If the crayons are replaced after they are drawn, the probability of drawing any particular color is 1/7.Also the result of each draw is independent of any previous result. Therefore the solution to the first part of the question is found from:P(violet\ then\ orange)=\frac{1}{7}\times\frac{1}{7}=\frac{1}{49}
If the crayons are not replaced after they are drawn, the probability of drawing violet on the first pick is 1/7. Then there are only six crayons left, so the probability of drawing orange is 1/6. The solution to the second part of the question is found from:
P(violet\ then\ orange)=\frac{1}{7}\times\frac{1}{6}=\frac{1}{42}
Does this help
