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The probablility that I can get a crumpled up piece of paper into the recycling bin is 211/243.  So, if I shoot 6 pieces of paper, then the probablity I can get at least two into the recyccling bin is
\(\left( \frac{211}{243} \right)^6 + 6 \left( \frac{211}{243} \right)^5 \left( \frac{22}{243} \right) + 15 \left( \frac{211}{243} \right)^4 \left( \frac{22}{243} \right)^2 + 25 \left( \frac{211}{243} \right)^3 \left( \frac{22}{243} \right)^3 + 6 \left( \frac{211}{243} \right)^2 \left( \frac{22}{243} \right)^4,\)
because I get two into the recycling bin if I get two, three, four, five or six.  This works out to about 77.91%.

If the probability that I will get at least one of five into the bin is 211/243, 

then the probability that I miss all five is 32/243.

But, 32/243  =  (2/3)⁵.

This means that the probability of missing one is 2/3 and the probability of making one is 1/3.

[miss x miss x miss x miss x miss  =  (2/3) x (2/3) x (2/3) x (2/3) x (2/3)  =  32/243]

So, if you take 6 shots and make at least two:

₆C₂·(2/3)²·(1/3)⁴ + ₆C₃·(2/3)³·(1/3)³ + ₆C₄·(2/3)⁴·(1/3)² + ₆C₅·(2/3)⁵·(1/3)¹ + ₆C₆·(2/3)⁶·(1/3)⁰  = 

First, we compute the probability that any given shot will make the recycling bin. The chance that at least one of five shots will make the recycling bin is 211/243, so none of five shots will make the recycling bin with probability 32/243. If the probability a given shot doesn't make the recycling bin is p that means that p^5 = 32/243 so p = \sqrt[5]32/243 = 2/3.

 Therefore, the probability any given shot will make the recycling bin is 1/3

Now, we can compute the desired probability by considering the complement—that is, by looking at the probability that no shot or exactly one shot will make the recycling bin. Since the probability I make any given shot is uniform, the chance none of my six shots make it is (2/3)^6=64/729

 If I make exactly one shot, there are 6 possibilities for which shot this could be; the chance I make that shot is 

 and the chance I don't make the other five shots is (2/3)^5=32/243

That means the probability I make at least two shots is 

 minus the sum of these two probabilities, or 473/729

Three very "smart" people get 3 different answers to the same question! What does this tell you about "probability"? It tells you that "Probability is a curse upon the world of Mathematics" !! By the way, geno3141's answer =0.9821673525.

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 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?

Edit:

NOTE: I have given the correct answer to the WRONG question.  There is another post below where I have answered the question that was actually asked    

If I shoot 5 bits the prob that they all miss is   

\( P(all\;fail\;on\; 5\; tries)=5C0*(F)^5=F^5\\ P(don't \;all\;fail\;with\; 5\; tries)=1-F^5=\frac{211}{243}\\ so\\ F^5=\frac{243-211}{243}=\frac{32}{243}\\~\\ F=\frac{2}{3}\qquad S=\frac{1}{3} \)

\( P(none\; in)=6C0\;* \; (\frac{1}{3})^0\;*\;(\frac{2}{3})^6\\ P(none\; in)=1\;* \; 1\;*\;(\frac{2}{3})^6\\ P(none\; in)=(\frac{2}{3})^6\\ P(none\; in)=\frac{64}{729} \\~\\ P(one\; in)=6C1\;* \; (\frac{1}{3})^1\;*\;(\frac{2}{3})^5\\ P(one\; in)=6\;* \; (\frac{1}{3})\;*\;(\frac{32}{243})\\ P(one\; in)=2\;*\;(\frac{32}{243})\\ P(one\; in)=\frac{64}{243}\\ \\~\\ P(two\; in)=6C2\;* \; (\frac{1}{3})^2\;*\;(\frac{2}{3})^4\\ P(two\; in)=15\;* \; (\frac{1}{9})\;*\;(\frac{16}{81})\\ P(two\; in)=5\;* \; (\frac{1}{3})\;*\;(\frac{16}{81})\\ P(two\; in)= \frac{80}{243}\\ \\~\\ P(\text{no more than 2 in)}=\frac{64}{729}+\frac{64}{243}+ \frac{80}{243}\\~\\ P(\text{no more than 2 in)}=\frac{64+3*64+3*80}{729}\\ P(\text{no more than 2 in)}=\frac{496}{729}\\ \)

\(P(\text{no more than 2 in)}\approx 0.68\)

There could easily be careless mistakes, maybe even careless logic mistakes, but the underlying idea is sound.

My answer almost agrees with the second guest, and much of the logic is in agreement with Gino.

So we all had similar ideas.

Coding:
P(all\;fail\;on\; 5\; tries)=5C0*(F)^5=F^5\\
P(don't \;all\;fail\;with\; 5\; tries)=1-F^5=\frac{211}{243}\\
so\\
F^5=\frac{243-211}{243}=\frac{32}{243}\\~\\
F=\frac{2}{3}\qquad S=\frac{1}{3}

P(none\; in)=6C0\;* \; (\frac{1}{3})^0\;*\;(\frac{2}{3})^6\\
P(none\; in)=1\;* \; 1\;*\;(\frac{2}{3})^6\\
P(none\; in)=(\frac{2}{3})^6\\
P(none\; in)=\frac{64}{729}
\\~\\
P(one\; in)=6C1\;* \; (\frac{1}{3})^1\;*\;(\frac{2}{3})^5\\
P(one\; in)=6\;* \; (\frac{1}{3})\;*\;(\frac{32}{243})\\
P(one\; in)=2\;*\;(\frac{32}{243})\\
P(one\; in)=\frac{64}{243}\\
\\~\\
P(two\; in)=6C2\;* \; (\frac{1}{3})^2\;*\;(\frac{2}{3})^4\\
P(two\; in)=15\;* \; (\frac{1}{9})\;*\;(\frac{16}{81})\\
P(two\; in)=5\;* \; (\frac{1}{3})\;*\;(\frac{16}{81})\\
P(two\; in)= \frac{80}{243}\\

\\~\\
P(\text{no more than 2 in)}=\frac{64}{729}+\frac{64}{243}+ \frac{80}{243}\\~\\
P(\text{no more than 2 in)}=\frac{64+3*64+3*80}{729}\\
P(\text{no more than 2 in)}=\frac{496}{729}\\

Hi Melody: Believe or not, if you reverse gene3141's power expansion, he gets EXACTLY  the same answer  as Guest #3, or: 473 / 729. Here is geno3141's calculation with "reversed powers" of 2/3 and 1/3 as follows:

6nCr2*(2/3)^4*(1/3)^2 + 6nCr3*(2/3)^3*(1/3)^3 + 6nCr4*(2/3)^2*(1/3)^4 + 6nCr5*(2/3)^1*(1/3)^5 + 6nCr6*(2/3)^0*(1/3)^6  = 473 / 729