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Mathematics 20 Online
OpenStudy (anonymous):

suppose that you are dealt five cards at random from a standard deck. what is the probability that AT LEAST four cards will be spades?

OpenStudy (anonymous):

at least four means four or five so we have to compute each of these separately

OpenStudy (anonymous):

would it be the hypergeometric function but just computed for four and five and then add the two probabilities together?

OpenStudy (anonymous):

5 is easiest, it is \[\frac{13}{52}\times\frac{12}{51}\times \frac{11}{50}\times \frac{10}{49}\times \frac{9}{48}\]

OpenStudy (anonymous):

yes you have to add them up

OpenStudy (anonymous):

i wrote it out for 5, you can also do it for 4 right?

OpenStudy (anonymous):

yeah just when i do the same thing i add the two probabilities up together?

OpenStudy (anonymous):

right, and yes you can use hypergeomtric, i just wrote it out like a bonehead. if you use hypergeometric you will have a ton of cancelation and get what i got i am sure

OpenStudy (anonymous):

okay sounds good thank you!

OpenStudy (anonymous):

think for 4 it will be \[\frac{\dbinom{13}{4}\dbinom{39}{1}}{\dbinom{52}{5}}\] if i am not mistaken

OpenStudy (anonymous):

yw

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