Two cards are drawn at random from a standard deck of $52$ cards and not replaced. What is the probability that the first card is a face card and the second card is a seven? What is the answer to this maths problem and how is the answer done?
[Math] Two cards are drawn at random from a standard deck of $52$ cards and not replaced.
card-gamesprobability
Related Solutions
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:
- The first card is a diamond and the second card is a heart.
- Both cards are hearts.
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.
Your mistake is in thinking that there is a $\tfrac34$ chance to get at least $1$ red; this would be the probability when drawing with replacement. But without replacement, the color of the one card affects the possibilities for the color of the other card.
If you count the number of ways to draw two face cards, you will find there are $\tbinom{12}{2}=66$ ways. The number of ways to draw no red cards is $\binom{6}{2}=15$, so the number of ways to draw at least one red card is $66-15=51$. This means the probability of at least one of the two drawn cards being red is $\tfrac{51}{66}\neq\tfrac{3}{4}$. Your second computation also reflects this.
Best Answer
There 12 face cards, so the probability that the first card is a face card is $\frac{12}{52}$. Then there are 51 cards left. If the first card was a face card, then 4 sevens are still in the deck. So the probability that the first card is a face card and the second card a seven is $\frac{12}{52} \times \frac{4}{51} = \frac{4}{221}$.