Conditional Probability — Card Question

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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2
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Suppose you have two cards: one is painted black on both sides and the other is painted black on one side and orange on the other You select a card at random and view one side.



You notice it is
black. What is the probability the other side is orange?



What I have done is following:



Card 1: $B_1$, $B_2$



Card 2: $B_1$, $O_2$



Now, we want to calculate $P(O_2|B_1)$, which is $$P(O_2|B_1)=dfracP(B_1cap O_2)P(B_1)$$



$P(B_1cap O_2)=dfrac12$ as there is only $1$ card (out of $2$) giving us black and orange. $P(B_1)=dfrac34$.



Therefore, the resulting conditional probability is $dfrac23$



However, the solution of this question telling me $dfrac13$ is the correct answer (without explanation, just a number).



What's wrong here?



Thank you!






probability conditional-probability







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  • Problem is you consider you are seeing face $B_1$ if the card is the black card. You could see face $B_2$...
    – Nicolas FRANCOIS
    2 hours ago














up vote
2
down vote

favorite












Suppose you have two cards: one is painted black on both sides and the other is painted black on one side and orange on the other You select a card at random and view one side.



You notice it is
black. What is the probability the other side is orange?



What I have done is following:



Card 1: $B_1$, $B_2$



Card 2: $B_1$, $O_2$



Now, we want to calculate $P(O_2|B_1)$, which is $$P(O_2|B_1)=dfracP(B_1cap O_2)P(B_1)$$



$P(B_1cap O_2)=dfrac12$ as there is only $1$ card (out of $2$) giving us black and orange. $P(B_1)=dfrac34$.



Therefore, the resulting conditional probability is $dfrac23$



However, the solution of this question telling me $dfrac13$ is the correct answer (without explanation, just a number).



What's wrong here?



Thank you!






probability conditional-probability







share|cite|improve this question





















  • Problem is you consider you are seeing face $B_1$ if the card is the black card. You could see face $B_2$...
    – Nicolas FRANCOIS
    2 hours ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Suppose you have two cards: one is painted black on both sides and the other is painted black on one side and orange on the other You select a card at random and view one side.



You notice it is
black. What is the probability the other side is orange?



What I have done is following:



Card 1: $B_1$, $B_2$



Card 2: $B_1$, $O_2$



Now, we want to calculate $P(O_2|B_1)$, which is $$P(O_2|B_1)=dfracP(B_1cap O_2)P(B_1)$$



$P(B_1cap O_2)=dfrac12$ as there is only $1$ card (out of $2$) giving us black and orange. $P(B_1)=dfrac34$.



Therefore, the resulting conditional probability is $dfrac23$



However, the solution of this question telling me $dfrac13$ is the correct answer (without explanation, just a number).



What's wrong here?



Thank you!






probability conditional-probability







share|cite|improve this question













Suppose you have two cards: one is painted black on both sides and the other is painted black on one side and orange on the other You select a card at random and view one side.



You notice it is
black. What is the probability the other side is orange?



What I have done is following:



Card 1: $B_1$, $B_2$



Card 2: $B_1$, $O_2$



Now, we want to calculate $P(O_2|B_1)$, which is $$P(O_2|B_1)=dfracP(B_1cap O_2)P(B_1)$$



$P(B_1cap O_2)=dfrac12$ as there is only $1$ card (out of $2$) giving us black and orange. $P(B_1)=dfrac34$.



Therefore, the resulting conditional probability is $dfrac23$



However, the solution of this question telling me $dfrac13$ is the correct answer (without explanation, just a number).



What's wrong here?



Thank you!






probability conditional-probability




probability conditional-probability






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asked 2 hours ago









JacobsonRadical

36811




36811











  • Problem is you consider you are seeing face $B_1$ if the card is the black card. You could see face $B_2$...
    – Nicolas FRANCOIS
    2 hours ago
















  • Problem is you consider you are seeing face $B_1$ if the card is the black card. You could see face $B_2$...
    – Nicolas FRANCOIS
    2 hours ago















Problem is you consider you are seeing face $B_1$ if the card is the black card. You could see face $B_2$...
– Nicolas FRANCOIS
2 hours ago




Problem is you consider you are seeing face $B_1$ if the card is the black card. You could see face $B_2$...
– Nicolas FRANCOIS
2 hours ago










1 Answer
1






active

oldest

votes

















up vote
3
down vote













$P(B_1cap O_2)=frac14$, since to view a black side when the other side is orange requires first selecting the black/orange card and then viewing the black side on that card, both of which have a $frac12$ chance of occurring.






share|cite|improve this answer




















  • Oh! Now I understand. Thank you so much!
    – JacobsonRadical
    1 hour ago










  • Just for curiosity, can we solve this question by using Bayes Formula?
    – JacobsonRadical
    1 hour ago










  • @JacobsonRadical You could, and the working would be essentially identical.
    – Parcly Taxel
    1 hour ago










  • thank you so much!
    – JacobsonRadical
    1 hour ago










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote













$P(B_1cap O_2)=frac14$, since to view a black side when the other side is orange requires first selecting the black/orange card and then viewing the black side on that card, both of which have a $frac12$ chance of occurring.






share|cite|improve this answer




















  • Oh! Now I understand. Thank you so much!
    – JacobsonRadical
    1 hour ago










  • Just for curiosity, can we solve this question by using Bayes Formula?
    – JacobsonRadical
    1 hour ago










  • @JacobsonRadical You could, and the working would be essentially identical.
    – Parcly Taxel
    1 hour ago










  • thank you so much!
    – JacobsonRadical
    1 hour ago














up vote
3
down vote













$P(B_1cap O_2)=frac14$, since to view a black side when the other side is orange requires first selecting the black/orange card and then viewing the black side on that card, both of which have a $frac12$ chance of occurring.






share|cite|improve this answer




















  • Oh! Now I understand. Thank you so much!
    – JacobsonRadical
    1 hour ago










  • Just for curiosity, can we solve this question by using Bayes Formula?
    – JacobsonRadical
    1 hour ago










  • @JacobsonRadical You could, and the working would be essentially identical.
    – Parcly Taxel
    1 hour ago










  • thank you so much!
    – JacobsonRadical
    1 hour ago












up vote
3
down vote










up vote
3
down vote









$P(B_1cap O_2)=frac14$, since to view a black side when the other side is orange requires first selecting the black/orange card and then viewing the black side on that card, both of which have a $frac12$ chance of occurring.






share|cite|improve this answer












$P(B_1cap O_2)=frac14$, since to view a black side when the other side is orange requires first selecting the black/orange card and then viewing the black side on that card, both of which have a $frac12$ chance of occurring.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 hours ago









Parcly Taxel

36.8k137095




36.8k137095











  • Oh! Now I understand. Thank you so much!
    – JacobsonRadical
    1 hour ago










  • Just for curiosity, can we solve this question by using Bayes Formula?
    – JacobsonRadical
    1 hour ago










  • @JacobsonRadical You could, and the working would be essentially identical.
    – Parcly Taxel
    1 hour ago










  • thank you so much!
    – JacobsonRadical
    1 hour ago
















  • Oh! Now I understand. Thank you so much!
    – JacobsonRadical
    1 hour ago










  • Just for curiosity, can we solve this question by using Bayes Formula?
    – JacobsonRadical
    1 hour ago










  • @JacobsonRadical You could, and the working would be essentially identical.
    – Parcly Taxel
    1 hour ago










  • thank you so much!
    – JacobsonRadical
    1 hour ago















Oh! Now I understand. Thank you so much!
– JacobsonRadical
1 hour ago




Oh! Now I understand. Thank you so much!
– JacobsonRadical
1 hour ago












Just for curiosity, can we solve this question by using Bayes Formula?
– JacobsonRadical
1 hour ago




Just for curiosity, can we solve this question by using Bayes Formula?
– JacobsonRadical
1 hour ago












@JacobsonRadical You could, and the working would be essentially identical.
– Parcly Taxel
1 hour ago




@JacobsonRadical You could, and the working would be essentially identical.
– Parcly Taxel
1 hour ago












thank you so much!
– JacobsonRadical
1 hour ago




thank you so much!
– JacobsonRadical
1 hour ago

















 

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