Question: A die is thrown as long as necessary for a 1 or a 6 to turn up. Given that no 1 turned up at the first two throws, what is the probability that at least three throws will be necessary?
Answer: $\frac{16}{25}$
what I did: Since the first 2, nothing turned up I assumed the probability was $\frac{4}{6}$ and for the 3rd try I assume it will show up so the probability is $\frac{2}{6}$ I then subtracted it to $1$ to get the probability since it says at least three throws but it yields a wrong value. Thank you for any help
$$1-\frac{4}{6}\frac{4}{6}\frac{2}{6}=\frac{23}{27}$$
To solve this question let us first find what is the probability that we don't meet the condition in three throws. Since we know there was no 1 in first two throws(you didn't consider this) the probability of us failing in two moves is $$P(Failure)= \frac{4}{6-1}.\frac{4}{6-1} $$ Now, As we need at least three throws we just have to ensure that it fails for two throws.
Hence,
the required probability is,$$P(n)=P(Failure)=\frac{4.4}{5.5}=\frac{16}{25}=0.64$$