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Mathematics 23 Online
OpenStudy (mathmath333):

Probability question

OpenStudy (mathmath333):

\(\large \color{black}{\begin{align}& \normalsize \text{Find the probability that when a hand of 7 cards is drawn from a well}\hspace{.33em}\\~\\ & \normalsize \text{shuffled deck of 52 cards, it contains atleast 3 Kings.}\hspace{.33em}\\~\\ \end{align}}\)

OpenStudy (mathmath333):

Total number of possible hands = 52C7

ganeshie8 (ganeshie8):

How many kings are there in a full deck ?

OpenStudy (mathmath333):

4

OpenStudy (phi):

break the "good" hands into the problem of choosing 4 "not kings" and 3 kings plus prob of choosing 3 not kings, and 4 kings

OpenStudy (mathmath333):

what u mean ?

OpenStudy (phi):

how many ways can you choose 4 "not kings" from the deck?

OpenStudy (mathmath333):

48

OpenStudy (phi):

48C4

OpenStudy (mathmath333):

yes

OpenStudy (phi):

that gives you all the "4 card" hands with no kings now pick 3 kings from the king pile (4 cards) 4C3 multiply to get the number of hands with exactly 3 kings

OpenStudy (phi):

next, we want the number of hands with 4 kings

OpenStudy (mathmath333):

4C3\(\times \)48C4

OpenStudy (mathmath333):

4C4×48C3

OpenStudy (phi):

ok, now add that is the numerator (# of hands that match our criteria) as you know the denominator is 52C7

OpenStudy (mathmath333):

=\(\dfrac{4C4×48C3+4C3×48C4}{52C7}\)

OpenStudy (phi):

yes

OpenStudy (mathmath333):

thnks

OpenStudy (phi):

yw

OpenStudy (loser66):

Thank you very much @phi . I am always in trouble with probability. Your method is PERFECT. It is easy to understand what is going on. Thanks you so much.

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