# Probability homework problem help

1. Mar 14, 2012

### gtfitzpatrick

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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 cant be even?)

so my probability is 39/52 + 26/52 which obviously isnt right...

2. Mar 14, 2012

### tiny-tim

hi gtfitzpatrick!
but P((oddUblack)Ueven) = P(oddUeven) = 100%

3. Mar 14, 2012

### gtfitzpatrick

Re: probability

but then Uwith black so does that mean 0.5?

4. Mar 14, 2012

### gtfitzpatrick

Re: probability

are the laws distributive can i say p((oUe)Ub)?

5. Mar 14, 2012

### tiny-tim

are you sure you have the question right?

it asks for odd or black or even …

everything is odd or black or even​

6. Mar 14, 2012

### gtfitzpatrick

Re: probability

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?

7. Mar 14, 2012

### HallsofIvy

Staff Emeritus
Re: probability

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

8. Mar 14, 2012

### gtfitzpatrick

Re: probability

no they are both U, do you think its a mistake?

9. Mar 14, 2012

### tiny-tim

$Y\cap p!$

10. Mar 14, 2012

### gtfitzpatrick

Re: probability

lol very funny :)

11. Mar 14, 2012

### HallsofIvy

Staff Emeritus
Re: probability

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. Mar 14, 2012

### gtfitzpatrick

Re: probability

cool, thanks chaps