My question is about calculating the probability of the following event A: obtaining 3 cards of the same color when drawing 3 cards among a set of 52 cards (and without putting back the drawn card).
I tried resolving this problem by adding the probability of the two events: drawing 3 red cards and drawing 3 black cards which would give:
$P(A) =(\frac{26}{52}*\frac{25}{51}*\frac{24}{50}) + (\frac{26}{52}*\frac{25}{51}*\frac{24}{50})$
The first term of the addition is understood as follows: drawing a first red card $\frac{26}{52}$ then a second red card: $\frac{25}{51}$ and so on. Likewise, the second term is about drawing 3 black cards.
However, it seems the above is wrong. Can someone please explain what is wrong in my reasoning?
Best Answer
The result of your formula is right. If you want to use combinations, you can think the formula as:
\begin{align*} & P = {26 \choose 3}{2 \choose 1} \end{align*}
There are 52 cards, 26 of each color. You want to have 3 cards of the 26 of the same color. You then must multiply that with the number of combinations that can be formed with two colors choosing only one, which is the same as 2 combinations (black or red).
Using combinations may make it easier if you want to then calculate the probability of drawing three cards of the same suit. Simply replace the 26 with a 13 (because each suit has 13 cards for a total of 52 cards) and replace the 2 with a 4 (because there are 4 suits instead of only 2 colors).