1. ## Playing Card Probability

How do I go about solving this?

4 cards are drawn from a pack of 52 playing cards without replacment. What is the probability that two are black and two are red?

2. The number of ways of getting 2 red cards from the 26 is $\left(\begin{array}{cc}26\\2\end{array}\right) = 325$
And similiarly for black cards.

The total number of different ways of taking 4 cards from 52 is $\left(\begin{array}{cc}52\\4\end{array}\right) = 270725$

So the probability of getting 2 red cards, and 2 black cards, is;

$
\frac{ \left(\begin{array}{cc}26\\2\end{array}\right) \left(\begin{array}{cc}26\\2\end{array}\right)}{\l eft(\begin{array}{cc}52\\2\end{array}\right)}=\fra c{325*325}{270725}=0.3901$

3. Thank you very muck for your help, however I do not fully understand the method you used to solve it... as in I have not come across the bracket things before

4. Hello, Paulo1913

Does this version look familiar to you?

The number of ways of getting 2 red cards from the 26 is: . ${\color{blue}_{26}C_2} \,=\, 325$
And similiarly for black cards: . ${\color{blue}_{26}C_2} \,=\,325$

The total number of different ways of taking 4 cards from 52 is: . ${\color{blue}_{52}C_4} \:=\: 270,\!725$

So the probability of getting 2 red cards, and 2 black cards, is: . $\frac{(_{26}C_2)(_{26}C_2)}{_{52}C_4} \;=\; \frac{325\cdot325}{270,\!725} \;=\;0.3901$