Ask your own question, for FREE!
Mathematics 29 Online
OpenStudy (anonymous):

You pick 5 cards out of a deck of 52. What is the probability that we get at least one card of each suit?

OpenStudy (anonymous):

\[\frac{ \left(\begin{matrix}4 \\ 1\end{matrix}\right) \left(\begin{matrix}4 \\ 1\end{matrix}\right) \left(\begin{matrix}4 \\ 1\end{matrix}\right) \left(\begin{matrix}4 \\ 1\end{matrix}\right) \left(\begin{matrix}48 \\ 1\end{matrix}\right) }{ \left(\begin{matrix}52 \\ 5\end{matrix}\right) } \] is what i think it is, but i'm not sure

OpenStudy (anonymous):

I was wrong last time. But i'm certain now the answer is 4 (13C2) (13C1) (13C1) (13C1) / (52C5) = 2197/8330

OpenStudy (anonymous):

why is it 4 times all of all of that?

OpenStudy (anonymous):

Actually, that makes sense! Thank you!

OpenStudy (anonymous):

combination and permutation are really intimidating aren't they XD

OpenStudy (anonymous):

Yeah it's the worst lol!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!