There are 20 cards (4 colors with 5 cards types each, e.g. ace).
Each of the 4 players randomly receives 5 cards.What is the probability that one player has three aces after the drawing?
I came up with a solution, but it produces the wrong result (according to the end results our prof gave us.)
Facts:
- There are 4 aces in the deck
- There are 20 cards in the deck
- Each of the 4 players gets 5 cards
(# means amount)
#(possible cases) = #(draw 5 cards out of 20)
#(favorable cases) = #(draw 3 aces out of 4 aces) $\cdot$ #(draw 2 other cards out of remaining 16 cards)
Use the hypergeometric distribution formula:
$$P=\frac{\text{#favorable cases}}{\text{#possible cases}} = \frac{\binom{4}{3}\binom{16}{2}}{\binom{20}{5}} = 0.0396...$$
The solution our professor gave us says the end result is $0.1238$ though.
What did I do wrong? Did I forget to take anything into account?
I chose to do this exercise first, since the other probabilities to calculate seem even more complicated:
a) Each player has one ace.
b) Exactly one player has exactly two aces.
c) At least one player has exactly two aces.
d) (This one) One player has three aces.