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 "no 6's."
The probability of no 6's is: .