Question 1118089
Let's look at the four cases:

Let P(R,R) = probability of selecting red then another red

What value is P(R,R)?

On the first pick, there are 7 red and 11 total, so P(R) = 7/11
On the 2nd pick, there are 6 red and 10 total, so P(R,R) = (7/11)*(6/10) = 42/110
1.  P(R,R) = 42/110 <br>

Similarly, using W for white:
2.  P(R,W) = (7/11)*(4/10) = 28/110
3.  P(W,R) = (4/11)*(7/10) = 28/110
4.  P(W,W) = (4/11)*(3/10) = 12/110  <br>

Notice these cover all possible outcomes, as expected.  We know this because the numerators, when added, sum to 110  and 110/110=1  <br>


The probability of one of each color is just the sum of case 2, P(R,W),  and case 3, P(W,R):

         28/110 + 28/110 = {{{ highlight( 56/110 ) }}}   or about 0.51   

I hope this helps you.