Probability of 5 card hand

Question

You have a $5$ card hand from randomly shuffled standard deck of $52$ cards:

• $X$ - Event that hand exactly contains $1$ spade
• $Y$ - Event that hand exactly contains $1$ ace

How do I calculate $P(X \cup Y)$?

Answer 1

A hand which contains exactly one spade and exactly one ace is either a hand which is:

a) the spade ace (one such card) and 4 non spade non-ace cards (of which there are 52-13-3=36 of them) b) a spade which is not an ace and an ace which is not a spade + 3 non-ace non-spade cards

There are C(36,4) ways to choose the non-spade non-ace cards in the first case, so case (a) has 1 (ways to choose the spade ace) * C(36,4) (ways to choose the remaining cards).

Case (b) has 12 ways to choose the spade which is not an ace, 3 ways to choose an ace which is not a spade and C(36,3) ways to choose a non-ace non-spade card. Thus, case (b) has 12*3*C(36,3) ways.

There are C(52,5) ways to choose a hand, so the probability of exactly one spade and one ace is (# of ways in a + # of ways in b)/total number of ways to choose a hand.

Answer 2

Use the Hypergeometric distribution:

$$ P(1\text{ spade}) = \frac{\binom{13}{1}\binom{39}{4}}{\binom{52}{5}} = \frac{{13} \cdot {82,251}}{2,598,960} = 0.41 $$

$$ P(1\text{ ace}) = \frac{\binom{4}{1}\binom{48}{4}}{\binom{52}{5}} = \frac{{4} \cdot {194,580}}{2,598,960} = 0.30 $$

$$ P(1\text{ spade} \cap 1\text{ ace}) = \frac{\binom{4}{1}\binom{13}{1}\binom{35}{3}}{\binom{52}{5}} = \frac{{4} \cdot{13} \cdot {6,545}}{2,598,960} = 0.13 $$

Then the general addition rule:

$$ P(1\text{ spade} \cup 1\text{ ace}) = P(1\text{ spade} + P(1\text{ ace}) - P(1\text{ spade} \cap 1\text{ ace}) $$

Answer 3

$$P\left(X\cup Y\right)=P\left(X\right)+P\left(Y\right)-P\left(X\cap Y\right)$$

Here:

• $P\left(X\right)=\binom{13}{1}\binom{39}{4}\binom{52}{5}^{-1}$
• $P\left(Y\right)=\binom{4}{1}\binom{48}{4}\binom{52}{5}^{-1}$
• $P\left(X\cap Y\right)=\binom{1}{1}\binom{3}{0}\binom{12}{0}\binom{36}{4}\binom{52}{5}^{-1}+\binom{1}{0}\binom{3}{1}\binom{12}{1}\binom{36}{3}\binom{52}{5}^{-1}$

I deliberately do not simplify, in the hope that you can see the structure. The first term on RHS of $P(X\cap Y)$ is the probability that the ace of spades is in the hand (and no other aces or spades). The second term is the probability that there is exactly one ace wich is not a spade and exactly one spade in the hand. There is a split up $1+3+12+36=52$. Here $1$ corresponds with the ace of spades, $3$ with the aces that are not spades, $12$ with the spades that are not aces and $36$ with the cards that are not an ace and not a spade.

Answer 4

Use the Principle of Inclusion and Exclusion.

$\begin{align} \mathsf P(X) & = {4\choose 1}{48\choose 4}\big/{52\choose 5} \\[2ex] \mathsf P(Y) & = {13\choose 1}{39\choose 4}\big/{52\choose 5} \\[2ex] \mathsf P(X\cap Y) & =\bigg({1\choose 1}{36\choose 4}+{3\choose 1}{12\choose 1}{36\choose 3}\bigg)\big/{52\choose 5} \\[2ex] \mathsf P(X\cup Y) & =\mathsf P(X)+\mathsf P(Y)-\mathsf P(X\cap Y) \end{align}$

Remark: For the intersection of the events you have to consider two cases; whether the 1 card that is an ace is the same 1 card that is a spade, or whether they are two different cards.