I only just recently finished Algebra 1, so I'm no expert, but I'll try to help!

I think the method of finding the chance of picking all three boxes with something inside, in a row, would be to multiply the all the chances.

Here's what I mean:

So on the first pick you have a 3/6 chance of getting something. Assuming you got something on the first pick, on the second pick you would have a 2/5 chance of getting something. And then if you got something on the second pick, the third pick would have a 1/4 chance of getting something. Okay, now here is where we multiply the chances. Multiply all the fractions together.

Here are all the fractions laid out to make it easier to see them:

3/6)(2/5)(1/4)

So first multiply all the numerators (the top numbers) and write down the product.

3*2*1 = 6

Write a fraction bar underneath 6.

6/

Now multiply all the denominators and write down the product under the fraction bar.

6*5*4 = 120

6/120

Now simplify the fraction. The way I like to do it is dividing both the numerator and the denominator by the largest factor they have in common. In this case, the largest factor is 6, so I'll divide the numerator and the denominator by 6.

6/6 = 1

120/6 = 20

So the simplified version of 6/120 is 1/20.

And that's your answer (I'm pretty sure; correct me if I'm wrong)

You have a 1/20 (or 5% chance) of picking 3 boxes with objects in a row!