Playing Cards

Consider two cases.

Case 1: When the first card is red, the second card is black.

$\frac{26}{52} \times \frac{26}{51} = \frac{13}{51}$

Case 2: When the first card is black, the second is red.

$\frac{26}{52} \times \frac{26}{51} = \frac{13}{51}$

Add the probabilities for both cases:

$\frac{13}{51} + \frac{13}{51} = \frac{26}{51}$

So, $m + n = \boxed{77}$

After drawing 1st card,there are 51 cards left 26 of them being of color different from the color of 1st card. So probability =26/51. Ans=26+51=77.

Nice solutions guys, here is another solution:

With 52 cards, there are ${ C }_{ 52 }^{ 2 }$ ways to draw 2 cards from a deck.

Moreover, with these 52 cards, there are ${ C }_{ 26 }^{ 2 }$ ways to draw 2 cards with the same color, and there are 2 colors: red and black, so there are $2\times { C }_{ 26 }^{ 2 }$ ways in total

So the probability for us to draw 2 cards of 2 different colors is:

$\frac { { C }_{ 2 }^{ 52 }-{ 2\times C }_{ 2 }^{ 26 } }{ { C }_{ 2 }^{ 52 } } =\frac { 26 }{ 51 }$

Hence, the sum of the numerator and the denominator is $26+51=\boxed { 77 }$

Total possible ways of selecting 2 cards from 52 cards = 52C4 ways = $52\times 26$ ways.

Number of favorable cases(i.e, selecting one card each of two colors) = 26C1 $\times$ 26C1 = $26\times 26$ ways.

Probability(they are of different colors) = $\frac { 26\times 26 }{ 26\times 51 }$

Thus we get $\frac { A }{ B } =\frac { 26 }{ 51 }$ .

Thus A+B=77