Mixing
Bayes Theorem Confirmation
$begingroup$
Suppose a soccer team plays $3$ games for Division A, $3$ games for Division B, and $5$ games for Division C. The team wins $20%$ of their games in Division $A$, $50 %$ of their games in Division B and $70 %$ of their games in Division C.
The team won their most recent game. What is the probability that it was a game in Division C?
So it would be: $$P(text{Division C}| text{won}) = frac{P(text{won}| text{Division C})P(text{Division C})}{P(text{won})} = frac{(0.7)(frac{5}{11})}{(0.2)(frac{3}{11})+(0.5)(frac{3}{11})+(0.7)(frac{5}{11})}$$
Is this correct?
probability
edited Jan 16 at 6:32
David G. Stork
11k41432
asked Jan 16 at 4:55
DamienDamien
2,05831932
$endgroup$
add a comment |
$begingroup$
Suppose a soccer team plays $3$ games for Division A, $3$ games for Division B, and $5$ games for Division C. The team wins $20%$ of their games in Division $A$, $50 %$ of their games in Division B and $70 %$ of their games in Division C.
The team won their most recent game. What is the probability that it was a game in Division C?
So it would be: $$P(text{Division C}| text{won}) = frac{P(text{won}| text{Division C})P(text{Division C})}{P(text{won})} = frac{(0.7)(frac{5}{11})}{(0.2)(frac{3}{11})+(0.5)(frac{3}{11})+(0.7)(frac{5}{11})}$$
Is this correct?
probability
edited Jan 16 at 6:32
David G. Stork
11k41432
asked Jan 16 at 4:55
DamienDamien
2,05831932
$endgroup$
$begingroup$
how can you win 20% of games if 3 games were played? There seems to be something strange going on here...Also, ignoring that problem, I think the logic only makes sense if the "most recent game" is included in the stats given in the beginning. That is, if this "most recent game" was already counted in the initial statement of wins and losses.
$endgroup$
– BelowAverageIntelligence
Jan 16 at 5:10
$begingroup$
@BelowAverageIntelligence "20% of their games" does not refer to the actual win proportion out of the games, but rather the chance that the team wins a given game.
$endgroup$
– angryavian
Jan 16 at 5:16
$begingroup$
Looks fine to me
$endgroup$
– angryavian
Jan 16 at 5:17
$begingroup$
not sure what you mean by "the chance that the team wins a given game". We're not throwing dice here...
$endgroup$
– BelowAverageIntelligence
Jan 16 at 5:19
$begingroup$
@BelowAverageIntelligence Unfortunately, yes, that is how the problem is set up. The "most recent game" is modeled as a random draw from the 11 games, and each game's outcome is determined by an independent coin flip with win probability depending on which division the game is in. I agree that this is not explicit from the wording, but this is how Bayes theorem exercises are usually set up.
$endgroup$
– angryavian
Jan 16 at 5:39
add a comment |
$begingroup$
Suppose a soccer team plays $3$ games for Division A, $3$ games for Division B, and $5$ games for Division C. The team wins $20%$ of their games in Division $A$, $50 %$ of their games in Division B and $70 %$ of their games in Division C.
The team won their most recent game. What is the probability that it was a game in Division C?
So it would be: $$P(text{Division C}| text{won}) = frac{P(text{won}| text{Division C})P(text{Division C})}{P(text{won})} = frac{(0.7)(frac{5}{11})}{(0.2)(frac{3}{11})+(0.5)(frac{3}{11})+(0.7)(frac{5}{11})}$$
Is this correct?
probability
edited Jan 16 at 6:32
David G. Stork
11k41432
asked Jan 16 at 4:55
DamienDamien
2,05831932
$endgroup$
Suppose a soccer team plays $3$ games for Division A, $3$ games for Division B, and $5$ games for Division C. The team wins $20%$ of their games in Division $A$, $50 %$ of their games in Division B and $70 %$ of their games in Division C.
The team won their most recent game. What is the probability that it was a game in Division C?
So it would be: $$P(text{Division C}| text{won}) = frac{P(text{won}| text{Division C})P(text{Division C})}{P(text{won})} = frac{(0.7)(frac{5}{11})}{(0.2)(frac{3}{11})+(0.5)(frac{3}{11})+(0.7)(frac{5}{11})}$$
Is this correct?
probability
probability
edited Jan 16 at 6:32
David G. Stork
11k41432
asked Jan 16 at 4:55
DamienDamien
2,05831932
edited Jan 16 at 6:32
David G. Stork
11k41432
asked Jan 16 at 4:55
DamienDamien
2,05831932
edited Jan 16 at 6:32
David G. Stork
11k41432
edited Jan 16 at 6:32
David G. Stork
11k41432
edited Jan 16 at 6:32
David G. Stork
11k41432
11k41432
asked Jan 16 at 4:55
DamienDamien
2,05831932
asked Jan 16 at 4:55
DamienDamien
2,05831932
asked Jan 16 at 4:55
DamienDamien
2,05831932
2,05831932
$begingroup$
how can you win 20% of games if 3 games were played? There seems to be something strange going on here...Also, ignoring that problem, I think the logic only makes sense if the "most recent game" is included in the stats given in the beginning. That is, if this "most recent game" was already counted in the initial statement of wins and losses.
$endgroup$
– BelowAverageIntelligence
Jan 16 at 5:10
$begingroup$
@BelowAverageIntelligence "20% of their games" does not refer to the actual win proportion out of the games, but rather the chance that the team wins a given game.
$endgroup$
– angryavian
Jan 16 at 5:16
$begingroup$
Looks fine to me
$endgroup$
– angryavian
Jan 16 at 5:17
$begingroup$
not sure what you mean by "the chance that the team wins a given game". We're not throwing dice here...
$endgroup$
– BelowAverageIntelligence
Jan 16 at 5:19
$begingroup$
@BelowAverageIntelligence Unfortunately, yes, that is how the problem is set up. The "most recent game" is modeled as a random draw from the 11 games, and each game's outcome is determined by an independent coin flip with win probability depending on which division the game is in. I agree that this is not explicit from the wording, but this is how Bayes theorem exercises are usually set up.
$endgroup$
– angryavian
Jan 16 at 5:39
add a comment |
$begingroup$
how can you win 20% of games if 3 games were played? There seems to be something strange going on here...Also, ignoring that problem, I think the logic only makes sense if the "most recent game" is included in the stats given in the beginning. That is, if this "most recent game" was already counted in the initial statement of wins and losses.
$endgroup$
– BelowAverageIntelligence
Jan 16 at 5:10
$begingroup$
@BelowAverageIntelligence "20% of their games" does not refer to the actual win proportion out of the games, but rather the chance that the team wins a given game.
$endgroup$
– angryavian
Jan 16 at 5:16
$begingroup$
Looks fine to me
$endgroup$
– angryavian
Jan 16 at 5:17
$begingroup$
not sure what you mean by "the chance that the team wins a given game". We're not throwing dice here...
$endgroup$
– BelowAverageIntelligence
Jan 16 at 5:19
$begingroup$
@BelowAverageIntelligence Unfortunately, yes, that is how the problem is set up. The "most recent game" is modeled as a random draw from the 11 games, and each game's outcome is determined by an independent coin flip with win probability depending on which division the game is in. I agree that this is not explicit from the wording, but this is how Bayes theorem exercises are usually set up.
$endgroup$
– angryavian
Jan 16 at 5:39
$begingroup$
how can you win 20% of games if 3 games were played? There seems to be something strange going on here...Also, ignoring that problem, I think the logic only makes sense if the "most recent game" is included in the stats given in the beginning. That is, if this "most recent game" was already counted in the initial statement of wins and losses.
$endgroup$
– BelowAverageIntelligence
Jan 16 at 5:10
$begingroup$
how can you win 20% of games if 3 games were played? There seems to be something strange going on here...Also, ignoring that problem, I think the logic only makes sense if the "most recent game" is included in the stats given in the beginning. That is, if this "most recent game" was already counted in the initial statement of wins and losses.
$endgroup$
– BelowAverageIntelligence
Jan 16 at 5:10
$begingroup$
@BelowAverageIntelligence "20% of their games" does not refer to the actual win proportion out of the games, but rather the chance that the team wins a given game.
$endgroup$
– angryavian
Jan 16 at 5:16
$begingroup$
@BelowAverageIntelligence "20% of their games" does not refer to the actual win proportion out of the games, but rather the chance that the team wins a given game.
$endgroup$
– angryavian
Jan 16 at 5:16
$begingroup$
Looks fine to me
$endgroup$
– angryavian
Jan 16 at 5:17
$begingroup$
Looks fine to me
$endgroup$
– angryavian
Jan 16 at 5:17
$begingroup$
not sure what you mean by "the chance that the team wins a given game". We're not throwing dice here...
$endgroup$
– BelowAverageIntelligence
Jan 16 at 5:19
$begingroup$
not sure what you mean by "the chance that the team wins a given game". We're not throwing dice here...
$endgroup$
– BelowAverageIntelligence
Jan 16 at 5:19
$begingroup$
@BelowAverageIntelligence Unfortunately, yes, that is how the problem is set up. The "most recent game" is modeled as a random draw from the 11 games, and each game's outcome is determined by an independent coin flip with win probability depending on which division the game is in. I agree that this is not explicit from the wording, but this is how Bayes theorem exercises are usually set up.
$endgroup$
– angryavian
Jan 16 at 5:39
$begingroup$
@BelowAverageIntelligence Unfortunately, yes, that is how the problem is set up. The "most recent game" is modeled as a random draw from the 11 games, and each game's outcome is determined by an independent coin flip with win probability depending on which division the game is in. I agree that this is not explicit from the wording, but this is how Bayes theorem exercises are usually set up.
$endgroup$
– angryavian
Jan 16 at 5:39
add a comment |
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$begingroup$
how can you win 20% of games if 3 games were played? There seems to be something strange going on here...Also, ignoring that problem, I think the logic only makes sense if the "most recent game" is included in the stats given in the beginning. That is, if this "most recent game" was already counted in the initial statement of wins and losses.
$endgroup$
– BelowAverageIntelligence
Jan 16 at 5:10
$begingroup$
@BelowAverageIntelligence "20% of their games" does not refer to the actual win proportion out of the games, but rather the chance that the team wins a given game.
$endgroup$
– angryavian
Jan 16 at 5:16
$begingroup$
Looks fine to me
$endgroup$
– angryavian
Jan 16 at 5:17
$begingroup$
not sure what you mean by "the chance that the team wins a given game". We're not throwing dice here...
$endgroup$
– BelowAverageIntelligence
Jan 16 at 5:19
$begingroup$
@BelowAverageIntelligence Unfortunately, yes, that is how the problem is set up. The "most recent game" is modeled as a random draw from the 11 games, and each game's outcome is determined by an independent coin flip with win probability depending on which division the game is in. I agree that this is not explicit from the wording, but this is how Bayes theorem exercises are usually set up.
$endgroup$
– angryavian
Jan 16 at 5:39