Problem: Jak rolls two die and wants the probability of rolling at least a 5 to be $\frac{1}{2}$. How many should Jak roll?
basically, I got two answers (same), but different approaches. Can someone tell me why one of them could be wrong?
Solution 1
The chance of getting a 5 is $\frac{1}{6}$ and rolling n dies or n times (same meaning in this situation) is $\frac{n}{6}$. Now I want the probability to be at least $\frac{1}{2}$
So I have $\frac{n}{6} > \frac{1}{2} \implies n > 3 \implies n = 4$
That is I want to roll four dices.
Solution 2
This one is a bit more formal.
It basically thinks about the probability of NOT getting that 5. So I have $\frac{5}{6}$. Now rolling n times, I get $(\frac{5}{6})^n$
Thus, the event of getting at least one five would be $1 - (\frac{5}{6})^n$ and we want this to be greater than 1/2.
$1 - (\frac{5}{6})^n > \frac{1}{2} \implies n > 3.80 \implies n = 4 $
Both gives me n = 4, but clearly the first one is informal. Could someone tell me why the first one could be wrong?