Five cards are drawn from a shuffled deck with $52$ cards. Find the probability that
a) four cards are aces
b) four cards are aces and the other is a king
c) three cards are tens and two are jacks
d) at least one card is an ace
My attempt:
a) $\left(13*12*\binom{4}{4}*\binom{4}{1}\right)/\binom{52}{5}$
b) same as (a)?
c) $\left(13 * 12 * \binom{4}{3} * \binom{4}{2}\right)/\binom{52}{5}$
d) $\left(13 * \binom{4}{1}\right)/\binom{52}{5}$
On
a) There are $ \binom{52}{5} = 2,598,960 $ ways of choosing 5 cards. There are $\binom{4}{4} = 1 $ way to select the 4 aces. So there are $\binom{48}{1}= 48 $ ways to select the remaining card. Thus there are a total of 48 ways to select 5 cards such that 4 of them are aces, and the probability is: $\frac{48}{2,598,960} = \frac{1}{54,145}$.
b) There are $\binom{4}{4} = 1 $ way to choose the 4 aces, and there are $\binom{4}{1} = 4 $ ways to choose a king. So there are $1\times4 = 4 $ ways to choose 5 cards such that 4 are aces and the other is a king card. The probability is: $\frac{4}{2,598,960} = \frac{1}{649,740} $.
c) There are $\binom{4}{3} = 4 $ ways to choose 3 ten cards, and there are $\binom{4}{2} = 6 $ ways of choosing 2 jacks. So there are $ 4\times 6 = 24 $ ways to choose 5 cards such that 3 are ten and 2 are jacks. The probability for this case is: $\frac{24}{2,598,960} = \frac{1}{108,290} $.
d) The probability of 5 non-ace cards is: $\frac{\binom{48}{5}}{\binom{52}{5}} = \frac{1,712,304}{2,598,960} = 0.6588 $, so the probability of getting 5 card at least one ace is: $1 - 0.6588 = 0.34$.
There are $\binom{52}{5}$ equally likely ways to choose $5$ cards. For solving all but the last problem, we count the number of "favourables" and divide by $\binom{52}{5}$.
a) There are $\binom{48}{1}$ $4$-Ace hands, for we have freedom only in choosing the non-Ace. Or, if you prefer, there are $\binom{12}{1}\binom{4}{1}$ such hands.
b) We can choose which King it is in $\binom{4}{1}$ ways.
c), The tens can be chosen in $\binom{4}{3}$ ways, and for each way the Jacks can be chosen in $\binom{4}{2}$ ways, for a total of $\binom{4}{3}\binom{4}{2}$.
d) Find first the probability of no Aces. We can choose a no Ace hand in $\binom{48}{5}$ ways. Calculate the probability of no Aces, and subtract this probability from $1$.