Here is the problem:
Player A chooses 5 cards from a deck of cards. Player B also chooses 5 cards from ANOTHER deck of cards. Player B wins if his cards match at least 3 cards of player A. What is the probability that the number of cards of player B matches that of player A's is
A) 0 B) 1 C) 2 D) 3 E) 4 F) 5
I attempted to solve A) this way: ((52C5)* (47C5)) / ((52C5)*(52C5)) = 0.5902 but I'm not sure if this is right.
Thanks in advance for any help!
Assuming that both players are using standard $52$-card decks, the probability that the set of cards $B$ draws has exactly $k$ cards in common with the set of cards $A$ draws is $$ \frac{{5\choose k}{47\choose 5-k}}{{52\choose 5}}$$ for $k=0,1,2,\dots,5$.
One way to see this is to number $B$'s cards $1$ to $52$, with cards $1$ to $5$ being the cards $A$ has drawn from his deck.
You solved part A) correctly!
For part B), we have $52\choose1$ ways of picking the card that $A$ and $B$ have in common, then $51\choose4$ ways of picking $A$'s other four cards, then $47\choose4$ ways of picking $B$'s other four cards (which have to be completely different from $A$'s). Thus, for part B), the probability is:
$$\frac{{52\choose1}\cdot{51\choose4}\cdot{47\choose4}}{{52\choose5}^2}$$
The rest of the parts proceed similarly.
The numerical results are:
A) $59.021\%$
B) $34.314\%$
C) $6.239\%$
D) $0.416\%$
E) $0.009\%$
F) $0.000038\%$
$\binom{A}{B}$$\binom{A}{B}$ especially relevant. – gen-ℤ ready to perish Sep 04 '17 at 06:49