Question

Five cards are dealt from a standard deck of 52 cards. What is the probability that three or four of the cards are aces? Round to the nearest ten-thousandth.

Answer 1

The probability that 3 of the 5 cards are aces is given by:
\[\large P(3\ aces)=\frac{\left(\begin{matrix}4 \\ 3\end{matrix}\right)\left(\begin{matrix}48 \\ 2\end{matrix}\right)}{\left(\begin{matrix}52 \\ 5\end{matrix}\right)}\]
The probability that 4 of the 5 cards are aces is given by:
\[\large P(4\ aces)=\frac{48}{\left(\begin{matrix}52 \\ 5\end{matrix}\right)}\]
The required probability is found from:
P(3 or 4 aces) = P(3 aces) + P(4 aces)

Answer 2

How did you get 3/5 or 4/5 cards? It says three of the four cards are aces

Answer 3

A standard deck of 52 cards has 4 aces. Therefore if five cards are dealt it is possible to get 3 aces or all 4 aces. The events '3 aces in 5 dealt cards' and '4 aces in 5 dealt cards' are mutually exclusive. Therefore the probability of getting 3 or 4 aces in 5 dealt cards is the the sum of P(3 aces in 5 dealt cards) and P(4 aces in 5 dealt cards).
Note that the question does not state that 3 of the 4 cards are aces. The question is asking for the probability of getting 3 aces or 4 aces in the 5 dealt cards.