There are 11 pens in a box. 8 are black and 3 are red. Two pens are taken out at random without replacement. Work out the probability that the two pens are the same colour.

There are two solutions here which must be found. Both 2 red pens and 2 black pens being selected.
First the red pensIf the total number of pens is 11 and number of red pens is 3, the probability of selecting a red pen on the first try is 3/11.We then remove one of the red pens and one of the total pens from the fraction,the probability of selecting a second red pen is now 2/10.We then find the probability of these two events happening in sequence by multiplying both probabilities,3/11 x 2/10 = 6/110
Repeat for the black pensIf the total number of pens is 11 and number of black pens is 8, the probability of selecting a black pen on the first try is 8/11.We then remove one of the black pens and one of the total pens from the fraction,the probability of selecting a second black pen is now 7/10.We then find the probability of these two events happening in sequence by multiplying both probabilities,8/11 x 7/10 = 56/110
The total probability is then found by adding up all of the possible solutions, which in this case is 6/110 + 56/110 = 62/110
62/110 is our answer and can be simplified to 31/55.