# A card is drawn from a standard deck. A second card is drawn, without replacing the first card. What is the probability that the first card is red and the second card is black?

Mar 1, 2015

In the standard deck of cards there are 52 cards, 26 red and 26 black.

SOLUTION 1

The first event, random drawing of a red card (event $A$), has a sample space of an entire deck with 52 different elementary events occurring with equal probabilities of $\frac{1}{52}$. Out of them only 26 events are "good" (that is, the card we randomly pick is red). Therefore, the probability of picking a red card equals to
$P \left(A\right) = \frac{26}{52} = \frac{1}{2}$.

The second event, random drawing of a black card from a deck of only 51 remaining cards (event $B$), is dependent on the results of the first event. If the first event (event $A$) occurs (red card is drawn), the deck for the second drawing contains 25 red and 26 black cards and the conditional probability of picking black card is
$P \left(B | A\right) = \frac{26}{51}$.

Now we can use the concept of conditional probability and a formula that describes the probability of a combined event through the probability of one and conditional probability of another:
$P \left(A \cdot B\right) = P \left(A\right) \cdot P \left(B | A\right)$

In our case $P \left(A\right) = \frac{1}{2}$ and $P \left(B | A\right) = \frac{26}{51}$

Therefore,
$P \left(A \cdot B\right) = \left(\frac{1}{2}\right) \cdot \left(\frac{26}{51}\right) = \frac{13}{51}$

SOLUTION 2

After shuffling there are 52! permutations of 52 cards. Each has the probability of occurrence equal to another, that is the probability of any sequence of cards is 1/(52!).

Let's count only those where there is a red card on the first place and black card on the second.
There are 26 red and 26 black cards. Therefore, we have $26 \cdot 26$ choices for a pair of two first cards. The sequence of other 50 cards in unimportant and any permutation among them is fine for us since we fixed the first two.

Therefore, we have 26*26*50! different permutations when the first card is red and the second is black.
The final probability, therefore, is

(26*26*50!)/(52!)=(26*26)/(52*51)=13/51

Informative lectures and solutions to many problems of the Theory of Probabilities for beginners can be found in the corresponding chapter on the Web site Unizor (free online course of advanced mathematics for teenagers).