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Mathematics 16 Online
OpenStudy (anonymous):

I need help with a probability question. Please help!

OpenStudy (anonymous):

yes?

OpenStudy (anonymous):

I will right the question

OpenStudy (anonymous):

kk

OpenStudy (anonymous):

P(x U Y)= P(X upside down "U" Y)= These are the formulas

OpenStudy (anonymous):

What is the probability of getting an ace or a diamond in a deck of cards?

OpenStudy (anonymous):

like the ace of diamonds or either of the two

OpenStudy (anonymous):

Both. Whati s the chance you will get an ace. What is the chance you will get a diamond.

OpenStudy (anonymous):

Can someone help?

OpenStudy (zarkon):

\[P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\]

OpenStudy (anonymous):

I don't understand what to do with that. can you explain more?

OpenStudy (zarkon):

what is the probability of an ace?

OpenStudy (anonymous):

There are four aces, one of them a diamond. there are 12 other diamonds. the number of possible successes is thus 4+12=16. There are 52 cards in all. successes/ possibles = 16/52=0.31

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