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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
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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
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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
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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.