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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?

 Jul 16, 2022
 #1
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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?

 Jul 16, 2022
 #2
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BuilderBoi, do you like to \(\text{su}\)\(\text{ck}\) \(\text{co}\)\(\text{ck}\)?

Guest Jul 16, 2022
 #3
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could you please explain further, and what is x in x/3

 Jul 20, 2022

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