What Is the Probability of Drawing an Odd, Black, or Even Card from a Deck?

gtfitzpatrick
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Homework Statement



a singlecard is drawn from a pack of 52 . what is the probability P((oddUblack)Ueven)

Homework Equations


The Attempt at a Solution



P(oddUblack) = 26/52 + 26/52 - 13/52 = 39/52

so is the next part mutually exclusive( if you draw and odd in the first part it can't be even?)

so my probability is 39/52 + 26/52 which obviously isn't right...
 
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hi gtfitzpatrick! :smile:
gtfitzpatrick said:
a singlecard is drawn from a pack of 52 . what is the probability P((oddUblack)Ueven)

but P((oddUblack)Ueven) = P(oddUeven) = 100% :confused:
 


but then Uwith black so does that mean 0.5?
 


are the laws distributive can i say p((oUe)Ub)?
 
are you sure you have the question right?

it asks for odd or black or even …

everything is odd or black or even​
 


thanks for help Tim.
The question is definatly P((ODD U BLACK) U EVEN)
I can see the way your thinking but do you think the brackets change it?
 


Is it possible that at least one of your "\cup" is supposed to be "\cap"?
 


no they are both U, do you think its a mistake?
 
Y\cap p! :biggrin:
 
  • #10


lol very funny :)
 
  • #11


If they are all \cup then the problem is trivial as tiny-tim has told you. E\cup U includes all cards so "adding" all black cards doesn't change anything.
 
  • #12


cool, thanks chaps
 
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