Data Analysis & Probability There are 9 stones in a bag. 4 stones are blue. 5 stones are green. Lisa takes a stone at random from the bad. She DOES NOT REPLACE IT. She then takes at random a second stone from the bag. Work out the probability that at least one of these two stones is blue. You may wish to complete the tree diagram to help. My tree diagram starts as 4/9 and 5/9 and branches out left: 3/8 blue - 5/8 green and right: 4/8 - 4/8 green. Is the equation to work out the probability of at least one of these two stones being blue: 1-[(5/9 x 4/8) + (4/9 x 3/8) + (2/9 x 1/8)] ?

Your tree diagram is correct. But since you want the probability of at least one blue stone, you want to add up the branches that gives you at least one blue stone. In your four branches:

A:$\frac{4}{9} \cdot \frac{3}{8}$ gives you first a blue stone then another blue stone.

B:$\frac{4}{9} \cdot \frac{5}{8}$ gives you first a blue stone then a green stone.

C:$\frac{5}{9} \cdot \frac{4}{8}$ gives you first a green stone then a blue stone.

D:$\frac{5}{9} \cdot \frac{4}{8}$ gives you first a green stone then another green stone.

So, A,B,C are branches that give you at least one blue stone. You want to add those up. Or equivalently, subtract the probability of branch D from the total probability 1.