Reverse probability problem.












0












$begingroup$


Suppose there are two cards A and B. Both of the cards have a yellow side and a green side. When tossed in the air the probability of the yellow side facing up is %31 for Card A and 35% for Card B. A person randomly chooses one of the cards and tosses it in the air 100 times. The card lands with the yellow side facing up 35 times out of the 100 tosses. What is the probability that it is card B? No additional information is provided.






probability inverse bayesian







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$endgroup$












  • $begingroup$
    It is expected that you provide the additional information. What have you tried? Where are you stuck?
    $endgroup$
    – John Douma
    Jan 14 at 23:29










  • $begingroup$
    this is the problem. There is no additional info. I don't know where to start honestly. I'm leaning towards 100% because there is no additional info but I'm pretty sure its wrong
    $endgroup$
    – memokerobi
    Jan 14 at 23:31










  • $begingroup$
    First find the probability that card A will land with yellow side up $35$ times in $100$ tosses. Also find the probability that card B will land with yellow side up $35$ times in $100$ tosses. Can you do this?
    $endgroup$
    – Daniel Mathias
    Jan 14 at 23:43










  • $begingroup$
    I thought about it but there are so many different combinations that can result in 35 yellow side landings. Do you know an easier method of calculating it?
    $endgroup$
    – memokerobi
    Jan 14 at 23:47


















0












$begingroup$


Suppose there are two cards A and B. Both of the cards have a yellow side and a green side. When tossed in the air the probability of the yellow side facing up is %31 for Card A and 35% for Card B. A person randomly chooses one of the cards and tosses it in the air 100 times. The card lands with the yellow side facing up 35 times out of the 100 tosses. What is the probability that it is card B? No additional information is provided.






probability inverse bayesian







share|cite|improve this question









$endgroup$












  • $begingroup$
    It is expected that you provide the additional information. What have you tried? Where are you stuck?
    $endgroup$
    – John Douma
    Jan 14 at 23:29










  • $begingroup$
    this is the problem. There is no additional info. I don't know where to start honestly. I'm leaning towards 100% because there is no additional info but I'm pretty sure its wrong
    $endgroup$
    – memokerobi
    Jan 14 at 23:31










  • $begingroup$
    First find the probability that card A will land with yellow side up $35$ times in $100$ tosses. Also find the probability that card B will land with yellow side up $35$ times in $100$ tosses. Can you do this?
    $endgroup$
    – Daniel Mathias
    Jan 14 at 23:43










  • $begingroup$
    I thought about it but there are so many different combinations that can result in 35 yellow side landings. Do you know an easier method of calculating it?
    $endgroup$
    – memokerobi
    Jan 14 at 23:47
















0












0








0





$begingroup$


Suppose there are two cards A and B. Both of the cards have a yellow side and a green side. When tossed in the air the probability of the yellow side facing up is %31 for Card A and 35% for Card B. A person randomly chooses one of the cards and tosses it in the air 100 times. The card lands with the yellow side facing up 35 times out of the 100 tosses. What is the probability that it is card B? No additional information is provided.






probability inverse bayesian







share|cite|improve this question









$endgroup$




Suppose there are two cards A and B. Both of the cards have a yellow side and a green side. When tossed in the air the probability of the yellow side facing up is %31 for Card A and 35% for Card B. A person randomly chooses one of the cards and tosses it in the air 100 times. The card lands with the yellow side facing up 35 times out of the 100 tosses. What is the probability that it is card B? No additional information is provided.






probability inverse bayesian




probability inverse bayesian






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 14 at 23:23









memokerobimemokerobi

103




103












  • $begingroup$
    It is expected that you provide the additional information. What have you tried? Where are you stuck?
    $endgroup$
    – John Douma
    Jan 14 at 23:29










  • $begingroup$
    this is the problem. There is no additional info. I don't know where to start honestly. I'm leaning towards 100% because there is no additional info but I'm pretty sure its wrong
    $endgroup$
    – memokerobi
    Jan 14 at 23:31










  • $begingroup$
    First find the probability that card A will land with yellow side up $35$ times in $100$ tosses. Also find the probability that card B will land with yellow side up $35$ times in $100$ tosses. Can you do this?
    $endgroup$
    – Daniel Mathias
    Jan 14 at 23:43










  • $begingroup$
    I thought about it but there are so many different combinations that can result in 35 yellow side landings. Do you know an easier method of calculating it?
    $endgroup$
    – memokerobi
    Jan 14 at 23:47




















  • $begingroup$
    It is expected that you provide the additional information. What have you tried? Where are you stuck?
    $endgroup$
    – John Douma
    Jan 14 at 23:29










  • $begingroup$
    this is the problem. There is no additional info. I don't know where to start honestly. I'm leaning towards 100% because there is no additional info but I'm pretty sure its wrong
    $endgroup$
    – memokerobi
    Jan 14 at 23:31










  • $begingroup$
    First find the probability that card A will land with yellow side up $35$ times in $100$ tosses. Also find the probability that card B will land with yellow side up $35$ times in $100$ tosses. Can you do this?
    $endgroup$
    – Daniel Mathias
    Jan 14 at 23:43










  • $begingroup$
    I thought about it but there are so many different combinations that can result in 35 yellow side landings. Do you know an easier method of calculating it?
    $endgroup$
    – memokerobi
    Jan 14 at 23:47


















$begingroup$
It is expected that you provide the additional information. What have you tried? Where are you stuck?
$endgroup$
– John Douma
Jan 14 at 23:29




$begingroup$
It is expected that you provide the additional information. What have you tried? Where are you stuck?
$endgroup$
– John Douma
Jan 14 at 23:29












$begingroup$
this is the problem. There is no additional info. I don't know where to start honestly. I'm leaning towards 100% because there is no additional info but I'm pretty sure its wrong
$endgroup$
– memokerobi
Jan 14 at 23:31




$begingroup$
this is the problem. There is no additional info. I don't know where to start honestly. I'm leaning towards 100% because there is no additional info but I'm pretty sure its wrong
$endgroup$
– memokerobi
Jan 14 at 23:31












$begingroup$
First find the probability that card A will land with yellow side up $35$ times in $100$ tosses. Also find the probability that card B will land with yellow side up $35$ times in $100$ tosses. Can you do this?
$endgroup$
– Daniel Mathias
Jan 14 at 23:43




$begingroup$
First find the probability that card A will land with yellow side up $35$ times in $100$ tosses. Also find the probability that card B will land with yellow side up $35$ times in $100$ tosses. Can you do this?
$endgroup$
– Daniel Mathias
Jan 14 at 23:43












$begingroup$
I thought about it but there are so many different combinations that can result in 35 yellow side landings. Do you know an easier method of calculating it?
$endgroup$
– memokerobi
Jan 14 at 23:47






$begingroup$
I thought about it but there are so many different combinations that can result in 35 yellow side landings. Do you know an easier method of calculating it?
$endgroup$
– memokerobi
Jan 14 at 23:47












1 Answer
1






active

oldest

votes


















2












$begingroup$

This is a problem of conditional probability (further reading: https://en.wikipedia.org/wiki/Conditional_probability).



We know that the yellow side showed up 35 times after 100 tosses with a randomly chosen card and want to calculate the probability that the chosen card is card B (let $mathbb{P}$ denote the probability measure):



begin{equation*}
begin{split}
& mathbb{P}(text{card B} | 35text{ times yellow (random choice)}) = frac{mathbb{P}(text{card B} cap 35text{ times yellow (random choice)})}{mathbb{P}(35text{ times yellow (random choice of cards)})}\
& = frac{mathbb{P}(35*text{ yellow (random choice)} | text{card B})cdotmathbb{P}(text{card B})}{mathbb{P}(35*text{ yellow (random choice)})}
end{split}
end{equation*}



Note that $mathbb{P}(35text{ times yellow after random choice of cards})=frac{1}{2}cdotmathbb{P}(35 text{ times yellow with card A}) + frac{1}{2}cdotmathbb{P}(35 text{ times yellow with card B})$.



We have (using Binomial Distribution)
begin{align}tag{1}
&mathbb{P}(35 text{ times yellow with card A})= {100choose35}cdot 0.31^{35}cdot 0.69^{100-35}approx0.0578.\ tag{2}
&text{Similarly, } mathbb{P}(35 text{ times yellow with card B})approx0.0834.
end{align}



Thus $mathbb{P}(35text{ times yellow after random choice of cards})approx0.0706$.




The final answer is thus $displaystylemathbb{P}(text{card B} | 35text{ times yellow})approxfrac{0.0834cdotfrac12}{0.0706}approx0.5906$.







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is correct if we assume that a priori both cards have the same probability. That is not stated.
    $endgroup$
    – leonbloy
    Jan 14 at 23:55












  • $begingroup$
    @leonbloy You are right, I assumed that "A person randomly chooses one of the cards" means that both cards can be chosen with a probability of 50 %. It could have been stated a little clearer
    $endgroup$
    – Maximilian Janisch
    Jan 14 at 23:59










  • $begingroup$
    Yes the probability of drawing both cards is 50%. Apologies for not stating clearly
    $endgroup$
    – memokerobi
    Jan 15 at 4:04












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

This is a problem of conditional probability (further reading: https://en.wikipedia.org/wiki/Conditional_probability).



We know that the yellow side showed up 35 times after 100 tosses with a randomly chosen card and want to calculate the probability that the chosen card is card B (let $mathbb{P}$ denote the probability measure):



begin{equation*}
begin{split}
& mathbb{P}(text{card B} | 35text{ times yellow (random choice)}) = frac{mathbb{P}(text{card B} cap 35text{ times yellow (random choice)})}{mathbb{P}(35text{ times yellow (random choice of cards)})}\
& = frac{mathbb{P}(35*text{ yellow (random choice)} | text{card B})cdotmathbb{P}(text{card B})}{mathbb{P}(35*text{ yellow (random choice)})}
end{split}
end{equation*}



Note that $mathbb{P}(35text{ times yellow after random choice of cards})=frac{1}{2}cdotmathbb{P}(35 text{ times yellow with card A}) + frac{1}{2}cdotmathbb{P}(35 text{ times yellow with card B})$.



We have (using Binomial Distribution)
begin{align}tag{1}
&mathbb{P}(35 text{ times yellow with card A})= {100choose35}cdot 0.31^{35}cdot 0.69^{100-35}approx0.0578.\ tag{2}
&text{Similarly, } mathbb{P}(35 text{ times yellow with card B})approx0.0834.
end{align}



Thus $mathbb{P}(35text{ times yellow after random choice of cards})approx0.0706$.




The final answer is thus $displaystylemathbb{P}(text{card B} | 35text{ times yellow})approxfrac{0.0834cdotfrac12}{0.0706}approx0.5906$.







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is correct if we assume that a priori both cards have the same probability. That is not stated.
    $endgroup$
    – leonbloy
    Jan 14 at 23:55












  • $begingroup$
    @leonbloy You are right, I assumed that "A person randomly chooses one of the cards" means that both cards can be chosen with a probability of 50 %. It could have been stated a little clearer
    $endgroup$
    – Maximilian Janisch
    Jan 14 at 23:59










  • $begingroup$
    Yes the probability of drawing both cards is 50%. Apologies for not stating clearly
    $endgroup$
    – memokerobi
    Jan 15 at 4:04
















2












$begingroup$

This is a problem of conditional probability (further reading: https://en.wikipedia.org/wiki/Conditional_probability).



We know that the yellow side showed up 35 times after 100 tosses with a randomly chosen card and want to calculate the probability that the chosen card is card B (let $mathbb{P}$ denote the probability measure):



begin{equation*}
begin{split}
& mathbb{P}(text{card B} | 35text{ times yellow (random choice)}) = frac{mathbb{P}(text{card B} cap 35text{ times yellow (random choice)})}{mathbb{P}(35text{ times yellow (random choice of cards)})}\
& = frac{mathbb{P}(35*text{ yellow (random choice)} | text{card B})cdotmathbb{P}(text{card B})}{mathbb{P}(35*text{ yellow (random choice)})}
end{split}
end{equation*}



Note that $mathbb{P}(35text{ times yellow after random choice of cards})=frac{1}{2}cdotmathbb{P}(35 text{ times yellow with card A}) + frac{1}{2}cdotmathbb{P}(35 text{ times yellow with card B})$.



We have (using Binomial Distribution)
begin{align}tag{1}
&mathbb{P}(35 text{ times yellow with card A})= {100choose35}cdot 0.31^{35}cdot 0.69^{100-35}approx0.0578.\ tag{2}
&text{Similarly, } mathbb{P}(35 text{ times yellow with card B})approx0.0834.
end{align}



Thus $mathbb{P}(35text{ times yellow after random choice of cards})approx0.0706$.




The final answer is thus $displaystylemathbb{P}(text{card B} | 35text{ times yellow})approxfrac{0.0834cdotfrac12}{0.0706}approx0.5906$.







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is correct if we assume that a priori both cards have the same probability. That is not stated.
    $endgroup$
    – leonbloy
    Jan 14 at 23:55












  • $begingroup$
    @leonbloy You are right, I assumed that "A person randomly chooses one of the cards" means that both cards can be chosen with a probability of 50 %. It could have been stated a little clearer
    $endgroup$
    – Maximilian Janisch
    Jan 14 at 23:59










  • $begingroup$
    Yes the probability of drawing both cards is 50%. Apologies for not stating clearly
    $endgroup$
    – memokerobi
    Jan 15 at 4:04














2












2








2





$begingroup$

This is a problem of conditional probability (further reading: https://en.wikipedia.org/wiki/Conditional_probability).



We know that the yellow side showed up 35 times after 100 tosses with a randomly chosen card and want to calculate the probability that the chosen card is card B (let $mathbb{P}$ denote the probability measure):



begin{equation*}
begin{split}
& mathbb{P}(text{card B} | 35text{ times yellow (random choice)}) = frac{mathbb{P}(text{card B} cap 35text{ times yellow (random choice)})}{mathbb{P}(35text{ times yellow (random choice of cards)})}\
& = frac{mathbb{P}(35*text{ yellow (random choice)} | text{card B})cdotmathbb{P}(text{card B})}{mathbb{P}(35*text{ yellow (random choice)})}
end{split}
end{equation*}



Note that $mathbb{P}(35text{ times yellow after random choice of cards})=frac{1}{2}cdotmathbb{P}(35 text{ times yellow with card A}) + frac{1}{2}cdotmathbb{P}(35 text{ times yellow with card B})$.



We have (using Binomial Distribution)
begin{align}tag{1}
&mathbb{P}(35 text{ times yellow with card A})= {100choose35}cdot 0.31^{35}cdot 0.69^{100-35}approx0.0578.\ tag{2}
&text{Similarly, } mathbb{P}(35 text{ times yellow with card B})approx0.0834.
end{align}



Thus $mathbb{P}(35text{ times yellow after random choice of cards})approx0.0706$.




The final answer is thus $displaystylemathbb{P}(text{card B} | 35text{ times yellow})approxfrac{0.0834cdotfrac12}{0.0706}approx0.5906$.







share|cite|improve this answer









$endgroup$



This is a problem of conditional probability (further reading: https://en.wikipedia.org/wiki/Conditional_probability).



We know that the yellow side showed up 35 times after 100 tosses with a randomly chosen card and want to calculate the probability that the chosen card is card B (let $mathbb{P}$ denote the probability measure):



begin{equation*}
begin{split}
& mathbb{P}(text{card B} | 35text{ times yellow (random choice)}) = frac{mathbb{P}(text{card B} cap 35text{ times yellow (random choice)})}{mathbb{P}(35text{ times yellow (random choice of cards)})}\
& = frac{mathbb{P}(35*text{ yellow (random choice)} | text{card B})cdotmathbb{P}(text{card B})}{mathbb{P}(35*text{ yellow (random choice)})}
end{split}
end{equation*}



Note that $mathbb{P}(35text{ times yellow after random choice of cards})=frac{1}{2}cdotmathbb{P}(35 text{ times yellow with card A}) + frac{1}{2}cdotmathbb{P}(35 text{ times yellow with card B})$.



We have (using Binomial Distribution)
begin{align}tag{1}
&mathbb{P}(35 text{ times yellow with card A})= {100choose35}cdot 0.31^{35}cdot 0.69^{100-35}approx0.0578.\ tag{2}
&text{Similarly, } mathbb{P}(35 text{ times yellow with card B})approx0.0834.
end{align}



Thus $mathbb{P}(35text{ times yellow after random choice of cards})approx0.0706$.




The final answer is thus $displaystylemathbb{P}(text{card B} | 35text{ times yellow})approxfrac{0.0834cdotfrac12}{0.0706}approx0.5906$.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 14 at 23:52









Maximilian JanischMaximilian Janisch

48612




48612








  • 1




    $begingroup$
    This is correct if we assume that a priori both cards have the same probability. That is not stated.
    $endgroup$
    – leonbloy
    Jan 14 at 23:55












  • $begingroup$
    @leonbloy You are right, I assumed that "A person randomly chooses one of the cards" means that both cards can be chosen with a probability of 50 %. It could have been stated a little clearer
    $endgroup$
    – Maximilian Janisch
    Jan 14 at 23:59










  • $begingroup$
    Yes the probability of drawing both cards is 50%. Apologies for not stating clearly
    $endgroup$
    – memokerobi
    Jan 15 at 4:04














  • 1




    $begingroup$
    This is correct if we assume that a priori both cards have the same probability. That is not stated.
    $endgroup$
    – leonbloy
    Jan 14 at 23:55












  • $begingroup$
    @leonbloy You are right, I assumed that "A person randomly chooses one of the cards" means that both cards can be chosen with a probability of 50 %. It could have been stated a little clearer
    $endgroup$
    – Maximilian Janisch
    Jan 14 at 23:59










  • $begingroup$
    Yes the probability of drawing both cards is 50%. Apologies for not stating clearly
    $endgroup$
    – memokerobi
    Jan 15 at 4:04








1




1




$begingroup$
This is correct if we assume that a priori both cards have the same probability. That is not stated.
$endgroup$
– leonbloy
Jan 14 at 23:55






$begingroup$
This is correct if we assume that a priori both cards have the same probability. That is not stated.
$endgroup$
– leonbloy
Jan 14 at 23:55














$begingroup$
@leonbloy You are right, I assumed that "A person randomly chooses one of the cards" means that both cards can be chosen with a probability of 50 %. It could have been stated a little clearer
$endgroup$
– Maximilian Janisch
Jan 14 at 23:59




$begingroup$
@leonbloy You are right, I assumed that "A person randomly chooses one of the cards" means that both cards can be chosen with a probability of 50 %. It could have been stated a little clearer
$endgroup$
– Maximilian Janisch
Jan 14 at 23:59












$begingroup$
Yes the probability of drawing both cards is 50%. Apologies for not stating clearly
$endgroup$
– memokerobi
Jan 15 at 4:04




$begingroup$
Yes the probability of drawing both cards is 50%. Apologies for not stating clearly
$endgroup$
– memokerobi
Jan 15 at 4:04


















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