I know this is not typically a hard one but I have a question about it:
A fair die has sides labeled w/ 1, 2, 3, 4, 5, 6. If the die is rolled 4 times, what is the probability that on at least 1 roll, the die will show 6?
Now, I'd like to know why you HAVE to solve by looking at it from the probability of NOT getting a six. In other words...why can't you add 1/6 4 times? When is it key to go with probability of not getting what you are trying to get, rather than solving for the probability of getting it?
the die rolling is a independent event so the total probability will be the multipication of probability of each event.
and the case you are refering where using 1/6 as the probabiltiy of getting 6 .....1/6*1/6*1/6*1/6 .. is the probabiltiy of getting 6 in each event not getting at least one six.
but in the question the probability of at least one 6 is asked .....so in 4 rolls you may have 1 six or 2 six or 3 six ...or all 4 six ......so the best way is to 1 - probability of None= probability of at least one.....otherwise you have to consider the cases all the case
1six3Nonesix+2six2Nonesix...and so on
hope it helps....