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Mathematics 17 Online
OpenStudy (anonymous):

when draw a 13 card hand from a 52 deck card, what is the probability of having 6 cards from the same suit

OpenStudy (anonymous):

help please?

OpenStudy (anonymous):

5.5%

OpenStudy (anonymous):

5.59

OpenStudy (anonymous):

really?

OpenStudy (anonymous):

52-13= 39

OpenStudy (anonymous):

binompdf(13,.25,6) = 5.59224

OpenStudy (anonymous):

.0559 but 5.59 percent

OpenStudy (anonymous):

there are \(\binom{52}{13}\) way to select your 13 card lets do it for one suit, say \(\diamondsuit\) out of the 13 you select, there are \(\binom{13}{6}\) ways to get 6 diamonds, and \(\binom{39}{7}\) ways to select the other 7 cards

OpenStudy (anonymous):

I think she was asking the probability

OpenStudy (anonymous):

so for all four suits my guess is \[\frac{4\binom{13}{6}\binom{39}{7}}{\binom{52}{13}}\]

OpenStudy (anonymous):

oh idk then

OpenStudy (anonymous):

it could be binompdf(13,.25,6) but i have no idea what that means or how to compute it

OpenStudy (anonymous):

i get a different number http://www.wolframalpha.com/input/?i= \frac{4\binom{13}{6}*\binom{39}{7}}{\binom{52}{13}}

OpenStudy (anonymous):

Can you guys show me the entire procedure?

OpenStudy (anonymous):

i wrote out the method above and i think it is \[\frac{4\binom{13}{6}\binom{39}{7}}{\binom{52}{13}}\] each of these numbers in the fraction are large and a pain to compute, i would use a calculator

OpenStudy (anonymous):

Is (13 6) the same as 13 nCr 6?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

I believe it is 5.59% Go onto a ti-83/84 and hit 2nd VARS and go down to binompdf Then the first number is the number of trials 13 second number is the initial percent .25 third is howmany you want to succeed 6

OpenStudy (anonymous):

@ryderpeppler i see what you are doing, but these are not bernoulli trials, you can't use the binomial distribution

OpenStudy (anonymous):

what you are calculating is the probability of getting exactly 6 successes out of 13 trials, where the probability of each success is .25 but you cannot use that here, because the trials are not independent

OpenStudy (anonymous):

ah so the method I used doesnt account for there being less cards and a lower probability after the first drawn card?

OpenStudy (anonymous):

Then why is my percentage lower than yours?

OpenStudy (anonymous):

right

OpenStudy (anonymous):

Thank you so much!

OpenStudy (anonymous):

Well it'll either be 16.6 (17 with sig figs) or 5.59 (6 with sig figs) percent

OpenStudy (anonymous):

this is an entirely different question you do not have only two possible outcomes, "success or failure"

OpenStudy (anonymous):

also there are four different suits to choose from

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