Jane had a box filled with 8 red pens, 4 blue pens and 6 green pens. She took out two pens at random, one after the other, without replacement. The tree diagram below gives probabilities for the first and second pen.

1. Find the probability that the first pen is green.

There are $6$ green pens out of $18$ pens.

$p=\frac{6}{18}=\frac{1}{3}$

2. Find the probability that the second pen is blue, given that the first pen was blue.

If the first pen was blue, there are $3$ blue pens left out of $17$ pens.

$q=\frac{3}{17}$

1. Calculate the probability of getting one blue pen and one green pen in either order.

One blue and one green can happen in two ways:

$P(B\text{ then }G)+P(G\text{ then }B)$

$=\frac{2}{9}\times\frac{6}{17}+\frac{1}{3}\times\frac{4}{17}$

$=\frac{12}{153}+\frac{4}{51}$

$=\frac{12}{153}+\frac{12}{153}=\frac{24}{153}=\frac{8}{51}$

1. List the possible ordered outcomes where the two pens are of different colours.

Different colours can occur as:

$RB, RG, BR, BG, GR, GB$

2. Calculate and add the probabilities of all different-colour outcomes.

3. Simplify the probability.

$\frac{104}{153}=\frac{34}{51}$