Every time I use a piece of scrap paper, I crumple it up and try to shoot it inside the recycling bin across the room. I'm pretty good at it: If I shoot $5$ pieces of paper at the recycling bin, at least one of them will make it inside the recycling bin with probability $\frac{211}{243}$. If I shoot $6$ pieces of paper at the recycling bin, what's the probability at least two of them make it inside the recycling bin?
+2666 Note that because \(243 = 3^5\), we know that the probability that you make a shot is \({ x \over 3}\) (because he makes 5 shots)
The probability of making at least 1 shot is the same as \(1 - \text{probability of failing every time}\).
From the given info, we know that the probability of failing every time is \(1 - {211 \over 2443} = {32 \over 243}\)
This means that \(x = \sqrt[5]{32} = 2\), so the probability of making any free throw is \(2 \over 3\).
Can you take it from here?
