# One card is selected at random from a deck of cards. What is the probability that the card selected is a spade or a queen?

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Nov 10, 2017

$P \left(\text{Spade" uu "Queen}\right) = \frac{4}{13}$

#### Explanation:

Whenever you solve a probability question involving two conditions, and you are being asked to find the probability that either will occur for a given action, you are looking for what is known as a "union probability". Formally speaking, if we say $A$ represents "the card is a Spade", and $B$ represents "the card is a Queen", then we are looking for the probability of "the card is a Spade or a Queen", or symbolically:

$P \left(A \cup B\right)$

The trick here is that these two possible events are not disjoint events; in other words, it can be possible to pull a single card and have it be a Spade and a Queen at the same time. The formula for determining $P \left(A \cup B\right)$ takes this into consideration:

$P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$

(This is read as "the probability of A union B is equal to the probability of A plus the probability of B minus the probability of the intersection of A and B".)

If we consider $P \left(A\right)$ (the probability the card is a Spade), in a standard deck of 52 cards there are exactly 13 cards which are Spades. Thus, $P \left(A\right) = \frac{13}{52} = \frac{1}{4}$. (This is intuitive, because there are 4 suits of cards with the same values/ranks within them and we're only interested in one of those four suits.)

If we consider $P \left(B\right)$ (the probability the card is a Queen), in a standard deck of 52 cards there are exactly 4 cards which are Queens (in suits of Hearts, Spades, Clubs, and Diamonds). Thus, $P \left(B\right) = \frac{4}{52} = \frac{1}{13}$. (Again, this is intuitive, because there are 13 unique values of cards, of which there is only one Queen value.)

However, the probability $P \left(A \cap B\right)$ represents the probability the card is a Spade and a Queen at the same time. Of all 52 cards in the deck, there is only one Queen of Spades, thus $P \left(A \cap B\right) = \frac{1}{52}$.

Thus:

$P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$

$= \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$

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