}$
$c) 3cdot frac{1}{2}cdot frac{1}{2}cdot frac{1}{4}= frac{3}{16}$
$d) frac{1}{2}cdot frac{1}{2}cdot frac{1}{2}= frac{1}{8}$
Therefore $P(W)=frac{27}{64}$
Now what ist $P(Wcap R)$?
$Wcap R$ means that no white marbles and no red marbles show up. That is when only green marbles show up-as you´ve already said: $ ggg$
$P(Wcap R)=frac{1}{4}cdot frac{1}{4}cdot frac{1}{4}= frac{1}{64}$
You can check your final result by using the converse probability:
$P(textrm{"Do not see all three colors"})=1-P(textrm{"See all three colors"})$
answered Jan 27 at 18:02
callculuscallculus
18.5k31428
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is how the principle of inclusion-exclusion looks with three events:
$begin{align*}
Pr(Wcup Rcup G) &= Pr(W)+ Pr(R)+ Pr(G)\
&quad-Pr(Wcap R)-Pr(Wcap G)-Pr(Gcap R)\
&qquad+ Pr(Wcap Rcap G)
end{align*}$
It’s up to you to compute each of the terms on the RHS.
answered Jan 26 at 22:09
Laars HeleniusLaars Helenius
6,3111423
$endgroup$
$begingroup$
I understand that P(W), P(G) and P(R) represent the probabilities that white, red and green do not show up, but what would P(W $cup$ R) represent? I assume it would be the outcomes that W and R have in common meaning only having green appear, so P(W $cup$ R) = $frac{2*2}{4*4}$
$endgroup$
– jerrythesphere
Jan 27 at 17:33
$begingroup$
@jerrythesphere For $P(Wcap R)$ you just have to multiply the probabilities that the i-th draw is green. The probablitly to pick a green marble at the i-th draw is $P(g)=frac{1}{1+1+2}=frac14$. Similar for $P(w)$ and $P(r)$. The sum of these probabilities has to be $1$.
$endgroup$
– callculus
Jan 27 at 18:12
add a comment |
$begingroup$
Here is how the principle of inclusion-exclusion looks with three events:
$begin{align*}
Pr(Wcup Rcup G) &= Pr(W)+ Pr(R)+ Pr(G)\
&quad-Pr(Wcap R)-Pr(Wcap G)-Pr(Gcap R)\
&qquad+ Pr(Wcap Rcap G)
end{align*}$
It’s up to you to compute each of the terms on the RHS.
answered Jan 26 at 22:09
Laars HeleniusLaars Helenius
6,3111423
$endgroup$
$begingroup$
I understand that P(W), P(G) and P(R) represent the probabilities that white, red and green do not show up, but what would P(W $cup$ R) represent? I assume it would be the outcomes that W and R have in common meaning only having green appear, so P(W $cup$ R) = $frac{2*2}{4*4}$
$endgroup$
– jerrythesphere
Jan 27 at 17:33
$begingroup$
@jerrythesphere For $P(Wcap R)$ you just have to multiply the probabilities that the i-th draw is green. The probablitly to pick a green marble at the i-th draw is $P(g)=frac{1}{1+1+2}=frac14$. Similar for $P(w)$ and $P(r)$. The sum of these probabilities has to be $1$.
$endgroup$
– callculus
Jan 27 at 18:12
add a comment |
$begingroup$
Here is how the principle of inclusion-exclusion looks with three events:
$begin{align*}
Pr(Wcup Rcup G) &= Pr(W)+ Pr(R)+ Pr(G)\
&quad-Pr(Wcap R)-Pr(Wcap G)-Pr(Gcap R)\
&qquad+ Pr(Wcap Rcap G)
end{align*}$
It’s up to you to compute each of the terms on the RHS.
answered Jan 26 at 22:09
Laars HeleniusLaars Helenius
6,3111423
$endgroup$
Here is how the principle of inclusion-exclusion looks with three events:
$begin{align*}
Pr(Wcup Rcup G) &= Pr(W)+ Pr(R)+ Pr(G)\
&quad-Pr(Wcap R)-Pr(Wcap G)-Pr(Gcap R)\
&qquad+ Pr(Wcap Rcap G)
end{align*}$
It’s up to you to compute each of the terms on the RHS.
answered Jan 26 at 22:09
Laars HeleniusLaars Helenius
6,3111423
answered Jan 26 at 22:09
Laars HeleniusLaars Helenius
6,3111423
answered Jan 26 at 22:09
Laars HeleniusLaars Helenius
6,3111423
answered Jan 26 at 22:09
Laars HeleniusLaars Helenius
6,3111423
6,3111423
$begingroup$
I understand that P(W), P(G) and P(R) represent the probabilities that white, red and green do not show up, but what would P(W $cup$ R) represent? I assume it would be the outcomes that W and R have in common meaning only having green appear, so P(W $cup$ R) = $frac{2*2}{4*4}$
$endgroup$
– jerrythesphere
Jan 27 at 17:33
$begingroup$
@jerrythesphere For $P(Wcap R)$ you just have to multiply the probabilities that the i-th draw is green. The probablitly to pick a green marble at the i-th draw is $P(g)=frac{1}{1+1+2}=frac14$. Similar for $P(w)$ and $P(r)$. The sum of these probabilities has to be $1$.
$endgroup$
– callculus
Jan 27 at 18:12
add a comment |
$begingroup$
I understand that P(W), P(G) and P(R) represent the probabilities that white, red and green do not show up, but what would P(W $cup$ R) represent? I assume it would be the outcomes that W and R have in common meaning only having green appear, so P(W $cup$ R) = $frac{2*2}{4*4}$
$endgroup$
– jerrythesphere
Jan 27 at 17:33
$begingroup$
@jerrythesphere For $P(Wcap R)$ you just have to multiply the probabilities that the i-th draw is green. The probablitly to pick a green marble at the i-th draw is $P(g)=frac{1}{1+1+2}=frac14$. Similar for $P(w)$ and $P(r)$. The sum of these probabilities has to be $1$.
$endgroup$
– callculus
Jan 27 at 18:12
$begingroup$
I understand that P(W), P(G) and P(R) represent the probabilities that white, red and green do not show up, but what would P(W $cup$ R) represent? I assume it would be the outcomes that W and R have in common meaning only having green appear, so P(W $cup$ R) = $frac{2*2}{4*4}$
$endgroup$
– jerrythesphere
Jan 27 at 17:33
$begingroup$
I understand that P(W), P(G) and P(R) represent the probabilities that white, red and green do not show up, but what would P(W $cup$ R) represent? I assume it would be the outcomes that W and R have in common meaning only having green appear, so P(W $cup$ R) = $frac{2*2}{4*4}$
$endgroup$
– jerrythesphere
Jan 27 at 17:33
$begingroup$
@jerrythesphere For $P(Wcap R)$ you just have to multiply the probabilities that the i-th draw is green. The probablitly to pick a green marble at the i-th draw is $P(g)=frac{1}{1+1+2}=frac14$. Similar for $P(w)$ and $P(r)$. The sum of these probabilities has to be $1$.
$endgroup$
– callculus
Jan 27 at 18:12
$begingroup$
@jerrythesphere For $P(Wcap R)$ you just have to multiply the probabilities that the i-th draw is green. The probablitly to pick a green marble at the i-th draw is $P(g)=frac{1}{1+1+2}=frac14$. Similar for $P(w)$ and $P(r)$. The sum of these probabilities has to be $1$.
$endgroup$
– callculus
Jan 27 at 18:12
add a comment |
$begingroup$
The principle of inclusion and exclusion works for any number of events:
$P(Acup Bcup C)=P(A)+P(B)+P(C)-P(Acap B)-P(Acap C)-P(Bcap C)+P(Acap Bcap C)$$
answered Jan 26 at 22:09
saulspatzsaulspatz
17.1k31435
$endgroup$
add a comment |
$begingroup$
The principle of inclusion and exclusion works for any number of events:
$P(Acup Bcup C)=P(A)+P(B)+P(C)-P(Acap B)-P(Acap C)-P(Bcap C)+P(Acap Bcap C)$$
answered Jan 26 at 22:09
saulspatzsaulspatz
17.1k31435
$endgroup$
add a comment |
$begingroup$
The principle of inclusion and exclusion works for any number of events:
$P(Acup Bcup C)=P(A)+P(B)+P(C)-P(Acap B)-P(Acap C)-P(Bcap C)+P(Acap Bcap C)$$
answered Jan 26 at 22:09
saulspatzsaulspatz
17.1k31435
$endgroup$
The principle of inclusion and exclusion works for any number of events:
$P(Acup Bcup C)=P(A)+P(B)+P(C)-P(Acap B)-P(Acap C)-P(Bcap C)+P(Acap Bcap C)$$
answered Jan 26 at 22:09
saulspatzsaulspatz
17.1k31435
answered Jan 26 at 22:09
saulspatzsaulspatz
17.1k31435
answered Jan 26 at 22:09
saulspatzsaulspatz
17.1k31435
answered Jan 26 at 22:09
saulspatzsaulspatz
17.1k31435
17.1k31435
add a comment |
add a comment |
$begingroup$
Let $W,R,G$ the events, that the white, red and green marbles do not show up.
Then the number of combinations for W are:
$a) gggrightarrow 1, b) ggrrightarrow 3, c) rrgrightarrow 3, d) rrrrightarrow 1$
The corresponding probabilities are
$a) frac{1}{4}cdot frac{1}{4}cdot frac{1}{4}= frac{1}{64}$
$b) 3cdot frac{1}{4}cdot frac{1}{4}cdot frac{1}{2}= frac{3}{32}$
$c) 3cdot frac{1}{2}cdot frac{1}{2}cdot frac{1}{4}= frac{3}{16}$
$d) frac{1}{2}cdot frac{1}{2}cdot frac{1}{2}= frac{1}{8}$
Therefore $P(W)=frac{27}{64}$
Now what ist $P(Wcap R)$?
$Wcap R$ means that no white marbles and no red marbles show up. That is when only green marbles show up-as you´ve already said: $ ggg$
$P(Wcap R)=frac{1}{4}cdot frac{1}{4}cdot frac{1}{4}= frac{1}{64}$
You can check your final result by using the converse probability:
$P(textrm{"Do not see all three colors"})=1-P(textrm{"See all three colors"})$
answered Jan 27 at 18:02
callculuscallculus
18.5k31428
$endgroup$
add a comment |
$begingroup$
Let $W,R,G$ the events, that the white, red and green marbles do not show up.
Then the number of combinations for W are:
$a) gggrightarrow 1, b) ggrrightarrow 3, c) rrgrightarrow 3, d) rrrrightarrow 1$
The corresponding probabilities are
$a) frac{1}{4}cdot frac{1}{4}cdot frac{1}{4}= frac{1}{64}$
$b) 3cdot frac{1}{4}cdot frac{1}{4}cdot frac{1}{2}= frac{3}{32}$
$c) 3cdot frac{1}{2}cdot frac{1}{2}cdot frac{1}{4}= frac{3}{16}$
$d) frac{1}{2}cdot frac{1}{2}cdot frac{1}{2}= frac{1}{8}$
Therefore $P(W)=frac{27}{64}$
Now what ist $P(Wcap R)$?
$Wcap R$ means that no white marbles and no red marbles show up. That is when only green marbles show up-as you´ve already said: $ ggg$
$P(Wcap R)=frac{1}{4}cdot frac{1}{4}cdot frac{1}{4}= frac{1}{64}$
You can check your final result by using the converse probability:
$P(textrm{"Do not see all three colors"})=1-P(textrm{"See all three colors"})$
answered Jan 27 at 18:02
callculuscallculus
18.5k31428
$endgroup$
add a comment |
$begingroup$
Let $W,R,G$ the events, that the white, red and green marbles do not show up.
Then the number of combinations for W are:
$a) gggrightarrow 1, b) ggrrightarrow 3, c) rrgrightarrow 3, d) rrrrightarrow 1$
The corresponding probabilities are
$a) frac{1}{4}cdot frac{1}{4}cdot frac{1}{4}= frac{1}{64}$
$b) 3cdot frac{1}{4}cdot frac{1}{4}cdot frac{1}{2}= frac{3}{32}$
$c) 3cdot frac{1}{2}cdot frac{1}{2}cdot frac{1}{4}= frac{3}{16}$
$d) frac{1}{2}cdot frac{1}{2}cdot frac{1}{2}= frac{1}{8}$
Therefore $P(W)=frac{27}{64}$
Now what ist $P(Wcap R)$?
$Wcap R$ means that no white marbles and no red marbles show up. That is when only green marbles show up-as you´ve already said: $ ggg$
$P(Wcap R)=frac{1}{4}cdot frac{1}{4}cdot frac{1}{4}= frac{1}{64}$
You can check your final result by using the converse probability:
$P(textrm{"Do not see all three colors"})=1-P(textrm{"See all three colors"})$
answered Jan 27 at 18:02
callculuscallculus
18.5k31428
$endgroup$
Let $W,R,G$ the events, that the white, red and green marbles do not show up.
Then the number of combinations for W are:
$a) gggrightarrow 1, b) ggrrightarrow 3, c) rrgrightarrow 3, d) rrrrightarrow 1$
The corresponding probabilities are
$a) frac{1}{4}cdot frac{1}{4}cdot frac{1}{4}= frac{1}{64}$
$b) 3cdot frac{1}{4}cdot frac{1}{4}cdot frac{1}{2}= frac{3}{32}$
$c) 3cdot frac{1}{2}cdot frac{1}{2}cdot frac{1}{4}= frac{3}{16}$
$d) frac{1}{2}cdot frac{1}{2}cdot frac{1}{2}= frac{1}{8}$
Therefore $P(W)=frac{27}{64}$
Now what ist $P(Wcap R)$?
$Wcap R$ means that no white marbles and no red marbles show up. That is when only green marbles show up-as you´ve already said: $ ggg$
$P(Wcap R)=frac{1}{4}cdot frac{1}{4}cdot frac{1}{4}= frac{1}{64}$
You can check your final result by using the converse probability:
$P(textrm{"Do not see all three colors"})=1-P(textrm{"See all three colors"})$
answered Jan 27 at 18:02
callculuscallculus
18.5k31428
edited Jan 27 at 18:13
answered Jan 27 at 18:02
callculuscallculus
18.5k31428
answered Jan 27 at 18:02
callculuscallculus
18.5k31428
answered Jan 27 at 18:02
callculuscallculus
18.5k31428
18.5k31428
add a comment |
add a comment |
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2
$begingroup$
It is much easier to use the converse probability: probability not seeing all three colors is equal to One minus the probability seeing all three colors.
$endgroup$
– callculus
Jan 26 at 21:31
$begingroup$
I have to use inclusion-exclusion; this is a multi-part problem and it specifies doing it this way
$endgroup$
– jerrythesphere
Jan 26 at 22:04
$begingroup$
You have to show some effort. What result do you get if you follow my advice?
$endgroup$
– callculus
Jan 26 at 22:07
$begingroup$
OK. Now it´s clear that you have to use the inclusion-exclusion principle. But with my advice you can double check the result which you get with the inclusion-exclusion principle. You just have to start to do something.
$endgroup$
– callculus
Jan 26 at 22:19