Probability of getting only one ball of red color

Question

There are total 48 balls in a box. 12 balls of red color, 12 balls of green color, 12 balls of blue color, and 12 balls of yellow color. Two balls are chosen at random. What is the probability of getting only one ball of red color?

Answer 1

12/48 = 1/4 chance of red with 1 draw

not red on 2nd draw = 36/47 chance

multiply 36/47 x 1/4 =9/47 for chance of just 1st red then not red

chance of 3/4 of no red on 1st draw

12/47 for red on 2nd

3/4 x 12/47 =9/47

P(1 red)= P(red, not red) + P(not red, red) = 9/47 + 9/47 = 18/47

or another method to solve the problem:

P(0 red)= 3/4(35/47)= 105/188

P(2 red)= 1/4 x 11/47 = 11/188

P(1 red)= 1- P(0) - P(2) = 1- 105/188 - 11/188

= 1 -116/188=72/188

=18/47 = about 38.3%