Two cards are drawn at random (without replacement) from a regular deck of 52 cards. What is the probability that the first card is a red and the second card is heart?
Let $A$ be the event that a red card is drawn and $B$ be the event the second card drawn is a heart.
We have;
$n(S) = C(52,2)$
$P(A) = \frac{26}{52}$
How is $P(B)$ calculated? Is a solution even possible?
Best Answer
In a deck of cards, there are four suits: clubs, diamonds, hearts, and spades. Diamonds and hearts are red; clubs and spades are black. There are $13$ cards of each suit.
We want to find the probability that the first card is red and the second card is a heart when two cards are drawn without replacement from a standard deck.
There are two possibilities:
Let $H$ denote the event that a heart is drawn; let $D$ denote the event that a diamond is drawn.
The first card is a diamond and the second card is a heart: The probability of drawing a diamond on the first draw is $\Pr(D) = 13/52$. Of the $51$ cards that remain, $13$ are hearts. Hence, the probability of drawing a heart given that a diamond was selected on the first draw is $\Pr(H \mid D) = 13/51$. Hence, the probability that the first card is a diamond and the second card is a heart is $$\Pr(H \mid D)\Pr(D) = \left(\frac{13}{51}\right)\left(\frac{13}{52}\right)$$
Both cards are hearts: The probability of drawing a heart on the first draw is $\Pr(H) = 13/52$. Of the $51$ cards that remain, $12$ are hearts. Hence, the probability of drawing a heart given that a heart was drawn on the first draw is $\Pr(H \mid H) = 12/51$. Thus, the probability that both cards are hearts is $$\Pr(H \mid H)\Pr(H) = \left(\frac{12}{51}\right)\left(\frac{13}{52}\right)$$
Since these cases are mutually exclusive and exhaustive, the desired probability can be found by adding the probabilities for the two cases.