In a game played with a standard deck of cards, each face card has a value of $10$ points,, each ace has a value of $1$ point, and each number card has a value equal to its number. Two cards are drawn at random. One card is the queen of diamonds. What is the probability that the sum of the cards is greater than $18$?
The first card can be drawn from a pack of cards in $52$ ways.
The second card can be drawn from the pack of cards in $51$ ways ($\because \;$ cards are drawn without replacement).
Let $A$ be the event that the card drawn is a queen of diamonds.
Then $P\left(A\right) = \dfrac{1}{52}$
Let $B$ be the event of drawing a card from the cards with value $9$ ($4$ cards), value $10$ ($4$ cards), or the face cards ($11$ cards :-- $4$ kings, $3$ queens, $4$ jacks)
(This is done so that the sum of the two cards is greater than $18$.)
$\therefore \;$ Number of elements in $B = n \left(B\right) = 19$
$\therefore \;$ $P \left(B\right) = \dfrac{19}{51}$
$\left(A \cap B\right) \;$ is the event that a card is queen of diamonds AND the sum of the two cards is greater than $18$
$\therefore \;$ $P \left(A \cap B\right) = \dfrac{1}{52} \times \dfrac{19}{51}$
$\therefore \;$ P(when one card is queen of diamonds, then the sum of the cards is greater than $18$)
$= P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)} = \dfrac{\dfrac{1}{52} \times \dfrac{19}{51}}{\dfrac{1}{52}} = \dfrac{19}{51}$