Bayes Theorem - playing cards example

`P(R)=P(A)xxP(R|A)+` `P(B)xxP(R|B)+` `P(C)xxP(R|C)`

`=26/52xx2/10+` `13/52xx4/10+` `13/52xx8/10`

`=2/5`

`P(H|R)` `=(P(H)xxP(R|H))/(P(R))` `=(13/52xx8/10)/(4/10)` `=1/2`

`P(D|R)` `=(P(D)xxP(R|D))/(P(R))` `=(13/52xx4/10)/(4/10)` `=1/4`

`P(Bl|R)` `=(P(Bl)xxP(R|Bl))/(P(R))` `=(26/52xx2/10)/(4/10)` `=1/4`

Conclusion: The probabilities add to 1, so we can compute the probability of a specified event given the subsequent event did occur.

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