Hello, chemekydie!

A die has 6 sides numbered 1-6.

If 3 dices are thrown, what is the probability that

at least one of the dices shows the number six?

For this question, do I have to take into consideration that the other two dices are not sixes?

This is what I get:

for first die to be a 6

for second die to be a 6 if the first is not.

for the third die to be a 6 if the first two are not.

Adding them up I get:

Is my analysis correct?

Yes! . . . Good work!

There is a back-door approach to this problem.

The opposite of "at least one 6" is "6's."no

The probability of no 6's is: .

Therefore: .